If a,b,c are in A.P. , b,c,d are in G.P. and c,d,e are in H.P. , then a,c,e are in

English

Gujarati

Hindi

8. Sequences and Series

hard

If $a,\;b,\;c$ are in $A.P.$, $b,\;c,\;d$ are in $G.P.$ and $c,\;d,\;e$ are in $H.P.$, then $a,\;c,\;e$ are in

A

No particular order

B

$A.P.$

C

$G.P.$

D

$H.P.$

Solution

(c) $a,\;b,\;c$ are in $A.P.$ then $2b = a + c$…..$(i)$

$b,\;c,\;d$ are in $G.P. $ then ${c^2} = bd$….$(ii)$

$c,\;d,\;e$ are in $H.P.$ then $d = \frac{{2ce}}{{c + e}}$….$(iii)$

From $(ii)$, $c^2=bd={(a+c)\over 2}{2ce\over(c+e)}$

$ \Rightarrow $ ${c^2} = \frac{{ace + {c^2}e}}{{c + e}}$

$\Rightarrow {c^3} + {c^2}e = ace + {c^2}e$

$ \Rightarrow $ ${c^3} = ace$

$\Rightarrow {c^2} = ae$

Hence $a,\;c,\;e$ will be in $G.P.$

Standard 11

Mathematics

Share

Similar Questions

Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.

$STATEMENT-1$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are neither in $A.P$. nor in $G.P.$ and

$STATEMENT-2$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are in $H.P.$

normal

(IIT-2008)

Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$

medium

Let $a _1, a _2, a _3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text {th }}$ and $8^{\text {th }}$ terms be $2$ and the product of its $3^{\text {rd }}$ and $5^{\text {th }}$ terms be $\frac{1}{9}$. Then $6\left( a _2+\right.$ $\left.a_4\right)\left(a_4+a_6\right)$ is equal to

hard

(JEE MAIN-2023)

The sum of the series $5.05 + 1.212 + 0.29088 + …\,\infty $ is

easy

If $2(y – a)$ is the $H.M.$ between $y – x$ and $y – z$, then $x – a,\;y – a,\;z – a$ are in

hard