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16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of $7 ^{th}$ and $(m-1)^{th}$ numbers is 5 : 9. Find the value of m.

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Let A,B,C.........M be m numbers.

Then, $AP=1,A,B,C..........M,31$

Here we have,

$a=1,a_m_+_2=31,n=m+2$

$\Rightarrow \, \, a+(n-1)d=a_n$

$\Rightarrow \, \, 1+(m+2-1)d=31$

$\Rightarrow \, \, (m+1)d=30$

$\Rightarrow \, \, d=\frac{30}{m+1}$

Given : the ratio of $7 ^{th}$ and $(m-1)^{th}$ numbers is 5 : 9.

$\Rightarrow \, \, \frac{a+(7)d}{a+(m-1)d}=\frac{5}{9}$

$\Rightarrow \, \, \frac{1+7d}{1+(m-1)d}=\frac{5}{9}$

$\Rightarrow \, \, 9(1+7d)=5(1+(m-1)d)$

$\Rightarrow \, \, 9+63d=5+5md-5d$

Putting value of d from above,

$\Rightarrow \, \, 9+63(\frac{30}{m+1})=5+5m\left ( \frac{30}{m+1} \right )-5\left ( \frac{30}{m+1} \right )$

$\Rightarrow \, \9(m+1)+1890=5(m+1)+150m-150$

$\Rightarrow \, \9m+9+1890=5m+5+150m-150$

$\Rightarrow \, 1890+9-5+150=155m-9m$

$\Rightarrow \, 2044=146m$

$\Rightarrow \, m=14$

Thus, value of m is 14.

Posted by

seema garhwal

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