Question

If the ratio between the sums of n terms of two A.P.’s is $2n + 5:5n + 8$, then the ratio between their 16th term is:
A. $\dfrac{{67}}{{163}}$
B. $\dfrac{{37}}{{88}}$
C. $\dfrac{{57}}{{147}}$
D. $\dfrac{{68}}{{167}}$

Answer 1

Hint: In the above question the ratio between the sum of n terms of two A.P.’ is given, we have to find the ratio between 16th terms. So, we will form an expression of the sum of nth terms for both APs and equate it with the given ratios in the question.

Complete step-by-step answer:
Given,
Ratio of the sum of n terms of two A.P.’s is $2n + 5:5n + 8$.
Let the first $A.{P_1}$have the first term be ${a_1}$and the difference between the two consecutive terms be ${d_1}$.
Let the second $A.{P_2}$have the first term be ${a_2}$and the difference between the two consecutive terms be ${d_2}$.
Now, $\dfrac{{S_{n1}}}{{S_{n2}}} = \dfrac{{2n + 5}}{{5n + 8}}$
[Sum of n terms of an AP is given by $\dfrac{n}{2}\left[ {2a + (n - 1)} \right]d$ ]
$ \Rightarrow \dfrac{{\dfrac{n}{2}\left[ {2{a_1} + (n - 1)} \right]{d_1}}}{{\dfrac{n}{2}\left[ {2{a_2} + (n - 1)} \right]{d_2}}} = \dfrac{{2n + 5}}{{5n + 8}}$
$\therefore \dfrac{{{a_1} + 15{d_1}}}{{{a_2} + 15{d_2}}} = ?$
$ \Rightarrow \dfrac{{{a_1} + \left( {\dfrac{{n - 1}}{2}} \right){d_1}}}{{{a_2} + \left( {\dfrac{{n - 1}}{2}} \right){d_2}}} = \dfrac{{2n + 5}}{{5n + 8}}$ … (i)
To find the ratio of 16th term$\dfrac{{{a_1} + 15{d_1}}}{{{a_2} + 15{d_2}}}$$ = ?$
Here, $\dfrac{{n - 1}}{2} = 15$
    ⇒ n = 31
Putting the value of n = 31 in equation (i), we get;
$\dfrac{{{a_1} + \left( {\dfrac{{n - 1}}{2}} \right){d_1}}}{{{a_2} + \left( {\dfrac{{n - 1}}{2}} \right){d_2}}} = \dfrac{{2n + 5}}{{5n + 8}}$
$ \Rightarrow \dfrac{{{a_1} + 15{d_1}}}{{{a_2} + 15{d_2}}} = \dfrac{{\left( {2 \times 31} \right) + 5}}{{\left( {5 \times 31} \right) + 8}} = \dfrac{{67}}{{163}}$
Thus, the ratio of 16th term is $\dfrac{{67}}{{163}}$.

So, the correct answer is “Option A”.

Note: Arithmetic progression is a sequence of numbers such that the difference between the two consecutive terms is always constant. It means A.P has a common difference between two successive terms. In arithmetic progression we can obtain the second number by adding a fixed number to the first number. The fixed number which is added to form a sequence is considered a common difference. Formula to sum of n terms is $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)} \right]d$ where a is the first term of the sequence, d is the common difference, n is number of terms.
In these types of questions, always find the value of n which satisfies both sides of the equation.