Question

If $a_1,a_2,a_3\dots \dots \dots ,a_n$ are in $H.P.$ then $\frac{a_1}{a_2+a_3+\dots \dots \dots \dots +a_n},\frac{a_2}{a_1+a_3+\dots \dots \dots +a_n}\dots \dots \frac{a_n}{a_1+a_2+\dots \cdots +a_{n-1}}$ are in

• A. $A.P$
• B. $G.P$
• C. $H.P$
• D. None of these

Answer 1

The correct option is C $H.P$

$a_1,a_2,a_3.\dots ...$ are in HP

$\therefore \frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},..........$ are in $AP$

similarly if we multiply each term of this AP

by $(a_1+a_2+a_3+\cdots$ an then it will still be an $AP$ series

$\frac{a_1+a_2+\cdots +a_n}{a_1},\frac{a_1+a_2+\dots +a_n}{a_2},\cdots \cdot \frac{a_1+a_2+\cdots a_n}{a_n}$

are terms of $A\cdot P$

Now adding 1 to each term of this $AP$.

$1+\frac{a_1+a_2+\cdots a_n}{a_1},1+\frac{a_1+a_2+\cdots a_n}{a_2},\cdots \cdot 1+\frac{a_1+a_2+\cdots a_n}{a_n}$

are also in $AP$

Now subtracting 1 from each term will also give an $AP$

$\frac{a_2+a_3+\cdots +a_n}{a_1},\frac{a_1+a_3+\cdots +a_n}{a_2},\cdots \cdot \frac{a_1+a_2+\cdots a_{n-1}}{a_n}$

so reciprocal of this series will give $HP$

$\frac{a_1}{a_2+a_3+\cdots +a_n},\frac{a_2}{a_1+a_3+\cdots a_n},\cdots \cdot \cdot ,\frac{a_n}{a_1+a_2+\cdots a_{n-1}}$

is our given $HP$ series

Answer: option (c).