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 number is 5: 9. Find the value of m.

Answer 1

To solve this question, we have to remember some general points of arithmetic progression. If two numbers a and b are given and we have to insert n numbers between them such that $a,{A_1},{A_2},{A_3},........,{A_n},b$ is an A.P, then the common difference will be given as, $d = \dfrac{{b - a}}{{n + 1}}$. Then we will use the formula of nth term of an AP and then we will find the ratio between 7th and (m-1)th term which will be equal to 5:9. On solving further we will get the value of m.

Complete step-by-step answer:
Given Data,
Between 1 and 31, m numbers have been inserted such that the resulting sequence will be an A.P.
Ratio of 7th and (m – 1)th number is 5: 9
We have to insert the numbers between 1 and 31
i.e. $1,{A_1},{A_2},{A_3},........,{A_n},31$
So, we have a total of m + 2 terms, i.e. n = m + 2
The last term of the series, ${a_n} = 31$
We know that, ${n^{th}}$ term of an A.P. can be given as:
$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$, ……… (i)
Where a is the first term and d is the common difference.
Here our first term a = 1 and n = m + 2 and ${a_n} = 31$
putting these values in equation (i), we will get
$\Rightarrow 31 = 1 + \left( {m + 2 - 1} \right)d$
$\Rightarrow 31 - 1 = \left( {m + 1} \right)d$
$\Rightarrow \dfrac{{30}}{{\left( {m + 1} \right)}} = d$
The 1st term inserted = a + d = 1 + d.
The 2nd term inserted = a + 2d = 1 + 2d
The 7th term inserted = a + 7d = 1 + 7d
Similarly, (m-1)th term inserted = a + (m-1)d = 1 + (m-1)d
Now given that, the ratio of the 7th term and (m – 1) th term inserted is 5: 9, i.e.
$\Rightarrow \dfrac{{{A_7}}}{{{A_{m - 1}}}} = \dfrac{5}{9}$
$\Rightarrow \dfrac{{1 + 7d}}{{1 + \left( {m - 1} \right)d}} = \dfrac{5}{9}$
Now, put the value of d,
$\Rightarrow \dfrac{{1 + 7\dfrac{{30}}{{\left( {m + 1} \right)}}}}{{1 + \left( {m - 1} \right)\dfrac{{30}}{{\left( {m + 1} \right)}}}} = \dfrac{5}{9}$
$\Rightarrow \dfrac{{\dfrac{{m + 1 + 210}}{{\left( {m + 1} \right)}}}}{{\dfrac{{\left( {m + 1} \right) + \left( {30m - 30} \right)}}{{\left( {m + 1} \right)}}}} = \dfrac{5}{9}$
Solving this,
$\Rightarrow \dfrac{{m + 211}}{{31m - 29}} = \dfrac{5}{9}$
Now, we will do the cross-multiplication on both sides, we will get
$\Rightarrow 9\left( {m + 211} \right) = 5\left( {31m - 29} \right)$
Simplifying this by opening the brackets,
$\Rightarrow 9m + 1899 = 155m - 145$
Taking the coefficients of m one side and the others on other side,
$\Rightarrow 1899 + 145 = 155m - 9m$
$\Rightarrow 2044 = 146m$
$\Rightarrow \dfrac{{2044}}{{146}} = m$
$\Rightarrow 14 = m$
Hence the value of m is 14.

Note: In order to solve this type of questions the key is to know the concept of A.P and its formula to find the nth term of an A.P, as well as sum of n terms in an A.P etc. We inserted terms in between 1 and 31, which is why the number of terms became m + 2.
Also, it is very important where we realized the term ${A_7}$ is the 8th term of the progression and the term ${A_{m - 1}}$ is the mth term of the progression. We use these as the values of n while computing their values respectively. They take these values because the series starts after the number 1.