Question

The ratio between the sum of $n$ terms of two arithmetic progression is $\left( {7n + 1} \right):\,\left( {4n + 27} \right).$ find the ration of their 11th terms.

Answer 1

Hint:
Use the formula of sum of ${n^{th}}$ term of A.P i.e. ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where $a$ is the first term $d$ is the common difference. As the ratio is given so, we will let the first term and common difference uniquely and put it equal to the ratio obtained using the formula of the sum of nth terms of an A.P. As its given in the question to find the ratio of $11^{\text{th}}$ term and compare both the sides and then we will find the value of \[n\] and thus substitute it in the ratio.

Complete step by step solution:
Given: The ratio between the sum of $n$ terms of two arithmetic progression is $\left( {7n + 1} \right):\,\left( {4n + 27} \right).$
Let us take $a\& d$ as the first term and the common difference of first A.P. and $A\& D$ as the first term and the common difference of second A.P.
According to the question,
$
  \dfrac{{\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\dfrac{n}{2}\left[ {2A + \left( {n - 1} \right)D} \right]}} = \dfrac{{7n + 1}}{{4n + 27}} \\
  \dfrac{{2a + \left( {n - 1} \right)d}}{{2A\left( {n - 1} \right)D}} = \dfrac{{7n + 1}}{{4n + 27}}{\text{ }} \to {\text{equation}}\left( 1 \right) \\
 $
We have to find the ratio of $11^{\text{th}}$ terms
$
\dfrac{T_{11}}{{T_{11}}^{’}}= \dfrac{a+\left(11-1 \right)d}{A+\left(11-1 \right)D} \Rightarrow \dfrac{a+10d}{A+10D} \;\;\;\;\;\to \text{equation} (2) \\
\text{Divide equation (1) by (2)}
\dfrac{a+\left(\dfrac{n-1}{2} \right)d}{A+\left(\dfrac{n-1}{2} \right)D} = \dfrac{7n+1}{4n+27} \;\;\;\;\;\to \text{equation} (3) \\
$
On comparing equation $\left( 2 \right)$ by equation $\left( 3 \right)$we get
$\dfrac{{n - 1}}{2} = 10 \Rightarrow n = 21$

Note:
Make the equation in that way so that the comparison is possible to find the variable. Use of formula of a finite sum of an A.P. to form the ratio. Do the comparison of both the sides to obtain the value of \[n\] and then we have substituted it in the given ratio to obtain the 11th term. We have chosen the value of the first term and common difference uniquely to find the ratio.