Arithmetic Progression
The $4^{th}$ and the $15^{th}$ terms of an A.P are $22$ and $66$ respectively. Find the first term and the sum of the first $11$ terms.
Given: $4^{th}$ term of A.P $= t_4 = 22$
$15^{th}$ term of A.P $= t_{15} = 66$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$
where $a = $ first term of A.P; $\;$ $d = $ common difference of A.P
$\therefore \;$ Fourth term $= t_4 = a + 3d = 22$ $\;\;\; \cdots \; (1a)$
Fifteenth term $= t_{15} = a + 14 d = 66$ $\;\;\; \cdots \; (1b)$
Subtracting equations $(1a)$ and $(1b)$ we have,
$11 d = 44$ $\implies$ $d = 4$
Substituting the value of $d$ in equation $(1a)$ we have,
$a + 3 \times 4 = 22$ $\implies$ $a = 10$
$\therefore \;$ The first term of the A.P $= a = 10$
Now, sum of $n$ terms of A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$
$\therefore \;$ Sum of $11$ terms of A.P $= S_{11} = \dfrac{11}{2} \left[2 \times 10 + \left(11 - 1\right) \times 4\right]$
i.e. $\;$ $S_{11} = \dfrac{11}{2} \times (20 + 40) = 330$