If a,b,c are in A.P. , then frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} =

English

Gujarati

Hindi

8. Sequences and Series

easy

If $a,\;b,\;c$ are in $A.P.$, then $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = $

A

$1$

B

$2$

C

$3$

D

$4$

Solution

(d) If $a,\;b,\;c$ are in $A.P.$ $ \Rightarrow $$2b = a + c$

So, $\frac{{{{(a – c)}^2}}}{{({b^2} – ac)}} = \frac{{{{(a – c)}^2}}}{{\left\{ {{{\left( {\frac{{a + c}}{2}} \right)}^2} – ac} \right\}}}$

$ = \frac{{{{(a – c)}^2}4}}{{[{a^2} + {c^2} + 2ac – 4ac]}} = \frac{{4{{(a – c)}^2}}}{{{{(a – c)}^2}}} = 4$.

Trick : Put $a = 1,\;b = 2,\;c = 3$,

then the required value is $\frac{4}{1} = 4$.

Standard 11

Mathematics

Share

Similar Questions

Between $1$ and $31, m$ numbers have been inserted in such a way that the resulting sequence is an $A. P.$ and the ratio of $7^{\text {th }}$ and $(m-1)^{\text {th }}$ numbers is $5: 9 .$ Find the value of $m$

medium

If ${a^2},\;{b^2},\;{c^2}$ are in $A.P.$, then ${(b + c)^{ – 1}},\;{(c + a)^{ – 1}}$ and ${(a + b)^{ – 1}}$ will be in

hard

If $\frac{1}{{b – c}},\;\frac{1}{{c – a}},\;\frac{1}{{a – b}}$ be consecutive terms of an $A.P.$, then ${(b – c)^2},\;{(c – a)^2},\;{(a – b)^2}$ will be in

hard

The sums of $n$ terms of two arithmatic series are in the ratio $2n + 3:6n + 5$, then the ratio of their ${13^{th}}$ terms is

medium

If $x,y,z$ are in $A.P. $ and ${\tan ^{ – 1}}x,{\tan ^{ – 1}}y$ and ${\tan ^{ – 1}}z$ are also in $A.P.$, then

medium