**Calculating the Total Length of Wire Used in a Spiral**

The problem presented involves calculating the total length of wire used to form a spiral made up of successive semicircles with centers at alternating points A and B. The radius of each semicircle increases by 0.5, 1, 1.5, and so on, until reaching a maximum radius of 5.

To begin, we need to understand the relationship between the radius of each semicircle and the total length of wire used. Since each semicircle is half of a full circle, the length of wire used for each semicircle is proportional to the circumference of its corresponding full circle. The formula for the circumference of a circle is 2πr, where r is the radius.

The problem states that the spiral consists of three successive semicircles, and we are asked to find the total length of wire used to form this spiral. Let's denote the radii of the first, second, third, and thirteenth semicircles as r1, r2, r3, and r13, respectively.

We can write the length of wire used for each semicircle as πr, where π is a constant approximately equal to 22/7. Since we are dealing with semicircles, we need to divide this by 2 to get the actual length of wire used. Therefore, the total length of wire used for all three semicircles can be expressed as:

πr1 + πr2 + πr3 + ... + πr13

**Finding the Sum of an AP Series**

To calculate this sum, we need to find a pattern or relationship between the terms in the series. Looking at the radii, we notice that they form an arithmetic progression (AP) with a common difference of 0.5.

The first term (a1) is 0.5, and the common difference (d) is also 0.5. We are dealing with 13 terms in total, but since the semicircles alternate between points A and B, we can simplify the problem by considering only every other radius.

Let's consider the sum of the radii up to the thirteenth term:

r1 + r2 + ... + r13

We can use the formula for the sum of an AP series to solve this:

Sn = n/2 [2a + (n - 1)d]

where Sn is the sum, a is the first term, d is the common difference, and n is the number of terms.

Plugging in the values we have:

13/2 [2(0.5) + (13 - 1)(0.5)]

= 6.5 [1 + 12(0.5)]

= 6.5 [1 + 6]

= 6.5[7]

= 44.5

**Finding the Total Length of Wire Used**

Now that we have the sum of the radii, we can find the total length of wire used by substituting this value back into our original expression:

πr1 + πr2 + πr3 + ... + πr13

= π(r1 + r2 + ... + r13)

= π(44.5)

Substituting the value of π as 22/7, we get:

(22/7) × 44.5

≈ 11

Therefore, the total length of wire used to form the spiral is approximately 143 units.

**The Spiral Formed by Successive Semicircles**

To visualize this spiral, let's imagine a series of semicircles with increasing radii, alternating between points A and B. The first semicircle has a radius of 0.5, the second one has a radius of 1, and so on.

As we move from left to right along the spiral, each successive semicircle increases in size by approximately 10% (from 0.5 to 1 to 1.5). This creates a visually striking pattern that tapers off towards the end.

**Conclusion**

In this article, we have explored the concept of using an arithmetic progression (AP) series to solve a problem involving the total length of wire used in a spiral made up of successive semicircles with alternating centers at points A and B. By recognizing the AP relationship between the radii and applying the sum formula for an AP series, we were able to calculate the total length of wire used to form this spiral.

This example demonstrates the power of mathematics in solving real-world problems that may seem complex at first glance.

"WEBVTTKind: captionsLanguage: enhello students welcome to paramar academy from the last video video sequence okay first half particularly what is an ap series what is the first term of that sequence what is the common difference how to find the value corresponding to a position and how to find the position corresponding to a value now in the second part path part we will discuss how to find the sum of first and terms of series means we can also we can find the sum of first 10 terms or we will have to find the sum of first 100 terms and so on we can also find the sum like from the from term 4 to from 100 for that will be find the sum of first 400 terms from that we will subtract the sum of first three terms this is how we can find the sum of term four to term hundred and similarly we can find another sums so let's get start sum of the first and terms of an ap now here we will discuss how to find the sum of first particular number of terms of a ap sequence so sum of n terms means s1 that is means sum of one term only that is a one only s2 that means sum of the first two terms that is a one plus a two as three means sum of first three terms that is a one plus a2 plus a3 as 4 means sum of first photon that is a1 a2 plus a3 plus a4 similarly sum of n terms means sn means sum of atoms that is a1 plus a2 plus a3 plus a4 up to so on up to nth term so this is called the sum of n terms now we are having the particular formula to find this sum or also we can find the sum from the particular number for example i have to find the sum of third to a term that is a3 plus a4 a5 up to zone up to a a8 so i can easily find it using s8 that is the sum of first atoms minus s2 that is sum of first two terms get deducted so using this i can easily get the sum of terms from a3 to a eight now the formula to find the sum of first and terms is that is s n is equal to n upon 2 2 a plus n minus 1 d this is a formula using that formula we will easily find the sum of first and terms of an ap sequence so we can also change it like this and upon 2 we can write 2a as a plus a plus n minus 1 times d now look at here sn is equal to n upon 2 this is a now this is a plus and upon my n minus 1 d that is equal to a n so i can write it as also a n so this is the another formula to find the sum of n terms when the nth term is also given so now let's start with the questions the first question is find the sum of ap 2 7 12 up to 10 terms here in this question the ap is given that is 2 comma 7 comma 12 and we have to find the sum of first tan terms so what is given to us so the first term of this ap is 2 the common difference d is equal to 7 minus 2 that is equal to 5 the number of terms 10 tons that is n is given we have to find the sum of first tan terms and now the main thing we have to find that is s 10 that is the sum of 10 terms now we will use the formula and we know that s n is equal to n upon 2 2a plus n minus 1 time d so now substitute the values that is s 10 is equal to 10 upon 2 to a that is 2 plus n that is 10 minus 1 d is 5 so now s 10 is equal to that is 5 2 4 plus 10 minus 1 9 9 into 5 45 so this is 45 so s 10 is equal to 5 into 45 plus 4 that is 49 so as 10 is equal to 245 that means sum of the first 10 terms of this ap sequence is 245 now i will repeat it again so the ap is given that is 2 comma 7 comma 12 up to so on now we have to find the sum of first tan term of this sequence how to find we are simply having the formula sn is equal to n upon 2 into 2a plus n minus 1 times of d now for that we will find the values the a that is the first term is 2 the that is the common difference is 7 minus 2 again 5 and number of terms already given how many terms we have to use or we have to find the sum of how many terms that is 10 so n is 10 given now we have to find s 10 that is sum of tan terms so using that formula i will substitute the values as 10 is equal to and upon 2 that is n is 10 10 upon 2 into 2 a plus n minus 1 d 2 a is replaced by its value that is 2 plus and is replaced by value that is 10 10 minus 1 and d is 5 so after calculation i will get some that is 245 so that's how we can find the sum of n terms of an ap the next one is find the sum of 34 plus 32 plus 30 up to so on plus 10 again the sequence is given we have to find the term find the sum first of all we will check is this ap or not so very easy the difference between these two terms that is second term minus first term 32 minus 34 that is 2 difference between third term and second term again that is 30 minus 32 that is minus 2 so clearly difference between the term is same that is minus 2 so this is an ap series now we have to find the sum of this series and 10 is the last term now look at here here the num we don't know the number of terms here this 10 is the value corresponding to any position 10 is the value that is 10 is the a n corresponding to some position we don't have the value of n so first of all we will find the value of n so a what is a that is 34 d that is minus 2 a n we have to find the value of n corresponding to 10 so a n is 10 first of all we will find the value of n using the basic formula that is a n is equal to a plus n minus 1 d means we have to find the sum of how many terms a n that is 10 is equal to a that is 34 plus and we have to find into d that is minus 2 so 10 minus 30 4 is equal to n minus 1 times minus 2 so that is minus 24 get divided by minus 2 that is equal to n minus 1 that is 12 is equal to n minus 1 1 gets added to this side that is that is equal to n so 13 is equal to n that means we have to find the sum of first 13 terms of this ap series so again using the formula to find us that to find the sum that is sn is equal to n upon 2 2a plus n minus 1 time d now put the values we have to find the sum of 13 terms n is 13 so that is 13 upon 2 2 a a is 34 plus and that is 13 minus 1 into d that is minus 2 now as 13 is equal to 13 upon 2 34 twos are 68 plus 13 minus 1 parallax into minus 2 that is equal to 13 upon 2 68 plus minus minus twelve twos are twenty four that is equal to thirteen upon two sixty eight minus twenty four that is that is forty four that is 44 now 13 into 22 that is 13 into 20 that is 286 so sum of the first 13 terms of this ap series is 286 in this question a n is given that is the last term for the sum is given first of all we will find the number of terms that the position of the 10 because we have to use that and in the formula we can also find the sum using the second formula that s n is equal to n upon 2 a plus a n that is equal to n that is 13 13 upon 2 a is 34 plus n n is a n that is a n is a n is here 10 a n is the last term we also say that that is 13 upon 2 into 44 that is same as this one equation 13 upon 2 into 44 13 upon 2 into 44 so using the both the formulas we will get the same answer but we we have to use this formula basic formula to find the value of n because in both these using both the formulas we need and here and here also so we have to use this formula first of all to find the value of n the next one is if a n is equal to 3 plus 4 and show that a 1 a 2 a 3 a 4 and so on a and form an arithmetic progression that is ap and also find the sum of first 40 for 25 terms means what is given to us we have given a generalized form of a sequence that a n is equal to 3 plus 4 and from that general form we have to find the separate separate uh values so for that first of all a n is equal to 3 plus 4 n so here we have to find a 1 a 2 a 3 a 4 and so on a n simply replace and by 1 here we will get a 1 is equal to 3 plus 4 into 1 that is equal to 3 plus 4 that is equal to 7 so a 1 is 7 similarly replacing a n by 2 we will get a 2 that is equal to 3 plus 4 into 2 that is 3 plus 8 that is equal to 11 replace n by three we'll get a three that is equal to three plus four into three that is three plus twelve four threes are twelve that is simply fifteen and a4 replace n by 4 we will get a 4 that is 3 plus 4 times 4 that is 3 plus 16 and equal to 19. so value of a1 is 7 a2 is 11 a3 is 15 and a4 is 19. now this is the ap so required ap is 7 comma 11 comma 15 comma 19 and so on so this is the required ap so this is the first part of this question now we also have to find the sum of first 25 terms that is n is equal to 25 because n is not the value n is the position and we have to find the sum of first 25 terms so what is given to us simply a is equal to 7 that is the first term the common difference that is equal to 11 minus 7 that is equal to 4 and n is 25 given we have to find the sum of 25 terms that is s and is equal to n upon 2 2a plus and minus 1 times difference so now put the values as 25 is equal to 25 upon 2 2 into a that is 7 plus n minus 1 that is 25 minus 1 into d that is 4 25 upon 2 7 2's are 14 plus 25 minus 1 20 24 into 4 this is 25 upon 2 14 plus 14 fours are 24 for the 96 so that is equal to 25 upon 2 into that is 1 10 so that is equal to 25 into 55 so 55 into 25 that is 550 550 and 225 1325 1325 that is the sum of first 25 terms of this ap so i will repeat it again first of all a n that is equal to 3 plus 4 and that is generalized form of the sequence is given and from that generalized form we have to find the separate terms now we have to find a1 a2 a3 a4 and so on and from that separate values we also have to find the sum of first 25 terms so simply a n is equal to 3 plus 4 n is given first four we have to find the separate terms so for that for finding the a one i will replace n by one so a one is equal to three plus four into one that is three plus four is seven to find the second term i simply replace n by 2 so i get 3 plus 4 into 2 that is 3 plus 8 that is 11 and so on i will find the a3 a4 and a5 now the second task is to find the sum of first 25 terms again this is the required ap that is 7 11 59 19 and so on from here the first term is 7 the common difference is 11 minus 4 7 that is 4 and the number of terms we have to find the sum that is 25 so now again using the formula to find the sum of an ap that is sn is equal to n upon 2 into 2a plus n minus 1 times d now i can find the sum of 25 terms so simply replace n by 25 a by 7 d by 4 and we will get the sum so that is equal to 25 upon 2 2 times of 7 plus 25 minus 1 into difference that is 4 so 25 upon 2 7 2 14 plus 25 minus 1 that is 24 into 4 so 25 by 2 that is 14 24 forza that is 96 96 plus 14 that is 110 that is cancel out 25 into 55 and the product is 1325 so some of the first 25 terms of the this ap sequence is 1325 the next one is if if the sum of first and terms of an ap is 4 and minus n square 40 is the first term that is s1 what is the sum of first two terms what is the second term and what is the similarly what is the third term and the another term so very very interesting and easy question in this question we are given the sum of first and terms that is sn is given 4 and minus n square this is given so our task is to find the first term means a 1 we have to find a 1 what is the sum of first 2 terms means we have to find a 1 plus a2 or simply s2 what is the second term that means we have to find a2 only similarly find third term that means we have to find a3 the tenth term we have to find a tan and the another term is we have to find a n so in this question first of all we will find a one then a one plus a two that is sum of two terms then the second term only then third term only tenth term and the and that term so very easy and interesting question so let's start first of all as we know that the first term sum of one term that is equal to first term only so there is no difference between first term and the sum of one term so replace n by one here we will get s1 that is equal to 4 into 1 minus 1 square that is equal to 4 minus 1 that is equal to 3 so this is also equal to a 1 so look at here the first term of the required ap is 3 or the sum of first term is also 3 our next task is to find what is the second term before that we have to find what sum of the sec two terms now simply replace n by two to find s2 s2 is equal to four into two minus 2 square so that is equal to 4 2's are 8 minus 2 square is 4 that is 4 so s2 is equal to 4 that is also equal to sum of first two terms so we are having the s2 we are having the a1 now our task is to find a2 so simply look at here this is the value of a1 that is 3 and a1 plus a2 is 4 replace a1 by its value here 4 is equal to a1 that is 3 plus a2 now 4 minus 3 that is a 2 that is 1 is equal to 8 2 so second term of the ap is 1 now we are having the second term also the next task is to find the third term before that look at here of an ap required ap we are having the first term that is three second term is one so here we will easily decide that d that is one second term minus first term that is minus two now we are having also having the value of d so next next task is to find the third term and we know by using the basic formula of ap sequence that third term is equal to a plus 2 times of d so that is equal to a plus 2 times of d now we are having the value of a that first term is 3 plus 2 and difference is minus 2 so this is equal to 3 minus 4 that is equal to minus 1 so third term of the sequence is minus 1 the next we have to find the tenth term again how to find tenth term simply a tan is equal to a plus nine times of d so replace the values a that is first term is 3 plus 9 d that is difference is minus 2 so that is equal to 3 9 2s are that is minus 18 and this is minus 15 so tenth term of the ap series is minus 50 now the last task is to find the and that's term now we have to find a n and we know the basic formula to find a n is a plus n minus 1 times of d so that is equal to a that is the first term 3 plus n minus 1 into d that is minus 2 that is equal to 3 2 is also that is minus 2 and minus 1 into minus 2 that is plus 2 so nth term of the given ap sequence is 1 5 3 plus 2 5 minus 2 n so this is a required answer so very this is very easy question but this question is break into many parts so first of all we will find s1 that is equal to a1 then we will find we have to find that's true that is sum of first two terms from that we have to find the second term only then we have to find third term tenth term and the nth term so very easy question this is a main key point you also have to find the because we have to use the here here and here also and it is not mentioned in the question so keep remembering when we are having the two terms of an ap we will easily find the common difference and that is very much helpful to find the another terms also now the next is how many terms of this ap 9 comma 17 comma 25 must be taken to give the sum 636 now again very easy and important question in this question we are given the ap sequence and also give us the sum of the terms now we have to find how many terms collaborate to find that sum means we have to find the and the how many terms we can use to find that sum so very easy simple so let's start a that is the first term is 9 common difference that is 17 minus 9 is equal to 8 ascent that is 636 and we have to find the corresponding value of n so we will easily find this number of terms using the formula of sn ascent is equal to and upon 2 2a plus n minus 1 times of d i put the values as n is 6 36 that is and upon 2 2 a that is two nines are 18 plus n minus 1 and we have to find into d that is it is equal to 636 equal to n upon 2 18 plus 8 and minus 8 that is equal to 6 36 is equal to n upon 2 18 minus 8 that is 10 plus 8 10 now 6 36 is equal to 2 get cancelled with inner factors that is 5 plus 4 n that is 6 36 is equal to 5 and plus 4 and square now this is simply a quadratic equation so 5 and 4 and square 4 4 and square plus 5 and minus 6 36 is equal to 0. now the whole calculation turns into a quadratic equation now we have to solve this quantity question using the middle term splitting method or the discriminate method to get the value of n so we will use the middle term speeding method now we have to break five and in two parts such that the product sum of minus plus or minus is equal to five and product is equal to product of first and last term so simply that speed is the position is 53 and 48 so that is 4 n square plus 53 and minus 48 minus 636 that is equal to zero the middle term splits into two parts now from the first two terms from first two terms and get common that is 4 and plus 53 and the from last two terms minus 12 get common again that is 4 and plus 53 now is equal to 0 now these two factors are same so that means our calculation is okay now from the whole equation 4n plus 53 get common and the next factor second factor is n minus 12 that is equal to zero and we know that if the product of two factors is equal to zero that the first one is zero or the second one is zero so from here we will get four n plus 53 is equal to zero or n minus 12 is equal to 0 here the n is 12 and from here and is minus 53 upon 4 we will get two values of n but n is equal to minus 15 3 upon 4 get rejected because n is always a natural number because position is always a natural number at the first position second position third position fourth position position can't be negative and the position can't be a decimal number you can't say that a position is 2.5 or position is 3.5 so simply this value get rejected so n is equal to 12 this means 12 terms of this ap must be taken to get the sum 636 so this is how we can find the number of terms to find the particular sum the next one is find the sum of first 15 multiples of 8 now in this question we have to find the sum of first 15 multiples of it first 15 multiple means we have to use first 15 terms that are multiplied multiple of eight so simply first multiple of eight is clearly that is eight itself the second is 16 third is 24 and so on we have to find up to 15 terms so what is given to us that a is 8 difference is also 6 minus 8 that is 60 minus 8 is 8 and n is 15 we have to find the sum so what is how to find the sum s n is equal to n upon 2 2a plus and minus times of a common difference so we have to find the sum that is n upon 2 means 15 upon 2 2 a that is 2 is multiplied by 8 plus 15 and is 15 that is 15 minus 1 into difference and difference is also 8. that is equal to 15 upon 2 2 8 are 16 15 minus 1 14 into 8 that is equal to 15 upon 2 16 plus that is 1 1 2 14 here's a 1 12 that is equal to 15 upon 2 into 128 now that is equal to 15 multiplied by 64. so the sum is that is 640 and 320 that is 960 so sum of first 15 multiples of 8 is 960 that is 8 plus 16 plus 24 plus 32 up to first 15 multiples the whole sum is 960. the next one is a sum of rupees 700 is to be used to give seven cash prizes to students of a school for their overall academic pro performance if each price is 20 rupees less than its preceding price find the value of each cash price so very interesting question what is given to us so need a statement carefully a sum of 700 means the sum is 700 so sn is given 700 rupees to be used to give seven cash prizes so number of prices are seven so that means n is equal to seven student of school for the each their overall academic performance if each price is 20 rupees less means the difference between the cost of price is 20 that is difference is 20 difference is 20 less than its perceiving price find the value of each cash price we have to find the value of each cash price so very interesting and easy question simply we will find the use the formula faster first of all sn is equal to n upon 2 2 a plus n minus 1 times of d so what is s n that is 700 is equal to n that is 7 upon 2 2a you can't have the value of a so this is okay 7 minus 1 into d that is 20 so that is 700 into two seven by two turns into two by seven that is equal to two a plus seven minus one that is six six into twenty that is one twenty now seven hundred cancel out hundred that is hundred into two 200 is equal to 2a plus 120 now 120 gets subtracted from this side 200 minus 120 that is equal to 2 a 200 minus 120 is 80 is equal to 2 a and 80 upon 2 is equal to a that is 40 is equal to a so this is the value of a that is the cost price of first price so cost of first price is 40 cost of the next price is 20 more than that is 60 next is 80 next is 100 the next one is 120 next is 140 and the next last one is 160. so these are the seven cash prizes in which first cash price is 40 and the last one is 160 and look at here their sum is total sum is 170 40 600 180 280 400 and 450 and the 700 so total sum is 700 so this is how we can find the amount of each cash prize so the next one a spiral is made up of successive semicircles with centers alternatively at a b this starting center at a of radius 0.5 1.0 1.5 to 5 2.0 centimeter as shown in the total what is the total length of such spiral made of 3 13 constructive semi circles now very interesting questions x spiral form here look at here spiral is formed in which this is the first semicircle whose center is a and now this is the second semicircle whose center is b again this is the third semicircle whose center is again a and this is the first semicircle whose center is b so now alternatively the this is the next next semicircle whose center is a and this is the next semicircle of center b and so on the next one center a center b and so on using like that there are 13 semi circles and to form that spring and we have to find the length of total wire used here so look at here first of all the first radius radius one that is zero point five zero point five second radius is one third radius is 1.5 fourth radius is two and so on now as this is a semicircle so what is the length of wire used for semicircle that is equal to pi r because total circumference of the circle is 2 pi r and we have to deal with the semicircles only so length is equal to pi r so for that first for that first is the length is pi into r1 for the second one pi r2 for third one pi r3 and the 431 that is pi r 13 now from all pi get common we left with r1 plus r2 plus r3 up to so on r13 that is equal to pi r1 that is 0.5 r2 1 next our radius is 1.5 next is 2 or so on up to r 30 now look at here clearly the inner one in our form is the ap sequence 0.5 1 1.52 and the common difference is 0.5 so we have to solve it that is pi so what is the values are here what are the values are here a that is the first term is 0.5 common difference that is also 1 minus 0.5 and the number of terms are 13 because we are using the 13 semicircles so to find that sum we will use the formula that is n upon 2 2a plus n minus 1 times of difference so that is pi is 22 upon 7 and that is 13 upon 2 2 a double of 0.5 that is 1 plus n minus 1 that is 30 minus 1 12 into difference that is also 0.5 that is equal to 22 upon 7 into 13 upon 2 into 1 plus 12 0.50 that is 6 that is equal to 22 upon 7 into 13 by 2 into 6 plus 1 that is again 7 so overall that is equal to 22 upon 7 into 13 upon 2 into 7 7 get cancelled with 7 2 get cancelled with 22 that is 11 that is 11 into 13 that is 130 and 13 14 3 so total length of the wire used is 143 units so this is oh so this is how we can find the length of the wire used i will repeat it again a semicircle is formed using the semicircles with different radius first radius as 0.5 second radius zero one third radius will be one point five four three is true and so on thirteen semicircles are used to perform that spiral we have to find find the wire used to form that spiral so simply radius is zero point five one one point five four of two and so on so as spiral is made up of half semi semi circles so length of a semi circle is pi r one second semi circle is pi r two third is pi r three and so on pi r thirty now pi by get common from o will get pi into r1 plus r2 plus r3 up to shown r 13 and after substituting the values of radius clearly we will look at here this is a sum of and terms of an ap so now using the sum formula we will easily get that total 143 units of wire is used to form that spiral so this is all about the chapter that is ap series hope so you will like it please like share and subscribe our channel to watch more videos thank youhello students welcome to paramar academy from the last video video sequence okay first half particularly what is an ap series what is the first term of that sequence what is the common difference how to find the value corresponding to a position and how to find the position corresponding to a value now in the second part path part we will discuss how to find the sum of first and terms of series means we can also we can find the sum of first 10 terms or we will have to find the sum of first 100 terms and so on we can also find the sum like from the from term 4 to from 100 for that will be find the sum of first 400 terms from that we will subtract the sum of first three terms this is how we can find the sum of term four to term hundred and similarly we can find another sums so let's get start sum of the first and terms of an ap now here we will discuss how to find the sum of first particular number of terms of a ap sequence so sum of n terms means s1 that is means sum of one term only that is a one only s2 that means sum of the first two terms that is a one plus a two as three means sum of first three terms that is a one plus a2 plus a3 as 4 means sum of first photon that is a1 a2 plus a3 plus a4 similarly sum of n terms means sn means sum of atoms that is a1 plus a2 plus a3 plus a4 up to so on up to nth term so this is called the sum of n terms now we are having the particular formula to find this sum or also we can find the sum from the particular number for example i have to find the sum of third to a term that is a3 plus a4 a5 up to zone up to a a8 so i can easily find it using s8 that is the sum of first atoms minus s2 that is sum of first two terms get deducted so using this i can easily get the sum of terms from a3 to a eight now the formula to find the sum of first and terms is that is s n is equal to n upon 2 2 a plus n minus 1 d this is a formula using that formula we will easily find the sum of first and terms of an ap sequence so we can also change it like this and upon 2 we can write 2a as a plus a plus n minus 1 times d now look at here sn is equal to n upon 2 this is a now this is a plus and upon my n minus 1 d that is equal to a n so i can write it as also a n so this is the another formula to find the sum of n terms when the nth term is also given so now let's start with the questions the first question is find the sum of ap 2 7 12 up to 10 terms here in this question the ap is given that is 2 comma 7 comma 12 and we have to find the sum of first tan terms so what is given to us so the first term of this ap is 2 the common difference d is equal to 7 minus 2 that is equal to 5 the number of terms 10 tons that is n is given we have to find the sum of first tan terms and now the main thing we have to find that is s 10 that is the sum of 10 terms now we will use the formula and we know that s n is equal to n upon 2 2a plus n minus 1 time d so now substitute the values that is s 10 is equal to 10 upon 2 to a that is 2 plus n that is 10 minus 1 d is 5 so now s 10 is equal to that is 5 2 4 plus 10 minus 1 9 9 into 5 45 so this is 45 so s 10 is equal to 5 into 45 plus 4 that is 49 so as 10 is equal to 245 that means sum of the first 10 terms of this ap sequence is 245 now i will repeat it again so the ap is given that is 2 comma 7 comma 12 up to so on now we have to find the sum of first tan term of this sequence how to find we are simply having the formula sn is equal to n upon 2 into 2a plus n minus 1 times of d now for that we will find the values the a that is the first term is 2 the that is the common difference is 7 minus 2 again 5 and number of terms already given how many terms we have to use or we have to find the sum of how many terms that is 10 so n is 10 given now we have to find s 10 that is sum of tan terms so using that formula i will substitute the values as 10 is equal to and upon 2 that is n is 10 10 upon 2 into 2 a plus n minus 1 d 2 a is replaced by its value that is 2 plus and is replaced by value that is 10 10 minus 1 and d is 5 so after calculation i will get some that is 245 so that's how we can find the sum of n terms of an ap the next one is find the sum of 34 plus 32 plus 30 up to so on plus 10 again the sequence is given we have to find the term find the sum first of all we will check is this ap or not so very easy the difference between these two terms that is second term minus first term 32 minus 34 that is 2 difference between third term and second term again that is 30 minus 32 that is minus 2 so clearly difference between the term is same that is minus 2 so this is an ap series now we have to find the sum of this series and 10 is the last term now look at here here the num we don't know the number of terms here this 10 is the value corresponding to any position 10 is the value that is 10 is the a n corresponding to some position we don't have the value of n so first of all we will find the value of n so a what is a that is 34 d that is minus 2 a n we have to find the value of n corresponding to 10 so a n is 10 first of all we will find the value of n using the basic formula that is a n is equal to a plus n minus 1 d means we have to find the sum of how many terms a n that is 10 is equal to a that is 34 plus and we have to find into d that is minus 2 so 10 minus 30 4 is equal to n minus 1 times minus 2 so that is minus 24 get divided by minus 2 that is equal to n minus 1 that is 12 is equal to n minus 1 1 gets added to this side that is that is equal to n so 13 is equal to n that means we have to find the sum of first 13 terms of this ap series so again using the formula to find us that to find the sum that is sn is equal to n upon 2 2a plus n minus 1 time d now put the values we have to find the sum of 13 terms n is 13 so that is 13 upon 2 2 a a is 34 plus and that is 13 minus 1 into d that is minus 2 now as 13 is equal to 13 upon 2 34 twos are 68 plus 13 minus 1 parallax into minus 2 that is equal to 13 upon 2 68 plus minus minus twelve twos are twenty four that is equal to thirteen upon two sixty eight minus twenty four that is that is forty four that is 44 now 13 into 22 that is 13 into 20 that is 286 so sum of the first 13 terms of this ap series is 286 in this question a n is given that is the last term for the sum is given first of all we will find the number of terms that the position of the 10 because we have to use that and in the formula we can also find the sum using the second formula that s n is equal to n upon 2 a plus a n that is equal to n that is 13 13 upon 2 a is 34 plus n n is a n that is a n is a n is here 10 a n is the last term we also say that that is 13 upon 2 into 44 that is same as this one equation 13 upon 2 into 44 13 upon 2 into 44 so using the both the formulas we will get the same answer but we we have to use this formula basic formula to find the value of n because in both these using both the formulas we need and here and here also so we have to use this formula first of all to find the value of n the next one is if a n is equal to 3 plus 4 and show that a 1 a 2 a 3 a 4 and so on a and form an arithmetic progression that is ap and also find the sum of first 40 for 25 terms means what is given to us we have given a generalized form of a sequence that a n is equal to 3 plus 4 and from that general form we have to find the separate separate uh values so for that first of all a n is equal to 3 plus 4 n so here we have to find a 1 a 2 a 3 a 4 and so on a n simply replace and by 1 here we will get a 1 is equal to 3 plus 4 into 1 that is equal to 3 plus 4 that is equal to 7 so a 1 is 7 similarly replacing a n by 2 we will get a 2 that is equal to 3 plus 4 into 2 that is 3 plus 8 that is equal to 11 replace n by three we'll get a three that is equal to three plus four into three that is three plus twelve four threes are twelve that is simply fifteen and a4 replace n by 4 we will get a 4 that is 3 plus 4 times 4 that is 3 plus 16 and equal to 19. so value of a1 is 7 a2 is 11 a3 is 15 and a4 is 19. now this is the ap so required ap is 7 comma 11 comma 15 comma 19 and so on so this is the required ap so this is the first part of this question now we also have to find the sum of first 25 terms that is n is equal to 25 because n is not the value n is the position and we have to find the sum of first 25 terms so what is given to us simply a is equal to 7 that is the first term the common difference that is equal to 11 minus 7 that is equal to 4 and n is 25 given we have to find the sum of 25 terms that is s and is equal to n upon 2 2a plus and minus 1 times difference so now put the values as 25 is equal to 25 upon 2 2 into a that is 7 plus n minus 1 that is 25 minus 1 into d that is 4 25 upon 2 7 2's are 14 plus 25 minus 1 20 24 into 4 this is 25 upon 2 14 plus 14 fours are 24 for the 96 so that is equal to 25 upon 2 into that is 1 10 so that is equal to 25 into 55 so 55 into 25 that is 550 550 and 225 1325 1325 that is the sum of first 25 terms of this ap so i will repeat it again first of all a n that is equal to 3 plus 4 and that is generalized form of the sequence is given and from that generalized form we have to find the separate terms now we have to find a1 a2 a3 a4 and so on and from that separate values we also have to find the sum of first 25 terms so simply a n is equal to 3 plus 4 n is given first four we have to find the separate terms so for that for finding the a one i will replace n by one so a one is equal to three plus four into one that is three plus four is seven to find the second term i simply replace n by 2 so i get 3 plus 4 into 2 that is 3 plus 8 that is 11 and so on i will find the a3 a4 and a5 now the second task is to find the sum of first 25 terms again this is the required ap that is 7 11 59 19 and so on from here the first term is 7 the common difference is 11 minus 4 7 that is 4 and the number of terms we have to find the sum that is 25 so now again using the formula to find the sum of an ap that is sn is equal to n upon 2 into 2a plus n minus 1 times d now i can find the sum of 25 terms so simply replace n by 25 a by 7 d by 4 and we will get the sum so that is equal to 25 upon 2 2 times of 7 plus 25 minus 1 into difference that is 4 so 25 upon 2 7 2 14 plus 25 minus 1 that is 24 into 4 so 25 by 2 that is 14 24 forza that is 96 96 plus 14 that is 110 that is cancel out 25 into 55 and the product is 1325 so some of the first 25 terms of the this ap sequence is 1325 the next one is if if the sum of first and terms of an ap is 4 and minus n square 40 is the first term that is s1 what is the sum of first two terms what is the second term and what is the similarly what is the third term and the another term so very very interesting and easy question in this question we are given the sum of first and terms that is sn is given 4 and minus n square this is given so our task is to find the first term means a 1 we have to find a 1 what is the sum of first 2 terms means we have to find a 1 plus a2 or simply s2 what is the second term that means we have to find a2 only similarly find third term that means we have to find a3 the tenth term we have to find a tan and the another term is we have to find a n so in this question first of all we will find a one then a one plus a two that is sum of two terms then the second term only then third term only tenth term and the and that term so very easy and interesting question so let's start first of all as we know that the first term sum of one term that is equal to first term only so there is no difference between first term and the sum of one term so replace n by one here we will get s1 that is equal to 4 into 1 minus 1 square that is equal to 4 minus 1 that is equal to 3 so this is also equal to a 1 so look at here the first term of the required ap is 3 or the sum of first term is also 3 our next task is to find what is the second term before that we have to find what sum of the sec two terms now simply replace n by two to find s2 s2 is equal to four into two minus 2 square so that is equal to 4 2's are 8 minus 2 square is 4 that is 4 so s2 is equal to 4 that is also equal to sum of first two terms so we are having the s2 we are having the a1 now our task is to find a2 so simply look at here this is the value of a1 that is 3 and a1 plus a2 is 4 replace a1 by its value here 4 is equal to a1 that is 3 plus a2 now 4 minus 3 that is a 2 that is 1 is equal to 8 2 so second term of the ap is 1 now we are having the second term also the next task is to find the third term before that look at here of an ap required ap we are having the first term that is three second term is one so here we will easily decide that d that is one second term minus first term that is minus two now we are having also having the value of d so next next task is to find the third term and we know by using the basic formula of ap sequence that third term is equal to a plus 2 times of d so that is equal to a plus 2 times of d now we are having the value of a that first term is 3 plus 2 and difference is minus 2 so this is equal to 3 minus 4 that is equal to minus 1 so third term of the sequence is minus 1 the next we have to find the tenth term again how to find tenth term simply a tan is equal to a plus nine times of d so replace the values a that is first term is 3 plus 9 d that is difference is minus 2 so that is equal to 3 9 2s are that is minus 18 and this is minus 15 so tenth term of the ap series is minus 50 now the last task is to find the and that's term now we have to find a n and we know the basic formula to find a n is a plus n minus 1 times of d so that is equal to a that is the first term 3 plus n minus 1 into d that is minus 2 that is equal to 3 2 is also that is minus 2 and minus 1 into minus 2 that is plus 2 so nth term of the given ap sequence is 1 5 3 plus 2 5 minus 2 n so this is a required answer so very this is very easy question but this question is break into many parts so first of all we will find s1 that is equal to a1 then we will find we have to find that's true that is sum of first two terms from that we have to find the second term only then we have to find third term tenth term and the nth term so very easy question this is a main key point you also have to find the because we have to use the here here and here also and it is not mentioned in the question so keep remembering when we are having the two terms of an ap we will easily find the common difference and that is very much helpful to find the another terms also now the next is how many terms of this ap 9 comma 17 comma 25 must be taken to give the sum 636 now again very easy and important question in this question we are given the ap sequence and also give us the sum of the terms now we have to find how many terms collaborate to find that sum means we have to find the and the how many terms we can use to find that sum so very easy simple so let's start a that is the first term is 9 common difference that is 17 minus 9 is equal to 8 ascent that is 636 and we have to find the corresponding value of n so we will easily find this number of terms using the formula of sn ascent is equal to and upon 2 2a plus n minus 1 times of d i put the values as n is 6 36 that is and upon 2 2 a that is two nines are 18 plus n minus 1 and we have to find into d that is it is equal to 636 equal to n upon 2 18 plus 8 and minus 8 that is equal to 6 36 is equal to n upon 2 18 minus 8 that is 10 plus 8 10 now 6 36 is equal to 2 get cancelled with inner factors that is 5 plus 4 n that is 6 36 is equal to 5 and plus 4 and square now this is simply a quadratic equation so 5 and 4 and square 4 4 and square plus 5 and minus 6 36 is equal to 0. now the whole calculation turns into a quadratic equation now we have to solve this quantity question using the middle term splitting method or the discriminate method to get the value of n so we will use the middle term speeding method now we have to break five and in two parts such that the product sum of minus plus or minus is equal to five and product is equal to product of first and last term so simply that speed is the position is 53 and 48 so that is 4 n square plus 53 and minus 48 minus 636 that is equal to zero the middle term splits into two parts now from the first two terms from first two terms and get common that is 4 and plus 53 and the from last two terms minus 12 get common again that is 4 and plus 53 now is equal to 0 now these two factors are same so that means our calculation is okay now from the whole equation 4n plus 53 get common and the next factor second factor is n minus 12 that is equal to zero and we know that if the product of two factors is equal to zero that the first one is zero or the second one is zero so from here we will get four n plus 53 is equal to zero or n minus 12 is equal to 0 here the n is 12 and from here and is minus 53 upon 4 we will get two values of n but n is equal to minus 15 3 upon 4 get rejected because n is always a natural number because position is always a natural number at the first position second position third position fourth position position can't be negative and the position can't be a decimal number you can't say that a position is 2.5 or position is 3.5 so simply this value get rejected so n is equal to 12 this means 12 terms of this ap must be taken to get the sum 636 so this is how we can find the number of terms to find the particular sum the next one is find the sum of first 15 multiples of 8 now in this question we have to find the sum of first 15 multiples of it first 15 multiple means we have to use first 15 terms that are multiplied multiple of eight so simply first multiple of eight is clearly that is eight itself the second is 16 third is 24 and so on we have to find up to 15 terms so what is given to us that a is 8 difference is also 6 minus 8 that is 60 minus 8 is 8 and n is 15 we have to find the sum so what is how to find the sum s n is equal to n upon 2 2a plus and minus times of a common difference so we have to find the sum that is n upon 2 means 15 upon 2 2 a that is 2 is multiplied by 8 plus 15 and is 15 that is 15 minus 1 into difference and difference is also 8. that is equal to 15 upon 2 2 8 are 16 15 minus 1 14 into 8 that is equal to 15 upon 2 16 plus that is 1 1 2 14 here's a 1 12 that is equal to 15 upon 2 into 128 now that is equal to 15 multiplied by 64. so the sum is that is 640 and 320 that is 960 so sum of first 15 multiples of 8 is 960 that is 8 plus 16 plus 24 plus 32 up to first 15 multiples the whole sum is 960. the next one is a sum of rupees 700 is to be used to give seven cash prizes to students of a school for their overall academic pro performance if each price is 20 rupees less than its preceding price find the value of each cash price so very interesting question what is given to us so need a statement carefully a sum of 700 means the sum is 700 so sn is given 700 rupees to be used to give seven cash prizes so number of prices are seven so that means n is equal to seven student of school for the each their overall academic performance if each price is 20 rupees less means the difference between the cost of price is 20 that is difference is 20 difference is 20 less than its perceiving price find the value of each cash price we have to find the value of each cash price so very interesting and easy question simply we will find the use the formula faster first of all sn is equal to n upon 2 2 a plus n minus 1 times of d so what is s n that is 700 is equal to n that is 7 upon 2 2a you can't have the value of a so this is okay 7 minus 1 into d that is 20 so that is 700 into two seven by two turns into two by seven that is equal to two a plus seven minus one that is six six into twenty that is one twenty now seven hundred cancel out hundred that is hundred into two 200 is equal to 2a plus 120 now 120 gets subtracted from this side 200 minus 120 that is equal to 2 a 200 minus 120 is 80 is equal to 2 a and 80 upon 2 is equal to a that is 40 is equal to a so this is the value of a that is the cost price of first price so cost of first price is 40 cost of the next price is 20 more than that is 60 next is 80 next is 100 the next one is 120 next is 140 and the next last one is 160. so these are the seven cash prizes in which first cash price is 40 and the last one is 160 and look at here their sum is total sum is 170 40 600 180 280 400 and 450 and the 700 so total sum is 700 so this is how we can find the amount of each cash prize so the next one a spiral is made up of successive semicircles with centers alternatively at a b this starting center at a of radius 0.5 1.0 1.5 to 5 2.0 centimeter as shown in the total what is the total length of such spiral made of 3 13 constructive semi circles now very interesting questions x spiral form here look at here spiral is formed in which this is the first semicircle whose center is a and now this is the second semicircle whose center is b again this is the third semicircle whose center is again a and this is the first semicircle whose center is b so now alternatively the this is the next next semicircle whose center is a and this is the next semicircle of center b and so on the next one center a center b and so on using like that there are 13 semi circles and to form that spring and we have to find the length of total wire used here so look at here first of all the first radius radius one that is zero point five zero point five second radius is one third radius is 1.5 fourth radius is two and so on now as this is a semicircle so what is the length of wire used for semicircle that is equal to pi r because total circumference of the circle is 2 pi r and we have to deal with the semicircles only so length is equal to pi r so for that first for that first is the length is pi into r1 for the second one pi r2 for third one pi r3 and the 431 that is pi r 13 now from all pi get common we left with r1 plus r2 plus r3 up to so on r13 that is equal to pi r1 that is 0.5 r2 1 next our radius is 1.5 next is 2 or so on up to r 30 now look at here clearly the inner one in our form is the ap sequence 0.5 1 1.52 and the common difference is 0.5 so we have to solve it that is pi so what is the values are here what are the values are here a that is the first term is 0.5 common difference that is also 1 minus 0.5 and the number of terms are 13 because we are using the 13 semicircles so to find that sum we will use the formula that is n upon 2 2a plus n minus 1 times of difference so that is pi is 22 upon 7 and that is 13 upon 2 2 a double of 0.5 that is 1 plus n minus 1 that is 30 minus 1 12 into difference that is also 0.5 that is equal to 22 upon 7 into 13 upon 2 into 1 plus 12 0.50 that is 6 that is equal to 22 upon 7 into 13 by 2 into 6 plus 1 that is again 7 so overall that is equal to 22 upon 7 into 13 upon 2 into 7 7 get cancelled with 7 2 get cancelled with 22 that is 11 that is 11 into 13 that is 130 and 13 14 3 so total length of the wire used is 143 units so this is oh so this is how we can find the length of the wire used i will repeat it again a semicircle is formed using the semicircles with different radius first radius as 0.5 second radius zero one third radius will be one point five four three is true and so on thirteen semicircles are used to perform that spiral we have to find find the wire used to form that spiral so simply radius is zero point five one one point five four of two and so on so as spiral is made up of half semi semi circles so length of a semi circle is pi r one second semi circle is pi r two third is pi r three and so on pi r thirty now pi by get common from o will get pi into r1 plus r2 plus r3 up to shown r 13 and after substituting the values of radius clearly we will look at here this is a sum of and terms of an ap so now using the sum formula we will easily get that total 143 units of wire is used to form that spiral so this is all about the chapter that is ap series hope so you will like it please like share and subscribe our channel to watch more videos thank you\n"