JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Sequences and Series

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Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference

d

. If the fourth term of GP is

3 r^{2}

, then

r^{2} - d

is equal to:

(A) $7 + 3$ ✓ Correct
(B) $7 - 3$
(C) $7 + 33$
(D) $7 - 73$

Answer: (A)

Solution

Let the increasing GP be

a, a r, a r^{2}

. Doubling the middle term makes

a, 2 a r, a r^{2}

an AP, so

2 (2 a r) = a + a r^{2}

, i.e.

r^{2} - 4 r + 1 = 0

, giving

r = 2 + 3

(the value greater than

1

). The fourth term

a r^{3} = 3 r^{2}

gives

a r = 3

, so

2 a r = 6

and

a = \frac{3}{r} = 6 - 33

. The common difference is

d = 2 a r - a = 6 - (6 - 33) = 33

. With

r^{2} = 7 + 43

r^{2} - d = 7 + 43 - 33 = 7 + 3

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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