SATHEE: Sequences and Series 3 Question 5

Sequences and Series 3 Question 5

5. Let $a, b$ and $c$ be the 7th, 11th and 13 th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\frac{a}{c}$ is equal to

(a) 2

(b) $\frac{7}{13}$

(d) $\frac{1}{2}$

(2019 Main, 9 Jan II)

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Answer:

Correct Answer: 5. (b)

Solution:

Let $A$ be the Ist term of $A P$ and $d$ be the common difference.

$∴ 7 th term = a = A + 6 d$

Then, $\Rightarrow$	$a + b + c = x b$ $a + a r + a r^{2} = x a r$
	$1 + r + r^{2} = x r$	$\dots$ (i) $[∵ a \neq 0]$
	$x = \frac{1 + r + r^{2}}{= 1 + r + \underset{―}{1}}$

$[∵ n$ th term $= A + (n - 1) d]$ 11 th term $= b = A + 10 d$

13 th term $= c = A + 12 d$

$∵ a, b, c$ are also in GP

$∴ b^{2} = a c$

$\Rightarrow (A + 10 d)^{2} = (A + 6 d) (A + 12 d)$

$\Rightarrow A^{2} + 20 A d + 100 d^{2} = A^{2} + 18 A d + 72 d^{2}$

$\Rightarrow 2 A d + 28 d^{2} = 0$

$\Rightarrow 2 d (A + 14 d) = 0$

$\Rightarrow d = 0$ or $A + 14 d = 0$

But $d \neq 0$

$[∵$ the series is non constant AP]

$\Rightarrow A = - 14 d$

$∴ a = A + 6 d = - 14 d + 6 d = - 8 d$

and $c = A + 12 d = - 14 d + 12 d = - 2 d$

$\Rightarrow \frac{a}{c} = \frac{- 8 d}{- 2 d} = 4$

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Sequences and Series 3 Question 5

5. Let $a, b$ and $c$ be the 7th, 11th and 13 th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\frac{a}{c}$ is equal to

Answer:

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Sequences and Series 3 Question 5

5. Let a,b and c be the 7th, 11th and 13 th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then ac is equal to

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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5. Let $a, b$ and $c$ be the 7th, 11th and 13 th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\frac{a}{c}$ is equal to