Question

Second and fourth term of an A.P. is 12 and 20 respectively. Find the sum of first 25 terms of that A.P.

Answer 1

We have t₂ = 12 and t₄ = 20 of the given A.P.
We know that the n^th term of an A. P. is t_n = a + (n – 1)d.
Thus, we have:
t₂ = 12 $\Rightarrow$a + (2 – 1)d = 12 $\Rightarrow$a + d = 12 …(1)
t₄ = 20$\Rightarrow$a + (4 – 1)d = 20 $\Rightarrow$ a + 3d = 20 …(2)
On subtracting (1) from (2), we get:
a + 3d – a – d = 20 – 12 = 8
$\Rightarrow$2d = 8
$\Rightarrow$d = $\frac{8}{2}=4$
From equation (1), we get:
a + 4 = 12
$\Rightarrow$a = 12 – 4 = 8

We know:
Sum of n terms of an A.P., $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

To find the sum of the first 25 terms, we need to take a = 8, d = 4 and n = 25.
Thus, we have:

$$\begin{gathered} S_{25}=\frac{25}{2}\left[2\times 8+\left(25-1\right)\times 4\right] \\ =\frac{25}{2}\left[16+24\times 4\right] \\ =\frac{25}{2}\times \left[16+96\right] \\ =\frac{25}{2}\times 112 \\ =1400 \end{gathered}$$