Question

The sums of n terms of three A.P.'s whose first term is 1 and common differences are 1, 2, 3 are $s_1,s_2,s_3$ respectively. The true relation is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We have $a_1=a_2=a_3=1$ and $d_1=1,d_2=2,d_3=3$.
Therefore, $S_1=\frac{n}{2}(n+1)$ .....(i)
$S_2=\frac{n}{2}\left[2n\right]$ ....(ii)
$S_3=\frac{n}{2}[3n-1]$ ....(iii)
Adding (i) and (iii),
$S_1+S_3=\frac{n}{2}\left[\right(n+1)+(3n-1\left)\right]=2\left[\frac{n}{2}\right(2n\left)\right]=2S_2$
Hence correct relation $S_1+S_3=2S_2$.