Question

If $S_1,S_2,S_3$ denote the sums of n terms of three A.P.s whose first terms are unity and common differences in H.P. prove that $n=\frac{2S_3{S}_1-S_1{S}_2-S_2{S}_3}{S_1-2S_2+S_3}$

Answer 1

$S_1=\frac{n}{2}[2\cdot 1+(n-1\left)d_1\right]\therefore \frac{2(s_1-2)}{n(n-1)}=d_1$
or $\frac{1}{d_1}=\frac{n(n-1)}{2(s_1-2)}$
Similarly for $\frac{1}{d_2}$ and $\frac{1}{d_3}$
Sinced $d_1,d_2,d_3$ are given to be in H.P. ,therefore $\frac{1}{d_1},\frac{1}{d_2},\frac{1}{d_3}$ are in A.P. or on cancelling $\frac{n(n-1)}{2}$ we can say that
$\frac{1}{S_1-n},\frac{1}{S_2-n},\frac{1}{S_3-n}$ are in A.P.
$\therefore \frac{1}{S_2-n}-\frac{1}{S_1-n}=\frac{1}{S_3-n}-\frac{1}{S_2-n}$
or $\therefore \frac{S_1-S_2}{S_1-n}=\frac{S_2-S_3}{S_3-n}\cdot$ cross multiply
or $(S_1-S_2)S_3-(S_2-S_3)S_1=n(S_1-2S_2+S_3)$
$\therefore$ $n=\frac{2S_3{S}_1-S_1{S}_2-S_2{S}_3}{S_1-2S_2+S_3}$

Hence proved.