Question ID: #776
Suppose $a, b, c$ are in A.P. and $a^{2}, 2b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a+b+c=1$ then $9(a^{2}+b^{2}+c^{2})$ is equal to
Solution:
Given $a, b, c$ are in A.P., let them be $b-d, b, b+d$.
Sum of terms:
$$ a+b+c = 1 $$
$$ (b-d) + b + (b+d) = 1 $$
$$ 3b = 1 \Rightarrow b = \frac{1}{3} $$
Given $a^2, 2b^2, c^2$ are in G.P.:
$$ (2b^2)^2 = a^2 c^2 $$
$$ 4b^4 = (ac)^2 $$
Taking the square root, $ac = 2b^2$ or $ac = -2b^2$.
Using $a = b-d$ and $c = b+d$, we have $ac = b^2 – d^2$.
**Case 1:** $b^2 – d^2 = 2b^2$
$$ -d^2 = b^2 $$
This gives $d^2 < 0$, which implies $d$ is imaginary. So, we reject this case.
**Case 2:** $b^2 – d^2 = -2b^2$
$$ d^2 = 3b^2 $$
Since we need $a^2 + b^2 + c^2$, we expand the terms:
$$ a^2 + b^2 + c^2 = (b-d)^2 + b^2 + (b+d)^2 $$
$$ = (b^2 – 2bd + d^2) + b^2 + (b^2 + 2bd + d^2) $$
$$ = 3b^2 + 2d^2 $$
Substitute $d^2 = 3b^2$ into the expression:
$$ \text{Sum} = 3b^2 + 2(3b^2) = 9b^2 $$
We need to find the value of $9(a^2 + b^2 + c^2)$:
$$ 9(9b^2) = 81b^2 $$
Substitute $b = \frac{1}{3}$:
$$ 81 \left(\frac{1}{3}\right)^2 = 81 \left(\frac{1}{9}\right) = 9 $$
Ans. (9)
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