If a1- a2- a3- - an are in H-P- then prove that a1a2 - a3 - - an- a2a2 - a3 - - an- a3a2 - a3 - - an are in H-P-

We are given that 1a_1,1a_2,1a_3,.....1a_n are in A.P. #160; ∴ #160; ∑ a_na_1,∑ a_na_2,∑ a_na_3,..... are alos in A.P. or 1 + a_2 + ...

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Byju's Answer

Standard XII

Mathematics

Greatest Binomial Coefficients

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}, . . . . ., a_{n}$ are in H.P. then prove that
$\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{2} + a_{3} + . . . a_{n}}$ are in H.P.

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Solution

We are given that $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, . . . . . \frac{1}{a_{n}}$ are in A.P.
$∴ \frac{\sum a_{n}}{a_{1}}, \frac{\sum a_{n}}{a_{2}}, \frac{\sum a_{n}}{a_{3}}, . . . . .$ are alos in A.P.
$o r 1 + \frac{a_{2} + a_{3} + . . . . + a_{n}}{a_{1}}, 1 + \frac{a_{1} + a_{3} + . . . . + a_{n}}{a_{2}}, . . . . .$ are in A. P.
$\frac{a_{2} + a_{3} + . . . . + a_{n}}{a_{1}}, \frac{a_{1} + a_{3} + . . . . + a_{n}}{a_{2}}, . . . . .$ are in A.P.
Take their reciprocal and hence given numbers are in H.P.

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If a1,a2,a3,.....,an are in H.P. then prove that a1a2+a3+...an,a2a2+a3+...an,a3a2+a3+...an are in H.P.

We are given that 1a1,1a2,1a3,.....1an are in A.P. ∴∑ana1,∑ana2,∑ana3,..... are alos in A.P.or1+a2+a3+....+ana1,1+a1+a3+....+ana2,..... are in A. P. a2+a3+....+ana1,a1+a3+....+ana2,..... are in A.P. Take their reciprocal and hence given numbers are in H.P.

If $a_{1}, a_{2}, a_{3}, . . . . ., a_{n}$ are in H.P. then prove that
$\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{2} + a_{3} + . . . a_{n}}$ are in H.P.