Question

. If the sum of first seven terms of an AP is 49 and that of seventeen terms is 289. Find the sum of

first n terms​

Answer 1

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

sum of first 7 terms = 49

S7= 7/2 (2a+(7-1)d)

49=7/2 (2a+(6)d)

49×2/7=(2a+6d)

14 = 2a+6d

14-6d = 2a

14-6d/2 = a

a=7-3d.............1

sum of first seventeen terms = 289

S17 = 17/2 (2a+(17-1)d)

289 = 17/2 (2a+(16)d)

289×2/7 = (2a+16d)

34 = 2a+16d

34-16d = 2a

34-16d/2 = a

a=17-8d...............2

from (1) and (2)

7-3d = 17-8d

8d-3d = 17-7

5d = 10

d= 10/5

d=2

a = 7-3d

a = 7-3×2

a = 7-6

a = 1

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

=n/2 (2×1 + (n-1) 2)

=n/2 (2 + (n-1) 2)

=n/2(2 + 2n - 2)

=n/2(0 + 2n)

=n/2 × 2n

=n×n

=n2