SATHEE: Sequences and Series 3 Question 15

Sequences and Series 3 Question 15

15.

Does there exist a geometric progression containing 27,8 and 12 as three of its term? If it exists, then how many such progressions are possible?

(1982, 2M)

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Answer:

Correct Answer: 15. (Yes, infinite)

Solution:

Let $27, 8, 12$ be three terms of a GP.

$\begin{aligned} \Rightarrow t_{m} = 27, t_{n} = 8 and t_{p} = 12 \\ A R^{m - 1} = 27, A R^{n - 1} = 8 \\ and A R^{p - 1} = 12 \\ ∴ R = {\frac{27}{8}}^{1 / (m - n)} and R = {\frac{8}{12}}^{1 / (n - p)} \\ \Rightarrow {\frac{27}{8}}^{1 / (m - n)} = {\frac{2}{3}}^{1 / (n - p)} \\ \Rightarrow 3^{3 / (m - n)} \cdot 3^{1 / (n - p)} = 2^{1 / (n - p)} \cdot 2^{3 / (m - n)} \\ \Rightarrow \frac{3^{\frac{3}{m - n} + \frac{1}{n - p}}}{2^{\frac{1}{n - p} + \frac{3}{m - n}}} = 1 \\ ∴ \frac{3}{m - n} + \frac{1}{n - p} = 0 and \frac{1}{n - p} + \frac{3}{m - n} = 0 \\ \Rightarrow 3 (n - p) = n - m and 2 n = 3 p - m \end{aligned}$

Hence, there exists infinite GP for which 27,8 and 12 as three of its terms.

Sequences and Series 3 Question 15

15.

Answer:

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Sequences and Series 3 Question 15

15.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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