Question

The middle term of arithmetic series 3, 7, 11...147, is

• A. 71
• B. 75
• C. 79
• D. 83

Answer 1

The correct option is B 75

3,7,11.....147
It is an arithmetic series whose
first term, a = 3
Last term, $x_n=147$
Common difference, d = 4
$x_n=a+(n-1)d$
$147=3+(n-1)\times 4$
$n-1=\frac{147-3}{4}$
$n-1=36,n=37$
Since number of term n is odd, so middle term is average of first and last term.
Hence, middle term = $x_{19}=\frac{3+147}{2}=75$

Alternative Approach:
3,7,11.....147
It is an arithmetic series whose
first term, a = 3
Last term, $x_n=147$
Common difference, d = 4
$x_n=a+(n-1)d$
$147=3+(n-1)\times 4$
$n-1=\frac{147-3}{4}$
$n-1=36,n=37$
The given series consists of 37 terms. Therefore, its middle term will be
$\frac{37+1}{2}=19thterm$
$x_{19}=3+(19-1)4$
$=3+18\times 4=75$
$\therefore$ The middle term of the given arithmetic series is 75.