Question

The sum of $n$ terms of two arithmetic progressions are in ratio $(3n+8):(7n+15).$ Find the ratio of their $\left(i\right)12th$ terms $\left(ii\right)15th$ terms.

• A. $\left(i\right)7:16$ $\left(ii\right)95:218$
• B. $\left(i\right)5:19$ $\left(ii\right)105:213$
• C. $\left(i\right)3:20$ $\left(ii\right)83:219$
• D. None of these
  

Answer 1

The correct option is A $\left(i\right)7:16$ $\left(ii\right)95:218$

We know that the sum of n terms of an arithmetic progression (AP) is given by: $\frac{n}{2}\times (2a+(n-1\left)d\right)$

where $a=\text{first term of the AP}$

$d=\text{common difference of the AP}$

Let,

$a_1,d_1=\text{first term and common difference of the first AP}$

$b_1,d_2=\text{first term and common difference of the second AP}$

Hence,

$\frac{(2a_1+(n-1\left)d_1\right)}{(2b_1+(n-1\left)d_2\right)}=\frac{(3n+8)}{(7n+15)}$ ---$\left(1\right)$

We need to find out ratios of ${12}^{th}$ terms of both APs

which is $\frac{a_1+11d_1}{b_1+11d_2}$ --(2).

Multiply and divide equation (2) with 2

which is $\frac{2a_1+22d_1}{2b_1+22d_2}$ --(3)

On comparing coefficients of common differences we get $n-1=22$

$\Rightarrow n=23$.

I.e., if we put $n=23$ in equ $\left(1\right)$ to get ratios of ${12}^{th}$ terms.

i) Ratios of ${12}^{th}$ terms $=\frac{\left(3\right(23)+8)}{\left(7\right(23)+15)}$

$=\frac{77}{176}$

$=\frac{7}{11}$

Ratios of ${15}^{th}$ terms =$\frac{a_1+14d_1}{b_1+14d_2}$ --(3).

Divide and multiply 2 with equation 3, which is $\frac{2a_1+28d_1}{2b_1+28d_2}$ --(4)

On comparing coefficients of common differences we get $n-1=28$

On substituting $n=29$ in equation (1). We get

iI) Ratios of ${15}^{th}$ terms $=\frac{\left(3\right(29)+8)}{\left(7\right(29)+15)}$

$=\frac{95}{218}$

Thus option A is correct.