Question

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

Answer 1

Hint: Here, the first number is 1 and the last number is 31, we are adding m numbers in between. Therefore total numbers are $n=m+2$. Since last term is given, find the value of d by the formula:

${{a}_{n}}=a+(n-1)d$. Now write the ratio of 7th and ${{(m-1)}^{th}}$ number which is given by:
$\dfrac{a+7d}{a+(m-1)d}=\dfrac{5}{9}$. From this find the value of m.


Complete step-by-step answer:

We are given that $m$ numbers are inserted between 1 and 31 with the resulting sequence in AP.
It is also given that the ratio of 7th and ${{(m-1)}^{th}}$ numbers is 5 : 9.
Now, we have to find the value of $m$.
Since, $m$ numbers are inserted between 1 and 31 we can say that:

Total number of terms in AP = $n=m+2$ (including 1,31 and m numbers in between)
Here the first term is given as:
$a=1$

The last term is given as:
${{a}_{n}}=31$

We know that the formula for the last term is:
${{a}_{n}}=a+(n-1)d$

Now, by substituting the values for $a,{{a}_{n}}$ and $n$, we get:
$\begin{align}
  & 31=1+(m+2-1)d \\
 & 31=1+(m+1)d \\
\end{align}$

Next, by taking 1 to the left side, 1 becomes -1, we obtain:
$\begin{align}
  & 31-1=(m+1)d \\
 & 30=(m+1)d \\
\end{align}$

Now, by cross multiplication we get the value of d, which is given by:
$d=\dfrac{30}{m+1}$ ….. (1)
The ratios of 7th term and ${{(m-1)}^{th}}$ number is given as 5 : 9

We know that the 7th number is the 8th term and is given as:
 $\begin{align}
  & {{a}_{8}}=a+(8-1)d \\
 & {{a}_{8}}=1+7d \\
\end{align}$.

Similarly, ${{(m-1)}^{th}}$ number is the ${{m}^{th}}$ term and is given as:
$\begin{align}
  & {{a}_{m}}=a+(m-1)d \\
 & {{a}_{m}}=1+(m-1)d \\
\end{align}$
Hence, we can say that:
$\dfrac{1+7d}{1+(m-1)d}=\dfrac{5}{9}$
By cross multiplication we get:
$\begin{align}
  & 9(1+7d)=5(1+(m-1)d) \\
 & 9+63d=5+5(m-1)d \\
\end{align}$

Now, take constants to one side and variables to the other side, 5 to the left side becomes -5 and 63d goes to the right side and becomes -63. Hence, we get:
$\begin{align}
  & 9-5=5(m-1)d-63d \\
 & 4=5d(m-1)-63d \\
\end{align}$

On the right side d is the common factor, so take outside d, we get:
$\begin{align}
  & 4=d(5(m-1)-63) \\
 & 4=d(5m-5-63) \\
 & 4=d(5m-68) \\
\end{align}$

Now, substitute the value of d in equation (1) we obtain:
$4=\dfrac{30}{m+1}(5m-68)$
 Next, by cross multiplication we get:
$\begin{align}
  & 4(m+1)=30(5m-68) \\
 & 4\times m+4\times 1=30\times 5m+30\times -68 \\
 & 4m+4=150m-2040 \\
\end{align}$

Now, take constants to one side and variables to the other, -2040 goes to left side and becomes 2040, $4m$ goes to right side and becomes $-4m$.
$\begin{align}
  & 4+2040=150m-4m \\
 & 2044=146m \\
\end{align}$
 Next, by cross multiplication we obtain:
$\dfrac{2044}{146}=m$
Hence, by cancellation we get:
$m=14$
Therefore, the value of $m=14$.

Note: Here, we are adding m numbers in between 1 and 31. So, while taking the ratio the 7th number and ${{(m-1)}^{th}}$numbers are taken among m numbers. Hence in AP they are 8th term and ${{m}^{th}}$ term. So don’t get confused that 7th and ${{(m-1)}^{th}}$ numbers are the 7th term and ${{(m-1)}^{th}}$ term, it may lead to wrong answers.