Question

Let ${s_1},{s_2},{s_3} \cdot \cdot \cdot \cdot$ and ${t_1},{t_2},{t_3} \cdot \cdot \cdot \cdot$ are two arithmetic sequence such that ${s_1} = {t_1} \ne 0$;${s_2} = 2{t_2}$ and $\sum\limits_{i = 1}^{10} {{s_i} = \sum\limits_{i = 1}^{15} {{t_i}} }$. Then the value of $\dfrac{{{s_2} - {s_1}}}{{{t_2} - {t_1}}}$ is
A) $\dfrac{8}{3}$
B) $\dfrac{3}{2}$
C) $\dfrac{{19}}{8}$
D) $2$

Answer 1

First assume the common difference and the first term of both the given series and then use the given data in the problem and the formulas given below to find out the relation between both the series, it gives the values of the common difference of both the series, which gives the desired result.
The ${n^{th}}$ term of the Arithmetic series is given as:
${a_n} = a + \left( {n - 1} \right)d$
The sum of the ${n^{th}}$ term is given as:
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

Complete step by step solution:
It is given in the problem that ${s_1},{s_2},{s_3} \cdot \cdot \cdot \cdot$ and ${t_1},{t_2},{t_3} \cdot \cdot \cdot \cdot$are two arithmetic sequences such that${s_1} = {t_1} \ne 0;{s_2} = 2{t_2}$and$\sum\limits_{i = 1}^{10} {{s_i} = \sum\limits_{i = 1}^{15} {{t_i}} }$.
We have to find the value of $\dfrac{{{s_2} - {s_1}}}{{{t_2} - {t_1}}}$, but we know that in the arithmetic sequence, the difference of two consecutive terms is constant and called as common difference, so we have to find the value ratio of the common difference of the series.
Assume that ${d_1}$ is a common difference and ${a_1}$ is the first term of the series ${s_1}, {s_2}, {s_3} \cdot \cdot \cdot \cdot$ and ${d_2}$ is a common difference and ${a_2}$ is the first term of the series ${t_1}, {t_2}, {t_3} \cdot \cdot \cdot \cdot$.
It is given to us that ${s_1} = {t_1} \ne 0$, it means that:
${a_1} = {a_2} \ne 0$ … (1)
It is also given that:
${s_2} = 2{t_2}$
Using the ${n^{th}}$ term formula of the arithmetic sequence, we have
${a_1} + {d_1} = 2\left( {{a_2} + {d_2}} \right)$
From equation (1):
$\Rightarrow {a_1} + {d_1} = 2\left( {{a_1} + {d_2}} \right)$
$\Rightarrow {a_1} + {d_1} = 2{a_1} + 2{d_2}$
$\Rightarrow {d_1} - {a_1} = 2{d_2}$
$\Rightarrow {d_2} = \dfrac{{{d_1} - {a_1}}}{2}$ … (2)
We have also given that:
$\sum\limits_{i = 1}^{10} {{s_i} = \sum\limits_{i = 1}^{15} {{t_i}} }$
Then using the formula of the sum of Arithmetic series, we have
$\Rightarrow \dfrac{{10}}{2}\left[ {2{a_1} + \left( {10 - 1} \right){d_1}} \right] = \dfrac{{15}}{2}\left[ {2{a_2} + \left( {15 - 1} \right){d_2}} \right]$
$\Rightarrow 2\left[ {2{a_1} + 9{d_1}} \right] = 3\left[ {2{a_2} + 14{d_2}} \right]$
Substitute the values of equation (1) and equation (2):
$\Rightarrow 2\left[ {2{a_1} + 9{d_1}} \right] = 3\left[ {2{a_1} + 14\left( {\dfrac{{{d_1} - {a_1}}}{2}} \right)} \right]$
$\Rightarrow 4{a_1} + 18{d_1} = 3\left[ {7{d_1} - 5{a_1}} \right]$
$\Rightarrow 4{a_1} + 18{d_1} = 21{d_1} - 15{a_1}$
$\Rightarrow 19{a_1} = 3{d_1}$
$\Rightarrow {d_1} = \dfrac{{19}}{3}{a_1}$ … (2)
Now, we put the value of ${d_1}$ in the equation (1):
$\Rightarrow {d_2} = \dfrac{{\dfrac{{19}}{3}{a_1} - {a_1}}}{2}$
$\Rightarrow {d_2} = \dfrac{{8{a_1}}}{3}$
Now, we divide ${d_1}$ by ${d_2}$:
$\Rightarrow \dfrac{{{d_1}}}{{{d_2}}} = \dfrac{{\dfrac{{19}}{3}{a_1}}}{{\dfrac{{8{a_1}}}{3}}}$
$\Rightarrow \dfrac{{{d_1}}}{{{d_2}}} = \dfrac{{19}}{8}$
As ${d_1}$ is the common difference of the series ${s_1}, {s_2}, {s_3} \cdot \cdot \cdot \cdot$ and ${d_2}$ is the common difference of the series ${t_1}, {t_2}, {t_3} \cdot \cdot \cdot \cdot$, so we have:
$\Rightarrow \dfrac{{{s_2} - {s_1}}}{{{t_2} - {t_1}}} = \dfrac{{19}}{8}$
This is the required result.

Hence, the option (C) is correct.

Note:
As given in the problem that given series are Arithmetic series it means that the common difference of the given series is constant, and the common difference is the difference of the two consecutive terms of the series.