If a1- a2- - are in H-P- and fk - -r - 1na1 - ai- then a1f1- a2f2- -anfn are in

The correct option is C H.P. Given f( k ) =[ ∑ #160; #160; _ r=1 ^ n a _ r ] a _ k = S _ n a _ k f( k ) #160; a _ k = S _ n a _ k ...

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Byju's Answer

Standard X

Mathematics

Formula for Sum of n Terms of an AP

If a1, a2, ...

Question

If $a_{1}, a_{2}, . . . .$ are in H.P. and $f (k) = n \sum r = 1 (a_{1} = a_{i})$ , then $\frac{a_{1}}{f (1)}, \frac{a_{2}}{f (2)}, . . . \frac{a_{n}}{f (n)}$ are in

A . P .

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G . P .

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H . P .

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None of these

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Solution

The correct option is C $H . P .$
Given $f (k) = [\sum_{r = 1}^{n} a_{r}] - a_{k} = S_{n} - a_{k} \frac{f (k)}{a_{k}} = \frac{S_{n}}{a_{k}} - 1$
Given $a_{1}, a_{2}, a_{3}, . . . . ., a_{n}$ are in HP
$\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, . . . ., \frac{1}{a_{n}}$ are in AP
Multiply each term with $S_{n}$
$\frac{S_{n}}{a_{1}}, \frac{S_{n}}{a_{2}}, \frac{S_{n}}{a_{3}}, . . . ., \frac{S_{n}}{a_{n}}$ are in AP
Subtracting $1$ from each term we get
$\frac{S_{n}}{a_{1}} - 1, \frac{S_{n}}{a_{2}} - 1, \frac{S_{n}}{a_{3}} - 1, . . . ., \frac{S_{n}}{a_{n}} - 1$ are in AP
$\frac{f (1)}{a_{1}}, \frac{f (2)}{a_{2}}, \frac{f (3)}{a_{3}}, . . . ., \frac{f (n)}{a_{n}}$ are in AP
$\frac{a_{1}}{f (1)}, \frac{a_{2}}{f (2)}, \frac{a_{3}}{f (3)}, . . . ., \frac{a_{n}}{f (n)}$ are in HP

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}

are in H.P and

f_{k} = \sum_{r = 1}^{n} a_{r} - a_{k}

; then

2^{α_{1}}, 2^{α_{2}}, 2^{α_{3}}, 2^{α_{4}}, . . . .

are in
{where

α_{1} = \frac{a_{1}}{f_{1}}, α_{2} = \frac{a_{2}}{f_{2}}, α_{3} = \frac{a_{3}}{f_{3}} . . . .

}

Let a $_{1}$ , a $_{2}$ , a $_{3}$ , a $_{4}$ and a $_{5}$ be such that a $_{1}$ , a $_{2}$ and a $_{3}$ are in A.P., a $_{2}$ , a $_{3}$ and a $_{4}$ are in G.P., and a $_{3}$ , a $_{4}$ and a $_{5}$ are in H.P. Then log $_{e}$ a $_{1}$ , log $_{e}$ $a_{3}$ and log $_{e}$ a $_{5}$ are in