Question

Suppose that all the terms of an arithmetic progression (A.P.) 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 A.P. is

• A. 7
• B. 8
• C. 9
• D. 11

Answer 1

Answer: (C)

$ \frac{ S _7}{ S _{11}}=\frac{6}{11} $
$ \frac{\frac{7}{2}[2 a+6 d]}{\frac{11}{2}[2 a+10 d]}=\frac{6}{11} $
Given $130 < a + 6d < 140$
$\frac{7(a+3 d)}{11(a+5 d)}=\frac{6}{11} $
$7 a+21 d=6 a+30 d $
$\Rightarrow 130 < 15d < 140$
Hence $d = 9$
$a = 81$
$ a=9 d$
Hence $d=9$
Alternative :
Let the AP be $a, a+d, a+2 d, \ldots \ldots$(2)
where $a, d \in N$
Given $\frac{S_7}{S_{11}}=\frac{6}{11}$ and $130< a+6 d< 140$
$\Rightarrow \frac{\frac{7}{2}\{2 a+6 d\}}{\frac{11}{2}\{2 a+10 d\}}=\frac{6}{11} $
$\Rightarrow \frac{14 a+42 d}{22 a+110 d}=\frac{6}{11} $
$\Rightarrow 154 a+462 d=132 a+660 d $
$\Rightarrow 22 a=198 d $
$\Rightarrow a=\frac{99 d}{11}=9 d $
$(2) \Rightarrow \therefore 130<9 d+6 d<140$
$\Rightarrow 8.6< d< 9.3$
$\therefore d=9$