This is the basic and conceptual question from Chapter name- Arithmetic Progression

Chapter number- 9

Exercise 9.6

In this question we have been given that the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289

Now we have to find the sum of n terms.

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Class:- 10th

Solutions of CBSE Mathematics

Question 22

We know sum of n terms of an A.P. is given by S

_{n}Â = n[2a + (n âˆ’ 1)d]/2.So we get, S

_{7}Â = 49=> 7[2a + (7 âˆ’ 1)d]/2 = 49

=> 7[a + 3d] = 49

=> a + 3d = 7 â€¦.. (1)

And also, S

_{17}Â = 289=> 17[2a + (17 âˆ’ 1)d]/2 = 289

=> 17[a + 8d] = 289

=> a + 8d = 17 â€¦.. (2)

On subtracting eq(1) from (2), we get,

=> a + 8d âˆ’ (a + 3d) = 17 âˆ’ 7

=> 5d = 10

=> d = 2

On putting d = 2 in (1), we get,

=> a + 3(2) = 7

=> a = 1

Here a = 1, d = 2, so sum of n terms would be,

S

_{n}Â = n[2(1) + (n âˆ’ 1)(2)]/2= n[2 + 2n âˆ’ 2]/2

= n[n] = n

^{2}Hence, the sum of n terms of the given A.P is n^{2}.