JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Sequences and Series

Let the first term

a

and the common ratio r of a geometric progression be positive integers. If the sum of its squares of first three terms is 33033, then the sum of these three terms is equal to

(A)231✓ Correct
(B)210
(C)220
(D)241

Answer: (A)

Solution

Let the three terms be

a, a r, a r^{2}

. Then

a^{2} (1 + r^{2} + r^{4}) = 33033 = 3 \cdot 7 \cdot 1 1^{2} \cdot 13

. For integer solutions

a^{2}

must be a perfect-square divisor of

33033

, and the only such divisor greater than

1

1 1^{2}

; hence

a = 11

and

1 + r^{2} + r^{4} = 273

. This gives

r^{4} + r^{2} - 272 = 0

, i.e.

(r^{2} - 16) (r^{2} + 17) = 0

, so

r^{2} = 16

and

r = 4

. The required sum is

a + a r + a r^{2} = 11 + 44 + 176 = 231

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option

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