Sequences And Series Ques 51
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: 51.$(9)$
Solution:
Formula:
An Arithmetic Progression (A.P.):
- Given, $\frac{S _7}{S _{11}}=\frac{6}{11}$ and $130<t _7<140$
$\Rightarrow \quad \frac{\frac{7}{2}[2 a+6 d]}{\frac{11}{2}[2 a+10 d]}=\frac{6}{11} \Rightarrow \frac{7(2 a+6 d)}{(2 a+10 d)}=6$
$\Rightarrow \quad a=9 d$ $\quad$ …….(i)
Also, $\quad 130<t _7<140$
$\Rightarrow \quad 130<a+6 d<140$
$\Rightarrow \quad 130<9 d+6 d<140$
$\Rightarrow \quad 130<15 d<140 \quad$ [from Eq. (i)]
$\Rightarrow \quad \frac{26}{3}<d<\frac{28}{3} \quad$ [since, $d$ is a natural number]
$\therefore \quad d=9$
BETA
