Let a and b be distinct positive real numbers. The 11th term of a GP (first term a, third term b) equals the pth term of

Question

Let a and b be distinct positive real numbers. The 11th term of a GP (first term a, third term b) equals the pth term of another GP (first term a, fifth term b). Find p:

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Given: Two geometric progressions are defined. In the first GP, the first term is $a$ and the third term is $b$ . In the second GP, the first term is $a$ and the fifth term is $b$ . The $11$ th term of the first GP equals the $p$ th term of the second GP.

Find: The value of $p$ .

Let the common ratio of the first GP be $r_1$ . Since its third term is $b$ ,

t_3 = a r_1^2 = b

So,

r_1 = \sqrt{\frac{b}{a}}

The $11$ th term of the first GP is

t_{11} = a r_1^{10} = a \left( \sqrt{\frac{b}{a}} \right)^{10} = \frac{b^5}{a^4}

Now let the common ratio of the second GP be $r_2$ . Since its fifth term is $b$ ,

T_5 = a r_2^4 = b

Hence,

r_2 = \left( \frac{b}{a} \right)^{\frac{1}{4}}

Therefore, the $p$ th term of the second GP is

T_p = a r_2^{p-1} = a \left( \frac{b}{a} \right)^{\frac{p-1}{4}}

Given that $t_{11} = T_p$ ,

\frac{b^5}{a^4} = a \left( \frac{b}{a} \right)^{\frac{p-1}{4}}

Dividing both sides by $a$ ,

\frac{b^5}{a^5} = \left( \frac{b}{a} \right)^{\frac{p-1}{4}}

So,

\left( \frac{b}{a} \right)^5 = \left( \frac{b}{a} \right)^{\frac{p-1}{4}}

Equating exponents,

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

Thus,

p - 1 = 20

p = 21

Therefore, the correct option is C.

Given: The first GP has first term a and third term b. The second GP has first term a and fifth term b. Find: The value of p for which the 11th term of the first GP equals the pth term of the second GP. From the first GP, ar1^2 = b r1^2 = ba Hence, r1^10 = (r1^2)^5 = ( ba)^5 So the 11th term is a ( ba)^5 From the second GP, ar2^4 = b r2^4 = ba Thus, r2^p-1 = (r2^4)^ p-14 = ( ba)^ p-14 So the pth term is a ( ba)^ p-14 Equating the two terms directly gives ( ba)^5 = ( ba)^ p-14 Therefore, 5 = p-14 which gives p = 21 Therefore, the correct option is C.

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