JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Sequences and Series

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The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is:

(A)36
(B)32
(C)24
(D)28✓ Correct

Answer: (D)

Solution

Write the three consecutive terms as

\frac{a}{r}, a, a r

. Their product is

a^{3} = 512

, so

a = 8

and the middle term is

8

. Adding

4

to the first two gives

\frac{8}{r} + 4, 12, 8 r

in arithmetic progression, so twice the middle equals the sum of the outer two:

24 = \frac{8}{r} + 4 + 8 r

. This becomes

8 r^{2} - 20 r + 8 = 0

, i.e.

2 r^{2} - 5 r + 2 = 0

, with

r = 2

r = \frac{1}{2}

. Either way the three original terms are

4, 8, 16

, whose sum is

28

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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