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

The sum of first n terms of this AP is n².

Step-by-step explanation:

The sum of first n terms of an AP is:

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

Given:

S₇ = 49

S₁₇ = 289

Compute the equations in terms of a and d as follows:

$S_{7}=\frac{7}{2}[2a+(7-1)d]\\49\times2=7[2a+6d]\\2a+6d=14...(i)$

$S_{17}=\frac{17}{2}[2a+(17-1)d]\\289\times2=17[2a+16d]\\2a+16d=34...(ii)$

Subtract (i) from (ii) as follows:

2a + 16d = 34

-2a - 6d =  -14

___________

10d = 20

d = 2

Compute the value of a as follows:

2a + 6d = 14

2a + 12 = 14

2a = 2

a = 1

Then the sum of first n terms is:

$S_{n}=\frac{n}{2}[(2\times 1)+(n-1)\times 2]\\=\frac{n}{2}[2+2(n-1)]\\=\frac{n}{2}[2+2n-2]\\=n^{2}$

Thus, the sum of first n terms of this AP is n².