Find the ${S_n}$ for the following arithmetic series described.
(i) $a = 5,n = 30,\ell = 121$
(ii) $a = 50,n = 25,d = - 4$

Hint: This question is based on AP which means Arithmetic Progression. There are 2 different formulas to find the sum of the given AP,
1. ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
2. ${S_n} = \dfrac{n}{2}\left[ {a + \ell } \right]$
Here ${S_n}$ is an addition of n terms, “a” is first term, and “d” is the difference between two consecutive terms. And “l” is the last term of AP.

Complete step by step answer:
 (i) According to given conditions
$a = 5$ which is first term, Number of terms $n = 30$, Last term $\ell = 121$
So, by using formula
$\Rightarrow{S_n} = \dfrac{n}{2}\left[ {a + \ell } \right]$
$\Rightarrow{S_{30}} = \dfrac{{30}}{2}\left[ {5 + 121} \right]$
$\Rightarrow = 15 \times 126 = 1890$
So ${S_n}$ for this series will be 1890.

(ii) According to given conditions
First term $a = 50$, Number of terms $n = 25$, Common difference $d = - 4$
So by using formula
$\Rightarrow{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
$\Rightarrow{S_{25}} = \dfrac{{25}}{2}\left[ {(2 \times 50) + (25 - 1)( - 4)} \right]$
$\Rightarrow = \dfrac{{25}}{2}\left[ {100 - 96} \right]$
$ = 50$
So, ${S_n}$ of this series will be 50.

Note: 1. We can solve question 1 by ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
$a = 5,$ $n = 30,$ $\ell = 121$
Find d using $\ell ({a_n}) = a + (n - 1)d$
$\Rightarrow121 = 5 + (30 - 1)d$
$\Rightarrow 116 = 29d$
$\Rightarrow d = 4$
Now ${S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
$\Rightarrow{S_{30}} = \dfrac{{30}}{2}\left[ {(2 \times 5) + (30 - 1)4} \right]$
$ = 1890$

2. We can solve question 2 by another method also ${S_n} = \dfrac{n}{2}[a + \ell ]$
$\Rightarrow{S_{25}} = \dfrac{{55}}{2}[50 - 46] = 50$.