Start free

Mathematics · Sequence and Series

JEE Main 2026 — 22 January, Evening Shift — Question 24

Suppose a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} are in A.P. and a2,2 b2,c2\mathrm{a}^{2}, 2 \mathrm{~b}^{2}, \mathrm{c}^{2} are in G.P. If a<b<c\mathrm{a}<\mathrm{b}<\mathrm{c} and a+b+c=1\mathrm{a}+\mathrm{b}+\mathrm{c}=1, then 9(a2+b2+c29\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\right. \mathrm{c}^{2} ) is equal to ____\_\_\_\_ .

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

a=b−d,c=b+d⇒b=13a = b - d, \quad c = b + d \Rightarrow b = \frac{1}{3} ⇒4b4=a2c2\Rightarrow 4b^4 = a^2 c^2 ⇒4b4=[(b−d)(b+d)]2\Rightarrow 4b^4 = \big[(b-d)(b+d)\big]^2 ⇒481=(19−d2)2\Rightarrow \frac{4}{81} = \left(\frac{1}{9} - d^2\right)^2 ⇒(19−d2)=±29\Rightarrow \left(\frac{1}{9} - d^2\right) = \pm \frac{2}{9} d2=13d^2 = \frac{1}{3} ⇒d=±13(as a<b<c)\Rightarrow d = \pm \frac{1}{\sqrt{3}} \quad \text{(as } a < b < c \text{)} ∴9(a2+b2+c2)\therefore 9(a^2 + b^2 + c^2) =9[(13−13)2+(13)2+(13+13)2]= 9\left[ \left(\frac{1}{3} - \frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3} + \frac{1}{\sqrt{3}}\right)^2 \right] =9(13+23)=3+6=9= 9\left(\frac{1}{3} + \frac{2}{3}\right) = 3 + 6 = 9

Answer key and solution verified before publishing.

Practise Sequence and Series

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Sequence and Series
Topic
Geometric Progression