# If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289.Find the sum of first n terms

7 = 7/2 [2a+(6)d]

49* 2/7 = 2a+6d

7*2 = 2a+6d

14 = 2a+6d........(1)

s17= 17/2 (2a+16d)

289*2/17 = 2a+16d

17*2 = 2a + 16d

34 = 2a + 16d .........(2)

subtracting eq. 2 from 1 we will get

20 = 10 d

so ,d = 2

putting d = 2 in eq. 1

14 = 2a + 6 * 2

14 = 2a + 12

14 - 12 = 2a

2a = 2

a = 1

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

=n/2[2(n)]

=n2

sn =n2

so sum of first natural number = n2

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YOU ARE RIGHT.................

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I HAVE THUMBS UP YOUR ANSWER NOW THUMBS UP MINE ALSO..........

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wow great u have helped me a lot mohil
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Corrrct! Helped me for my boards☺
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I done this in other method but it is also correct can you plz check it.. 7a+21d=49 17a+136d=289 Solving these we get, a=1 and d=2
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nis the sum of n terms of this particular a.p
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Here it is...

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!!
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If x square is equals to 2 by 3 and X = 2 - 3 are roots of the quadratic equation X square + 7 x + 20 20 find the values of a and b
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Please answer for this question
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Thanks, it helps alot
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I think it is correct

• -12
Check if it's right

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answer is here...

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Hope dis helped you a lot.....

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I hope u understand this question 👍👍

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All Answers correct
Great job!!
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S 7 = 7/2 [2a+(6)d] 49* 2/7 = 2a+6d 7*2 = 2a+6d 14 = 2a+6d........(1) s17= 17/2 (2a+16d) 289*2/17 = 2a+16d 17*2 = 2a + 16d 34 = 2a + 16d .........(2) subtracting eq. 2 from 1 we will get 20 = 10 d so ,d = 2 putting d = 2 in eq. 1 14 = 2a + 6 * 2 14 = 2a + 12 14 - 12 = 2a 2a = 2 a = 1 sn= n/2 [2a+ (n-1)d] =n/2[2(n)] =n2 sn =n2 so sum of first natural number = n2
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S 7 = 7/2 [2a+(6)d] 49* 2/7 = 2a+6d 7*2 = 2a+6d 14 = 2a+6d........(1) s17= 17/2 (2a+16d) 289*2/17 = 2a+16d 17*2 = 2a + 16d 34 = 2a + 16d .........(2) subtracting eq. 2 from 1 we will get 20 = 10 d so ,d = 2 putting d = 2 in eq. 1 14 = 2a + 6 * 2 14 = 2a + 12 14 - 12 = 2a 2a = 2 a = 1 sn= n/2 [2a+ (n-1)d] =n/2[2(n)] =n2 sn =n2 so sum of first natural number = n2
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equate the sn of 7 and 17
ans= n^2
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