Thank you for reporting, we will resolve it shortly
Q.
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 $n=$
Sequences and Series
Solution:
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} $
$\Rightarrow n=\frac{2 S_{3} S_{1}-S_{1} S_{2}-S_{2} S_{3}}{S_{1}-2 S_{2}+S_{3}}$