Question

The sum of ‘n’ terms of two arithmetic progressions is in ratio $5n+4:9n+6$. find the ratio of their ${18}^{\text{th}}$ terms:

Answer 1

Hint: In this problem, we will be using the concept of the sum of an arithmetic progression (A.P). In order to solve this question, we divide the sum of n terms of the two arithmetic progression(A.P) with each other and equate it with $5n+4:9n+6$.
Now we will try to get the equation in the form of the ${\text{n}}^{\text{th}}$ term of an A.P and will find the values for ‘n’. Upon putting that value of ‘n’ in $5n+4:9n+6$. We can calculate the ratio of the ${18}^{\text{th}}$ term of two given A.P.
Complete step by step solution: As mentioned in the question, there are two arithmetic progressions with different first terms and different common differences.
For the first A⋅P:-
Let the first term of A⋅P be = a, and the common difference be = d;
So, Sum of ‘n’ terms of an A⋅P is;
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
and the nth term of an A⋅P is;
${{a}_{n}}=a+\left( n-1 \right)d$
For the second A⋅P:-
Let the first term of A⋅P. be = A and the common difference be = D;
So, the sum of its ‘n’ terms will be;
${{S}_{n}}=\dfrac{n}{2}\left[ 2A+\left( n-1 \right)D \right]$
And the nth term of an A⋅P is;
${{A}_{n}}=A+\left( n-1 \right)D$
It’s given in the question that the ratio of the sum of ‘n’ terms of the two AP is $5n+4:\ 9n+6;$
$\Rightarrow \dfrac{\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]}{\dfrac{n}{2}\left[ 2\text{A}+\left( n-1 \right)\text{D} \right]}=\dfrac{5n+4}{9n+6}$
$\Rightarrow \dfrac{\left[ 2a+\left( n-1 \right)d \right]}{\left[ 2A+\left( n-1 \right)d \right]}=\dfrac{5n+4}{9n+6}$
Taking L.H.S;
When we take ‘2’ common in numerator and denominator;
\[=\dfrac{2\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{2\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}\]
$=\dfrac{\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}$
So, now;
\[\dfrac{\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}=\dfrac{5n+4}{9n+6}\] (1)
We need to find the ratio of the 18th term of Arithmetic progression:
\[=\dfrac{18\text{th}\ \text{term}\ \text{of}\ \text{1st}\ \text{A}\cdot \text{P}}{18\text{th}\ \text{term}\ \text{of}\ \text{2nd}\ \text{A}\cdot \text{P}}\]
$\dfrac{{{a}_{18}}\ \text{of}\ 1\text{st}\ \text{A}\cdot \text{P}}{{{A}_{18}}\ \text{of}\ 2\text{nd}\ \text{A}\cdot \text{P}}$
$=\dfrac{a+\left( 18-1 \right)d}{A+\left( 18-1 \right)D}$
$=\dfrac{a+17d}{A+17D}$ (2)
Comparing equation (2) with equation (1); $a+17d=a+\left( \dfrac{n-1}{2} \right)d$
$\Rightarrow 17=\dfrac{n-1}{2}$
$\Rightarrow n-1=17\times 2$
$\Rightarrow n-1=34$
$\Rightarrow n=34+1$
$\Rightarrow n=35$
Now, putting $n=35$ in equation (1);
\[\Rightarrow \dfrac{\left[ a+\left( \dfrac{n-1}{2} \right)d \right]}{\left[ A+\left( \dfrac{n-1}{2} \right)D \right]}=\dfrac{5n+4}{9n+6}\]
$\Rightarrow \dfrac{a+\left( \dfrac{35-1}{2} \right)d}{A+\left( \dfrac{35-1}{2} \right)D}=\dfrac{5\left( 35 \right)+4}{9\left( 35 \right)+6}$
$\Rightarrow \dfrac{a+\left( \dfrac{34}{2} \right)d}{A+\left( \dfrac{34}{2} \right)D}=\dfrac{175+4}{315+6}$
$\Rightarrow \dfrac{a+17d}{A+17D}=\dfrac{179}{321}$
Therefore, $\dfrac{18\text{th}\ \text{term}\ \text{of}\ 1\text{st}\ \text{A}\cdot \text{P}}{18\text{th}\ \text{term}\ \text{of}\ 2\text{nd}\ \text{A}\cdot \text{P}}=\dfrac{179}{321}$
Hence, the ratio of ${18}^{\text{th}}$ term of ${1}^{\text{st}}$ A⋅P and ${18}^{\text{th}}$ term of ${2}^{\text{nd}}$ A⋅P is 179: 321.

Note: The sum of the ‘n’ terms of any A⋅P is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ and ${n}^{\text{th}}$ term of an A⋅P is ${{a}_{n}}=a+\left( n-1 \right)d$ where ‘a’ is the first term of an A.P and ‘d’ is common difference of an A.P.