Question

The sum of first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms.

Answer 1

Let a be the first term and d be the common difference of the given AP.
Then we have:
$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ {S}_7=\frac{7}{2}\left[2a+6d\right]=7[a+3d] \\ {S}_{17}=\frac{17}{2}\left[2a+16d\right]=17[a+8d] \end{gathered}$$

However, S₇ = 49 and S₁₇ = 289
Now, 7[a + 3d] = 49
⇒ a + 3d = 7 ...(i)
Also, 17[a + 8d] = 289
​⇒ a + 8d = 17 ...(ii)

Subtracting (i) from (ii), we get:

5d = 10
⇒ d = 2

Putting d = 2 in (i), we get:
a + 6 = 7
⇒ a = 1
Thus, a = 1 and d = 2

∴ Sum of n terms of AP = $\frac{n}{2}\left[2\times 1+\left(n-1\right)\times 2\right]=n[1+(n-1\left)\right]=n^2$