Question

If $a_1,a_2,a_3$ are in H.P and $f_k={\sum }_{r=1}^n{a}_r-a_k$; then $2^{{\alpha }_1},2^{{\alpha }_2},2^{{\alpha }_3},2^{{\alpha }_4},....$ are in
{where ${\alpha }_1=\frac{a_1}{f_1},{\alpha }_2=\frac{a_2}{f_2},{\alpha }_3=\frac{a_3}{f_3}....$}

• A. A.P
• B. G.P
• C. H.P
• D. none of these

Answer 1

The correct option is D none of these

Given that, $a_1,a_2,...,a_n$ are in H.P.
$\Rightarrow \frac{1}{a_1},\frac{1}{a_2},...,\frac{1}{a_n}$ are in A.P.
$\Rightarrow \frac{a_1+a_2+...+a_n}{a_1},\frac{a_1+a_2+...+a_n}{a_2},...,\frac{a_1+a_2+...+a_n}{a_n}$ are in A.P.
$\Rightarrow \frac{a_1+a_2+...+a_n}{a_1}-1,\frac{a_1+a_2+...+a_n}{a_2}-1,...,\frac{a_1+a_2+...+a_n}{a_n}-1$ are in A.P.
$\Rightarrow \frac{a_2+a_3+...+a_n}{a_1},\frac{a_1+a_3+...+a_n}{a_2},...,\frac{a_1+a_2+...+a_{n-1}}{a_n}$ are in A.P.
$\Rightarrow \frac{f_1}{a_1},\frac{f_2}{a_2},...,\frac{f_n}{a_n}$ are in A.P.
$\Rightarrow {\alpha }_1,{\alpha }_2,...,{\alpha }_n$ are in H.P.
Therefore, $2^{{\alpha }_1},2^{{\alpha }_2},2^{{\alpha }_3},2^{{\alpha }_4},....$ are not in A.P, H.P or G.P
Ans: D