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.

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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’.
$$\Rightarrow \dfrac{5}{2} \left( 2a+\left( 5-1\right) d\right) +\dfrac{7}{2} \left( 2a+\left( 7-1\right) d\right) =167$$ [ 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,
$$\Rightarrow \dfrac{10}{2} \left( 2a+\left( 10-1\right) d\right) =235$$
$$\Rightarrow 5\left( 2a+9d\right) =235$$
$$\Rightarrow \left( 2a+9d\right) =\dfrac{235}{5}$$
$$\Rightarrow \left( 2a+9d\right) =47$$.......(2)
Equation (1) and (2) are the pair of linear equations, so we are going to solve it by substitution method.
From equation (1),
$$\Rightarrow 12a=167-31d$$
$$\Rightarrow a=\dfrac{167-31d}{12}$$
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,...

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