Find the sum of the terms of the A.P $\;$ $4, \; 9, \; 14, \; \cdots \; 89$. Also find the $4^{th}$ term from the end.
Given A.P is: $\;$ $4, \; 9, \; 14, \; \cdots \; 89$
First term $= a = 4$
Last term $= n^{th}$ term $= t_n = \ell = 89$
Common difference $= d = 5$
Let the number of terms in the given A.P be $= n$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$
i.e. $\;$ $89 = 4 + \left(n - 1\right) \times 5$
i.e. $\;$ $n - 1 = \dfrac{85}{5} = 17$ $\implies$ $n = 18$
$\therefore \;$ Number of terms in the given A.P is $= n = 18$
Sum of $n$ terms of an A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$
Here, $\;$ $S_{18} = \dfrac{18}{2} \left[2 \times 4 + \left(18 - 1\right) \times 5\right]$
i.e. $\;$ $S_{18} = 9 \left[8 + 85\right] = 837$
$n^{th}$ term from the end of an A.P $= \ell - \left(n - 1\right)d$
Here, $4^{th}$ term from the end of the A.P $= 89 - \left(4 - 1\right) \times 5 = 89 - 15 = 74$