Discussion Forum

Suppose that all the term of an arithmetic progression are natural numbers. If the ratio of the sum of 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

Answer:

Explanation:

Given , $\frac{S_{7}}{S_{11}}=\frac{6}{11}$    and 130< i₇ <140

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

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

$\Rightarrow $  a=9d        ........(i)

                   Also, 130 <i₇ <140

$\Rightarrow $     130<a+6d<140

$\Rightarrow $     130<9d+6d<140               [from Eq.(i)]

$\Rightarrow $     130<15d<140

$\Rightarrow $           $\frac{26}{3} <d<\frac{28}{3}$

                           [since , d is a natural  number]

$\therefore$    d=9