Question

If m arithmetic means are inserted between $1$ and $31$ so that the ratio of the $7th$ and $(m - 1) $th means is $5:9$, then the value of m is
A. $9$
B. $11$
C. $13$
D. $14$

Answer 1

Hint:- Take first term of an A.P. series as = $1$
And last term as b=$31$,
Find the common difference
By using formula,
 $d = \dfrac{{b - a}} {{n + 1}}$ .
Later find the $7$Th and (m-$1$) th term and take the ratio and equate it to the given value to form a linear equation. In the linear equation substitute the value of common difference to find the value of ‘m’.


Complete step by step by solution
As we know that to insert n numbers between a and b of the common difference
$d = \dfrac{{b - a}} {{n + 1}}$
Here we have to insert m numbers between $1$ and $31$
So, b=$31$, a= $1$
If m arithmetic numbers are inserted before $1$ and $31$
So that,
$\begin {gathered}
  1, {a_1},{a_2},{a_3}.................{a_n},31 \\
  Then, \\
  {a_1}, {a_2}, {a_3}................. {a_{n - 1}},{a_n} \\
  a,a + d,a + 2d.................a + nd,a + (n - 1)d \\
\end{gathered} $
Formula –
${a_n} = a + (n - 1)d$
In last term –
$\begin {gathered}
  n = n + 2 \\
  {a_n} = a + (n + 2 - 1) d \\
  {a_n} = a + (n - 1) d \\
\end{gathered} $
Given ration
$\dfrac{{A - 1}}{{{A_n} - 1}} = \dfrac{5}{a}$
Here, ${a_1} = 1, {a_2} = a + d$
So, ${A_7} = a + 7d$
${A_{n - 1}} = a + (n - 1)d$
And the numbers of terms to be inserted =n=m.
Therefore, $d = \dfrac{{31 - 1}}{{m + 1}}$
$d = \dfrac{{30}}{{m + 1}} $
Now,
$a = 1$,
$d = \dfrac{{30}}{{m + 1}} $
$b = 31$
We need to find $7$th and $(m - 1)$th numbers inserted
So, $7$Th term is $a + 7d = 1 + 7d$
And $(m - 1)$th term is
$a + (m - 1) d = 1 + (m - 1)d$
Now it is given that ratio of
$\begin {gathered}
  \dfrac{{7thnumber}}{{(m - 1)thnumber}} = \dfrac{5}{9} \\
  \dfrac{{1 + 7d}}{{1 + (m - 1)d}} = \dfrac{5}{9} \\
\end{gathered} $
So when we do cross multiplication we will get:
$(1 + 7d)9 = 5[1 + (m - 1)d]$
By further solving we get,
$ \Rightarrow 9 + 63d = 5 + 5d (m - 1) $
$ \Rightarrow 4 + 68d = 5dm$
Now when we put $d = \dfrac{{30}}{{m + 1}}$ in the above
Equation we will get:
$ \Rightarrow 4 + 68\left ( {\dfrac{{30}}{{m + 1}}} \right) = 5\left ( {\dfrac{{30}}{{m + 1}}} \right)m$
Now making the fractions common
\[ \Rightarrow \dfrac{{4(m + 1) + 68 \times 30}}{{m + 1}} = \dfrac{{5 \times 30 \times m}}{{m + 1}}\]
$ \Rightarrow 4(m + 1) + 2040 = 150m$
$m \ne - 1$
$ \Rightarrow 4m + 4 + 2040 = 150m$
$ \Rightarrow 2044 = 150m - 4m$
$ \Rightarrow 2044 = 146m$
$ \Rightarrow m = \dfrac{{2044}}{{146}} = 14$
$ \Rightarrow m = 14$
Hence, the value of m is $14$


Note –We must remember the equation to find common differences when inserting n elements in A.P.
It also should be noted that when we tried to find the $7$th and the $(m - 1) $ terms of
Inserted elements, we were actually finding the, $8$th and not $m + n$ elements of the A.P.