QUESTION

# Between 1 and 31 are inserted m arithmetic means so that the ratio of the 7th and (m-1)th is 5:9. Find the value of m.

Hint: Take first term of an A.P series as a=1 and last term as b=31, find the common difference by using the formula $d = \dfrac{{b - a}}{{n + 1}}$. Later find the 7th 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’.

As we know that to insert n numbers between a and b 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

And the number 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 7th and (m - 1)th numbers inserted.

So,7th 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

$\dfrac{{7th{\text{ number}}}}{{\left( {m - 1} \right){\text{th number}}}} = \dfrac{5}{9}$

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

So when we do cross multiplication we will get:

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

By further solving we get,

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

$\Rightarrow 9 + 63d = 5 + 5dm - 5d$

$\Rightarrow 4 + 68d = 5dm$

Now when we put $d{\text{ = }}\dfrac{{30}}{{m + 1}}$ 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,

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

$\Rightarrow 4\left( {m + 1} \right) + 2040 = 150m{\text{ , }}m \ne - 1$

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

$\Rightarrow 2044 = 150m - 4m$

$\Rightarrow 2044 = 146m$

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

$\Rightarrow m = 14$

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 7th and the (m - 1)th terms of inserted elements, we were actually finding the 8th and the mth elements of the A.P.