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

Question

Let $a$ and $b$ be be two distinct positive real numbers. Let ${11}^{}$ term of a GP, whose first term is $a$ and third term is $b$, is equal to ${p}^{}$ term of another GP, whose first term is $a$ and fifth term is $b$. Then $p$ is equal to

Accepted Answer

The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.

For the first GP, with first term $a$ and third term $b$:The third term is given by $t_3 = ar^2 = b$, leading to $$r^2 = \frac{b}{a}$$.The eleventh term is $$t_{11} = ar^{10} = a\left(\frac{b}{a}\right)^5$$.For the second GP, with first term $a$ and fifth term $b$:The fifth term is $T_5 = ar^4 = b$, yielding $$r^4 = \frac{b}{a}$$ and $$r = \left(\frac{b}{a}\right)^{1/4}$$.The $$p^{\text{th}}$$ term is thus $$T_p = ar^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$.Equating the eleventh term of the first GP to the $$p^{\text{th}}$$ term of the second GP gives: $$a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$.

Simplifying, we find that $$5 = \frac{p-1}{4}$$, leading to $p = 21$, which is the solution.

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