Question

If the sum of first 7 terms of AP is 49 and that of first 17 terms is 289, find the sum of

first n terms.

Answer 1

$s7 = \frac{7}{2} (2a + ( 6)d) \\ 49 \times \frac{2}{7} = 2 a+ 6d \\ 14 = 2(a + 3d) \\ a + 3d = 7...............(1) \\ s17 = \: \frac{17}{2} (2a + ( 16)d) \\ 289 \times \frac{2}{17} = 2(a + 8d) \\ 17 = a + 8d............(2) \\ sub \: 1 \: and2 \: we \: get \\ d = 2 \\ nd \: a \: = 1 \\ sn = \frac{n}{2} (2 \times 1 + (n - 1) \times 2) \\ = \frac{n}{2} (2 + 2n - 2) \\ = \frac{n}{2}(2n) \\ = {n}^{2}$

Answer 2

s7=49

s17=289

sn=?

s7=7/2[2a+6d]=49

s17=17/2[2a+16d]=289

7a+21d=49

17a+136d=289

17[7a+21d=49]

7[17a+136d=289]

119a+357d=833

119a+952d=2023

595d=1190

d=1190/595

d=2

substituting value of d,we get,

7a+21*2=49

7a=49-42

7a=7

a=7/7=1

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

sn=n/2[2+2n-2]

sn=n/2(2n)

sn=n*n=n^2

sn=n^2