JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Sequences and Series

Let a and b be be two distinct positive real numbers. Let

1 1^{th}

term of a GP, whose first term is

a

and third term is

b

, is equal to

p^{th}

term of another GP, whose first term is a and fifth term is b . Then p is equal to

Answer: (C)

Solution

For the first GP,

t_{1} = a

t_{3} = a r^{2} = b \Rightarrow r^{2} = \frac{b}{a}

, so

t_{11} = a r^{10} = a (\frac{b}{a})^{5}

.

For the second GP,

T_{1} = a

T_{5} = a R^{4} = b \Rightarrow R^{4} = \frac{b}{a}

, so

T_{p} = a R^{p - 1} = a (\frac{b}{a})^{\frac{p - 1}{4}}

.

Setting

t_{11} = T_{p}

, and since

a \neq = b

\frac{b}{a} \neq = 1

, the exponents must match:

5 = \frac{p - 1}{4} \Rightarrow p = 21

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option