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# 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 ${{\left( m-1 \right)}^{th}}$ number is $5:9$. Find the value of m.

Last updated date: 15th Jul 2024
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Hint: We first form the full A.P. series including 1 and 31. We use two unknowns of common difference and the value of m. From the given ratio and condition of 31, we get two equations of the variables. We solve them to find the value of m. We will be using the formula ${{t}_{n}}=a+\left( n-1 \right)d$.

The made-up A.P. sequence has two end terms as 1 and 31. There are m numbers in between those two numbers. This means the sequence consists of in total $\left( m+2 \right)$ terms, all of them being in A.P.

We define the terms of the A.P. as ${{t}_{n}}$, ${{n}^{th}}$ term of the A.P.

Now we know that if a be the first term and d be the common difference of an A.P. then ${{t}_{n}}$, ${{n}^{th}}$ term of the A.P. can be defined as ${{t}_{n}}=a+\left( n-1 \right)d$.

Here, a = first term of the A.P

n = no of terms

d = common difference

The first term is 1 which implies $a=1$

The last, ${{\left( m+2 \right)}^{th}}$ term is 31 which implies ${{t}_{m+2}}=31={{t}_{1}}+\left( m+2-1 \right)d$.

We got a linear equation of d which is $31=1+\left( m+1 \right)d$. We solve it to find the value of the common difference of the A.P.

$31=1+\left( m+1 \right)d$

$\Rightarrow \left( m+1 \right)d=30$

$\Rightarrow d=\dfrac{30}{m+1}...(i)$

We also have been given that ratio of ${{7}^{th}}$ and ${{\left( m-1 \right)}^{th}}$ number is $5:9$. These numberings are of the m terms that have been inserted between 1 and 31.

So, for the full series including 1 and 31, the ratio of ${{8}^{th}}$ and ${{m}^{th}}$ numbers is $5:9$.

So, the eighth term is ${{t}_{8}}={{t}_{1}}+\left( 8-1 \right)d=1+7d$

The ${{m}^{th}}$ term ${{t}_{m}}={{t}_{1}}+\left( m-1 \right)d=1+\left( m-1 \right)d$.

This means $\dfrac{1+7d}{1+\left( m-1 \right)d}=\dfrac{5}{9}$. We solve it to place the value of d from equation (i).

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

$\Rightarrow 63d-5\left( m-1 \right)d=5-9=-4$

$\Rightarrow d=\dfrac{4}{5m-5-63}=\dfrac{4}{5m-68}.....(ii)$

We got two values of d and we equate them from equation (i) and equation (ii).

$d=\dfrac{30}{m+1}=\dfrac{4}{5m-68}$

$\Rightarrow 30\left( 5m-68 \right)=4\left( m+1 \right)$

$\Rightarrow 150m-4m=4+2040$

$\Rightarrow 146m=2044$

$\Rightarrow m=\dfrac{2044}{146}=14$

Therefore, the value of $m$ is 14.

Note: We need to consider if the ratio is of the actual terms of the full series or in this problem the terms of the series excluding 1 and 31. As it’s in the same sentence of the previous line the terms are referring to the “m” terms only. The meaning and the solution both change if we consider the other case where the ratio is of the whole series.