Question

In an A.P., if ${S_6} + {S_7} = 167$ and ${S_{10}} = 235$, then find the A.P.

Answer 1

Here, we will find the A.P by using the formula for sum of $n$ terms of an A.P.. For this we will create two equations using the given information. Then, we will solve the equations to get the first term and common difference of the A.P. Using the first term and common difference, we will find the first few terms of the A.P.
Formula Used: The sum of $n$ terms of an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ is the first term of the A.P. and $d$ is the common difference.

Complete step by step solution:
First, we will use the formula for the sum of terms of an A.P. to get the sum of the first 6 and the sum of the first 7 terms.
The sum of $n$ terms of an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ is the first term of the A.P. and $d$ is the common difference.
Substituting $n = 6$ in the formula for sum of $n$ terms of an A.P., we get
${S_6} = \dfrac{6}{2}\left[ {2a + \left( {6 - 1} \right)d} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {S_6} = 3\left( {2a + 5d} \right)\\\ \Rightarrow {S_6} = 6a + 15d\end{array}$
Substituting $n = 7$ in the formula for sum of $n$ terms of an A.P., we get
${S_7} = \dfrac{7}{2}\left[ {2a + \left( {7 - 1} \right)d} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {S_7} = \dfrac{7}{2}\left( {2a + 6d} \right)\\\ \Rightarrow {S_7} = 7\left( {a + 3d} \right)\\\ \Rightarrow {S_7} = 7a + 21d\end{array}$
Now, it is given that ${S_6} + {S_7} = 167$.
We will substitute the values of the sums to get equation $\left( 1 \right)$.
Substituting ${S_6} = 6a + 15d$ and ${S_7} = 7a + 21d$ in the expression, we get
$\Rightarrow 6a + 15d + 7a + 21d = 167$
Adding the like terms, we get
$\Rightarrow 13a + 36d = 167 \ldots \ldots \ldots \left( 1 \right)$
Now, we will find the sum of the first 10 terms of the A.P.
Substituting $n = 10$ in the formula for sum of $n$ terms of an A.P., we get
${S_{10}} = \dfrac{{10}}{2}\left[ {2a + \left( {10 - 1} \right)d} \right]$
Simplifying the expression, we get
$\Rightarrow {S_{10}} = 5\left( {2a + 9d} \right)$
It is given that ${S_{10}} = 235$.
Substituting ${S_{10}} = 5\left( {2a + 9d} \right)$ in the expression, we get
$\Rightarrow 5\left( {2a + 9d} \right) = 235$
Dividing both sides of the equation by 5, we get
$\Rightarrow 2a + 9d = 47 \ldots \ldots \ldots \left( 2 \right)$
Now, we can observe that equations $\left( 1 \right)$ and $\left( 2 \right)$ are a pair of linear equations in two variables.
We will solve these two equations by elimination method to get the values of $a$ and $d$.
Multiplying both sides of equation $\left( 1 \right)$ by $\dfrac{1}{4}$, we get
$\begin{array}{l} \Rightarrow \left( {13a + 36d} \right)\dfrac{1}{4} = 167 \times \dfrac{1}{4}\\\ \Rightarrow \dfrac{{13a}}{4} + 9d = \dfrac{{167}}{4} \ldots \ldots \ldots \left( 3 \right)\end{array}$
Subtracting equation $\left( 2 \right)$ from equation $\left( 3 \right)$, we get

$\begin{array}{l}\dfrac{{13a}}{4}{\text{ }} + 9d = {\text{ }}\dfrac{{167}}{4}\\\\\underline {{\text{ }}2a{\text{ }} + 9d = {\text{ }}47} \\\\\dfrac{{13a}}{4} - 2a{\text{ }} = \dfrac{{167}}{4} - 47\end{array}$
Taking the L.C.M. on both sides, we get
$\Rightarrow \dfrac{{13a - 8a}}{4} = \dfrac{{167 - 188}}{4}$
Simplifying the expressions, we get
$\Rightarrow 5a = - 21$
Dividing both sides by 5, we get
$\therefore a=-\dfrac{{21}}{{5}}$
Therefore, the first term of the A.P. is $- \dfrac{{21}}{5}$.
Substituting $a = - \dfrac{{21}}{5}$ in equation $\left( 2 \right)$, we get
$\begin{array}{l} \Rightarrow 2\left( { - \dfrac{{21}}{5}} \right) + 9d = 47\\\ \Rightarrow - \dfrac{{42}}{5} + 9d = 47\end{array}$
Adding $\dfrac{{42}}{5}$ on both sides, we get
$\begin{array}{l} \Rightarrow - \dfrac{{42}}{5} + 9d + \dfrac{{42}}{5} = 47 + \dfrac{{42}}{5}\\\ \Rightarrow 9d = 47 + \dfrac{{42}}{5}\end{array}$
Taking the L.C.M. and adding the terms, we get
$\begin{array}{l} \Rightarrow 9d = \dfrac{{235 + 42}}{5}\\\ \Rightarrow 9d = \dfrac{{277}}{5}\end{array}$
Dividing both sides by 9, we get
$\therefore d=\dfrac{{277}}{{45}}$
Therefore, the common difference of the A.P. is $\dfrac{{277}}{{45}}$.
Now, we have the first term and the common difference of the A.P.
We know that each term of the A.P. is the sum of the previous term and the common difference.
Thus, we get
Second term of the A.P. $= - \dfrac{{21}}{5} + \dfrac{{277}}{{45}} = \dfrac{{ - 189 + 277}}{{45}} = \dfrac{{88}}{{45}}$
Similarly, we get
Third term of the A.P. $= \dfrac{{88}}{{45}} + \dfrac{{277}}{{45}} = \dfrac{{365}}{{45}} = \dfrac{{73}}{9}$

Therefore, we get the A.P. as $- \dfrac{{21}}{5},\dfrac{{88}}{{45}},\dfrac{{73}}{9}, \ldots \ldots$.

Note:
Note: We can also use a substitution method to simplify the two linear equations in two variables.
Rewriting equation $\left( 2 \right)$, we get
$\begin{array}{l} \Rightarrow 2a = 47 - 9d\\\ \Rightarrow a = \dfrac{{47 - 9d}}{2}\end{array}$
Substituting $a = \dfrac{{47 - 9d}}{2}$ in equation $\left( 1 \right)$, we get
$\Rightarrow 13\left( {\dfrac{{47 - 9d}}{2}} \right) + 36d = 167$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow \dfrac{{611 - 117d}}{2} + 36d = 167\\\ \Rightarrow \dfrac{{611 - 117d + 72d}}{2} = 167\\\ \Rightarrow 611 - 45d = 334\end{array}$
Rewriting the equation, we get
$\begin{array}{l} \Rightarrow 45d = 611 - 334\\\ \Rightarrow 45d = 277\end{array}$
Dividing both sides by 45, we get
$\therefore d=\dfrac{{277}}{{45}}$
Now, substituting $d = \dfrac{{277}}{{45}}$ in the equation $a = \dfrac{{47 - 9d}}{2}$, we get
$\begin{array}{l} \Rightarrow a = \dfrac{{47 - 9\left( {\dfrac{{277}}{{45}}} \right)}}{2}\\\ \Rightarrow a = \dfrac{{47 - \dfrac{{277}}{5}}}{2}\\\ \Rightarrow a = \dfrac{{\dfrac{{235 - 277}}{5}}}{2}\\\ \Rightarrow a = \dfrac{{ - 42}}{{10}}\\\ \Rightarrow a = \dfrac{{ - 21}}{5}\end{array}$
Thus, we get the first term and the common difference of the A.P. using substitution method to solve the linear equations in two variables.