Mathematics · Sequence and Series

JEE Main 2026 — 28 January, Morning Shift — Question 20

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is R−(a,b)\mathbb{R}-(\mathrm{a}, \mathrm{b}), then a2+b2\mathrm{a}^{2}+\mathrm{b}^{2} is equal to ____\_\_\_\_ .

Answer: 90

Numerical answer — enter this value.

Step-by-step solution

Let first three terms of G.P. are Ar,A,Ar\frac{\mathrm{A}}{\mathrm{r}}, \mathrm{A}, \mathrm{Ar}

Ar⋅A⋅Ar=27\frac{\mathrm{A}}{\mathrm{r}} \cdot \mathrm{A} \cdot \mathrm{Ar}=27

A=3\mathrm{A}=3 3(1r+1+r)=3+3(r+1r)3\left(\frac{1}{\mathrm{r}}+1+\mathrm{r}\right)=3+3\left(\mathrm{r}+\frac{1}{\mathrm{r}}\right)

We know, r+1r≥2\mathrm{r}+\frac{1}{\mathrm{r}} \geq 2 or

r+1r≤−2\mathrm{r}+\frac{1}{\mathrm{r}} \leq-2

S∈R−(−3,9)\mathrm{S} \in \mathrm{R}-(-3,9)

a2+b2=9+81=90\mathrm{a}^{2}+\mathrm{b}^{2}=9+81=90

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Mathematics
Chapter
Sequence and Series
Topic
Geometric Progression
In a G.P., if the product of the first three terms is 27 and the set… | JEE Main 2026 PYQ with Solution · DhiX AI