Suppose a, b, c are in A.P. and a^ 2 , 2 ~b^ 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 \_\_\_\_ .

Correct answer: 0 Since a, b, c are in A.P., we have 2b = a + c . With a + b + c = 1 , we get 3b = 1 , so b = (1)/(3) and a + c = (2)/(3) . Since a^2, 2b^2, c^2 are in G.P.: (2b^2)^2 = a^2 c^2 , giving 4b^4 = a^2c^2 . With b = (1)/(3) : a^2c^2 = (4)/(81) . For real values with a Thus a, c satisfy t^2 - (2)/(3)t - (2)/(9) = 0 , giving t = 1 ± 3 3 . So a = 1- 3 3 , c = 1+ 3 3 . Calculating: a^2 + c^2 = (1- 3 )^2 + (1+ 3 )^2 9 = 4-2 3 +4+2 3 9 = (8)/(9) . Therefore a^2 + b^2 + c^2 = (8)/(9) + (1)/(9) = 1 , and 9(a^2 + b^2 + c^2) = 9 .

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Sequences and Series

Suppose a, b, c are in A.P. and a^ 2 , 2 ~b^ 2 , c^ 2 are in G.P....

Suppose

\mathrm{a}, \mathrm{b}, \mathrm{c}

are in A.P. and

\mathrm{a}^{2}, 2 \mathrm{~b}^{2}, \mathrm{c}^{2} are in G.P. If

\mathrm{a}<\mathrm{b}<\mathrm{c}

and

\mathrm{a}+\mathrm{b}+\mathrm{c}=1

, then

\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

Since

a, b, c

are in A.P., we have

2b = a + c

.
With

a + b + c = 1

, we get

3b = 1

, so

b = \frac{1}{3}

and

a + c = \frac{2}{3}

.
Since

a^2, 2b^2, c^2

are in G.P.:

(2b^2)^2 = a^2 c^2

, giving

4b^4 = a^2c^2

.
With

b = \frac{1}{3}

a^2c^2 = \frac{4}{81}

.
For real values with

a < b < c

, we need

ac = -\frac{2}{9}

.
Thus

a, c

satisfy

t^2 - \frac{2}{3}t - \frac{2}{9} = 0

, giving

t = \frac{1 \pm \sqrt{3}}{3}

. So

a = \frac{1-\sqrt{3}}{3}

c = \frac{1+\sqrt{3}}{3}

.
Calculating:

a^2 + c^2 = \frac{(1-\sqrt{3})^2 + (1+\sqrt{3})^2}{9} = \frac{4-2\sqrt{3}+4+2\sqrt{3}}{9} = \frac{8}{9}

.
Therefore

a^2 + b^2 + c^2 = \frac{8}{9} + \frac{1}{9} = 1

, and

9(a^2 + b^2 + c^2) = 9

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