Question

# In an AP, if $S_{5}+S_{7}=167$ and $S_{10}=235$, then find the AP, where $S_{n}$ denotes the sum of first n terms.

Hint: In this question it is given if $S_{5}+S_{7}=167$ and $S_{10}=235$, then we have to find the AP, where $S_{n}$ denotes the sum of first n terms. So to find the solution we need to know the formula of summation of first n terms in an Arithmetic Progression,
i.e, $S_{n}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right)$...........(1)
Where ‘a’ is the first term of AP and ‘d’ is the common difference.
Complete step-by-step solution:
Let the first term of Ap is ‘a’ and the common difference ‘d’.
Given,
$S_{5}+S_{7}=167$
$\Rightarrow \dfrac{5}{2} \left( 2a+\left( 5-1\right) d\right) +\dfrac{7}{2} \left( 2a+\left( 7-1\right) d\right) =167$ [ by using formula (1)]
$\Rightarrow \dfrac{5}{2} \left( 2a+4d\right) +\dfrac{7}{2} \left( 2a+6d\right) =167$
$\Rightarrow 5\left( 2a+4d\right) +7\left( 2a+6d\right) =2\times 167$ [ Multiplying both side by 2]
$\Rightarrow 10a+20d+14a+42d=334$
$\Rightarrow 24a+62d=334$
$\Rightarrow 12a+31d=167$.....(1) [dividing both side by 2]
Now the second condition,
$S_{10}=235$
$\Rightarrow \dfrac{10}{2} \left( 2a+\left( 10-1\right) d\right) =235$
$\Rightarrow 5\left( 2a+9d\right) =235$
$\Rightarrow \left( 2a+9d\right) =\dfrac{235}{5}$
$\Rightarrow \left( 2a+9d\right) =47$.......(2)
Equation (1) and (2) are the pair of linear equations, so we are going to solve it by substitution method.
From equation (1),
$12a+31d=167$
$\Rightarrow 12a=167-31d$
$\Rightarrow a=\dfrac{167-31d}{12}$
Now substituting the value of ‘a’ in the equation (2), we get,
$2a+9d=47$
$\Rightarrow 2\left( \dfrac{167-31d}{12} \right) +9d=47$
$\Rightarrow \dfrac{167-31d}{6} +9d=47$
$\Rightarrow 167-31d+6\times 9d=6\times 47$ [multiplying both side by 6]
$\Rightarrow 167-31d+54d=282$
$\Rightarrow 23d=282-167$
$\Rightarrow 23d=115$
$\Rightarrow d=\dfrac{115}{23}$
$\Rightarrow d=5$
Now putting the value of ‘d’ in equation (2), we get,
$2a+9\times 5=47$
$\Rightarrow 2a+45=47$
$\Rightarrow 2a=47-45$
$\Rightarrow 2a=2$
$\Rightarrow a=1$
Therefore, the terms of the given AP is,
a, a+d, a+2d, a+3d,...
$\Rightarrow$ 1, 1+5, 1+2$\times 5$, 1+3$\times 5$,...
$\Rightarrow$ 1, 6, 11, 16, …
Note: So in order to find the A.P, you need to find the first term and common difference of the A.P , because if you know the first term ‘a’ and common difference ‘d’ then you can easily write the each term of A.P, i.e, a, a+d, a+2d, a+3d,...