Question

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 7th and (m-1)th numbers is 5:9. Find the value of m.

Answer 1

We know that to insert m numbers b/w a & b

common difference (d) = $\frac{b-1}{m+1}$

Here,

We have to insert m numbers b/w 1 & 31

So, b=31 & a=1

And, number of terms to be inserted = n= m

Therefore,

$d=\frac{31-1}{m+1}=\frac{30}{m+1}$

Now, $a=1,d=\frac{30}{m+1},b=31$

We need to find 7th and (m-1)th numbers inserted.

Given that

$\frac{a_7}{a_{m-1}}=\frac{5}{9}$

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

$(1+7d)9=5[1+(m-1\left)d\right]$

$9+63d=5+5d(m-1)$

$4+68d=5dm$

Putting $d=\frac{30}{m+1}$

$4+68\left(\frac{30}{m+1}\right)=5\left(\frac{30}{m+1}\right)m$

$\frac{4(m+1)+68\times 30}{m+1}=\frac{5\times 30\times m}{m+1}$

$4(m+1)+2040=150m$

$m=14$