RRB NTPC Mains or Stage 2 Model Paper 10 with solutions

(b) 3, 7, 11 ....... 147
It is an arithmetic series whose
first term, a = 3
last term, ${\mathit{x}}_{\mathit{n}}=147$
common difference, d = 4
$\mathit{x}\mathit{n}=\mathit{a}+\left(\mathit{n}\mathit{–}1\right)\mathit{d}$
$147=3+\left(\mathit{n}\mathit{–}1\right)\times 4$
$\mathit{n}\mathit{–}1=\frac{147-3}{4}$
n – 1 = 36, n = 37
The given series consists of 37 terms.
Therefore, its middle term will be
$\frac{37+1}{2}+=$19th term
${\mathit{x}}_{19}=3+\left(19\mathit{–}1\right)4$
$=3+18\times 4=75$
∴ The middle term of the given arithmetic series is 75.