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If a, b, c are in A.P. and $a^{2},b^{2},c^{2}$ are in G.P. such that $a< b< c$ and $a+ b+ c=\frac{3}{4}$ , then the value of a is :

Option 1)
$\frac{1}{4}-\frac{1}{4\sqrt{2}}$

Option 2)
$\frac{1}{4}-\frac{1}{3\sqrt{2}}$

Option 3)
$\frac{1}{4}-\frac{1}{2\sqrt{2}}$

Option 4)
$\frac{1}{4}-\frac{1}{\sqrt{2}}$

Answers (2)

best_answer

As we learned

Selection of terms -

If we have to take 3 terms in an AP, whose sum is known, we take them as a - d, a, a + d.

- wherein

Extension : If we have to take ( 2r + 1) terms we take them as

a - rd, a - (r - 1) d - - - - - - - -,a - d, a, a+d, - - - - - - - a + rd.

and

Geometric Progession (GP) -

A progression of non - zero terms, in which every term bears to the preceding term a constant ratio.

- wherein

eg 2, 4, 8, 16,- - - - - -

and

100, 10, 1, 1/10,- - - - - - -

Let b=a+d

Also, a+b+c= $\frac{3}{4}$

and a=b-d ; c=b+d.

Thus a+b+c= $\frac{3}{4}$ =3b $\Rightarrow$ b= $\frac{1}{4}$

Thus numbers are $\frac{1}{4}-d,\frac{1}{4},\frac{1}{4}+d$

Also $\left (\left (\frac{1}{4} \right )^{2} \right )^{2}=\left (\frac{1}{4}-d \right )^{2}\cdot \left (\frac{1}{4}+d \right )^{2}$

$\Rightarrow \: \frac{1}{256}=\left ( \frac{1}{16}-d^{2} \right )^{2}$

$\Rightarrow \: \pm \frac{1}{16}=\frac{1}{16}-d^{2}$

Thus either $d^{2}=0$ or $d^{2}=\frac{1}{8}$

Thus $d=\pm \frac{1}{2\sqrt{2}};\: but\: d> 0\: thus\: d=\frac{1}{2\sqrt{2}}$

$a=\frac{1}{4}-\frac{1}{2\sqrt{2}}$

Option 1)

$\frac{1}{4}-\frac{1}{4\sqrt{2}}$

Option 2)

$\frac{1}{4}-\frac{1}{3\sqrt{2}}$

Option 3)

$\frac{1}{4}-\frac{1}{2\sqrt{2}}$

Option 4)

$\frac{1}{4}-\frac{1}{\sqrt{2}}$

Posted by

Himanshu

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