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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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