- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
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
$6$
B
$7$
C
$8$
D
$9$
(IIT-2015)
Solution
Let seventh term be ' $a$ ' and common difference be ' $d$ ' Given $\frac{ S _7}{ S _{11}}=\frac{6}{11} \Rightarrow a =15 d$
Hence, $130 < 15 d < 140$
$\Rightarrow d=9$
Standard 11
Mathematics
