Question

If $S_1,S_2$ and $S_3$ denote the sums up to $n>1$ terms of three sequences in A.P. whose first terms are unity and common differences are in H.P. then $n=$

• A. $\frac{2S_3{S}_1+S_1{S}_2+S_2{S}_3}{S_1-2S_2+S_3}$
• B. $\frac{2S_3{S}_1-S_1{S}_2-S_2{S}_3}{S_1+2S_2+S_3}$
• C. $\frac{2S_3{S}_1-S_1{S}_2-S_2{S}_3}{S_1-2S_2+S_3}$
• D. None of these

Answer 1

Answer: (C)

Let the common difference of the three A.P.s be $d_1,d_2$ and $d_3$ Then, we have
$S_1=\frac{n}{2}\left[2.1+(n-1)d_1\right]$
$\Rightarrow d_1=\frac{2\left(S_1-n\right)}{n(n-1)}(1)$
Similarly, $d_2=\frac{2\left(S_2-n\right)}{n(n-1)}(2)$
and, $d_3=\frac{2\left(S_3-n\right)}{n(n-1)}(3)$
Since $d_1,d_2$ and $d_3$ are given to be in H.P, therefore,
$\frac{1}{d_2}-\frac{1}{d_1}=\frac{1}{d_3}-\frac{1}{d_2}$
$\Rightarrow \frac{1}{S_2-n}-\frac{1}{S_1-n}=\frac{1}{S_3-n}-\frac{1}{S_2-n}$
[Using results $(1),(2),(3)$ ]
$\Rightarrow \frac{S_1-S_2}{\left(S_2-n\right)\left(S_1-n\right)}=\frac{S_2-S_3}{\left(S_2-n\right)\left(S_3-n\right)}$
$\Rightarrow \frac{S_1-S_2}{S_1-n}=\frac{S_2-S_3}{S_3-n}$
$\Rightarrow n=\frac{2S_3{S}_1-S_1{S}_2-S_2{S}_3}{S_1-2S_2+S_3}$