JEE Main 2026 — Mathematics Algebra

Since $a, b, c$a,b,c are in A.P., we have $2b = a + c$2b=a+c.

With $a + b + c = 1$a+b+c=1, we get $3b = 1$3b=1, so $b = \frac{1}{3}$b=31​ and $a + c = \frac{2}{3}$a+c=32​.

Since $a^2, 2b^2, c^2$a2,2b2,c2 are in G.P.: $(2b^2)^2 = a^2 c^2$(2b2)2=a2c2, giving $4b^4 = a^2c^2$4b4=a2c2.

With $b = \frac{1}{3}$b=31​: $a^2c^2 = \frac{4}{81}$a2c2=814​.

For real values with $a < b < c$a<b<c, we need $ac = -\frac{2}{9}$ac=−92​.

Thus $a, c$a,c satisfy $t^2 - \frac{2}{3}t - \frac{2}{9} = 0$t2−32​t−92​=0, giving $t = \frac{1 \pm \sqrt{3}}{3}$t=31±3​​. So $a = \frac{1-\sqrt{3}}{3}$a=31−3​​, $c = \frac{1+\sqrt{3}}{3}$c=31+3​​.

Calculating: $a^2 + c^2 = \frac{(1-\sqrt{3})^2 + (1+\sqrt{3})^2}{9} = \frac{4-2\sqrt{3}+4+2\sqrt{3}}{9} = \frac{8}{9}$a2+c2=9(1−3​)2+(1+3​)2​=94−23​+4+23​​=98​.

Therefore $a^2 + b^2 + c^2 = \frac{8}{9} + \frac{1}{9} = 1$a2+b2+c2=98​+91​=1, and $9(a^2 + b^2 + c^2) = 9$9(a2+b2+c2)=9.