If a, b, c are in A.P.

Question

If  a,   b,   c  are in A.P. and  a2,  b2,  c2 are in G.P. such that

a < b < c and   a + b + c = $${3 \over 4},$$ then the value of a is :

Accepted Answer

$$ \because $$ $\,\,\,$$a, b, c are in A.P. then

a + c = 2b

also it is given that,

a + b + c = $${{3 \over 4}}$. . . .(1)$ \Rightarrow $$ $\,\,\,$$ 2b + b = $${{3 \over 4}}$$   $$ \Rightarrow $$ $\,\,\,$b =${{1 \over 4}}$$  . . . . .(2)

Again it is given that, a2, b2, c2 are in G.P. then

(b2)2 = a2c2   $$ \Rightarrow $ac =$ \pm $${{1 \over {16}}}$$      . . . . (3)

From (1), (2) and (3), we get;

$$a \pm {1 \over {16a}}$=${1 \over 2}$$ \Rightarrow $16a2$-$8a$ \pm $1 = 0

Case I : 16a2$-$8a + 1 = 0$ \Rightarrow $$ $\,\,\,$$a = $${1 \over 4}$$ (not possible as a < b)

Case II: 16a2 - 8a - 1 = 0

$$ \Rightarrow $$ $\,\,\,$a =${{8 \pm \sqrt {128} } \over {32}}$$ \Rightarrow $$ $\,\,\,$$ a = $${1 \over 4} \pm {1 \over {2\sqrt 2 }}$$

$$ \therefore $$ $\,\,\,$a =${1 \over 4} - {1 \over {2\sqrt 2 }}$($ \because $

Solutions · 1