Question

Find the sum of 51 terms of the AP whose second term is 2 and the 4th term is 8.

Answer 1

Let a be the first term and d be the common difference of the given AP.
Then T₂ = 2 and T₄ = 8

⇒ a + (2 - 1)d = 2 and a + (4 - 1)d = 8
i.e., a + d = 2 ...(i)
also, a + 3d = 8 ...(ii)
On subtracting (i) from (ii), we get:
2d = 6
⇒ d = 3
Putting d = 3 in (i), we get:
a = -1
∴ a = -1, d = 3 and n = 51

Now, sum of the first 51 terms = $\frac{n}{2}[2a+\left(n-1\right)d]$

$$\begin{gathered} =\left(\frac{51}{2}\right)\times \left[2\times \left(-1\right)+\left(51-1\right)\times 3\right] \\ =\left(\frac{51}{2}\right)\times \left[-2+150\right]=\left(\frac{51}{2}\right)\times 148=51\times 74=3774 \end{gathered}$$

Hence, the required sum is 3774.