The fourth term of an A.P is $11$ and the $8^{th}$ term exceeds twice the $4^{th}$ term by $5$. Find the sum of the first fifty terms of the A.P.
Let the first term of A.P $= a$
Let the common difference of A.P $= d$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$
Fourth term of A.P $= t_4 = a + 3d = 11$ (Given) $\;\;\; \cdots \; (1)$
Eighth term of A.P $= t_8 = a + 7d$
Given: $\;$ $t_8 = 2 \times t_4 + 5$
i.e. $\;$ $a + 7d = 2 \left(a + 3d\right) + 5$
i.e. $\;$ $a + 7d = 2a + 6d + 5$
i.e. $\;$ $a - d = -5$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously gives,
$4d = 16$ $\implies$ $d = 4$
Substituting the value of $d$ in equation $(2)$ gives $\;$ $a = -1$
Sum to $n$ terms of A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$
$\therefore \;$ Sum to $50$ terms of A.P $= S_{50} = \dfrac{50}{2} \left[2 \times \left(-1\right) + \left(50 - 1\right) \times 4\right]$
i.e. $\;$ $S_{50} = 25 \left[-2 + 196\right] = 4850$