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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Sequences and Series

In a G.P., if the product of the first three terms is 27 and the...

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}-(a, b), then a2+b2a^{2}+b^{2} is equal to ____\_\_\_\_.

Correct answer

0

Step-by-step solution

In the G.P., let the first three terms be ar,a,ar\frac{a}{r}, a, ar.
From the product condition: ar⋅a⋅ar=a3=27\frac{a}{r} \cdot a \cdot ar = a^3 = 27, so a=3a = 3.
The sum of first three terms is S=3r+3+3r=3(1r+1+r)S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + 1 + r).
For r>0r > 0, by AM-GM: 1r+r≥2\frac{1}{r} + r \geq 2, so S≥9S \geq 9.
For r<0r < 0: 1r+r≤−2\frac{1}{r} + r \leq -2.
So S≤−3S \leq -3.
Thus the possible values of SS are (−∞,−3]∪[9,∞)=R−(−3,9)(-\infty, -3] \cup [9, \infty) = \mathbb{R} - (-3, 9).
Therefore (a,b)=(−3,9)(a, b) = (-3, 9) and a2+b2=9+81=90a^2 + b^2 = 9 + 81 = 90.
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