Class 11

Math

Algebra

Sequences and Series

If $m$ arithmetic means are inserted between $1$ and $31$ so that the ratio of the $7_{th}$ and $(m−1)_{th}$ means is $5:9,$ then the value of $m$ is

- $9$
- $11$
- $13$
- $14$

1, $x_{1}$,$x_{2}$,....$x_{m}$, $31$ is an A.P. of $(m+2)$ terms with $1$ as the first term and $31$ as the last term.

$l=a+(n−1)d$

$31=1+(m+2−1)×d$

$d=m+130 $

Now $A_{m}−1A_{7} $ $=$ $95 $

$⇒T_{m}T_{8} $ $=$ $95 $

$⇒1+(m−1)d1+7d $ $=$ $95 $

$⇒1+(m−1)(m+130 )1+7(m+130 ) =95 $

$⇒m+1+30m−30m+1+210 =95 $

$⇒9m+1899=135m−145$

$⇒146m=2044$

$⇒m=14$