SATHEE: Sequences and Series 2 Question 20

Sequences and Series 2 Question 20

21.

Suppose that all the terms of an arithmetic progression are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6 : 11$ and the seventh term lies in between 130 and 140, then the common difference of this AP is

(2015 Adv.)

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Answer:

Correct Answer: 21. $(9)$

Solution:

Given, $\frac{S_{7}}{S_{11}} = \frac{6}{11}$ and $130 < t_{7} < 140$

$\Rightarrow \frac{\frac{7}{2} [2 a + 6 d]}{\frac{11}{2} [2 a + 10 d]} = \frac{6}{11} \Rightarrow \frac{7 (2 a + 6 d)}{(2 a + 10 d)} = 6$

$\Rightarrow a = 9 d$

Also, $130 < t_{7} < 140$

$\Rightarrow 130 < a + 6 d < 140$

$\Rightarrow 130 < 9 d + 6 d < 140$

$\Rightarrow 130 < 15 d < 140$

[from Eq. (i)]

$\Rightarrow \frac{26}{3} < d < \frac{28}{3}$ [since, $d$ is a natural number]

$∴ d = 9$

Sequences and Series 2 Question 20

21.

Answer:

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Sequences and Series 2 Question 20

21.

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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