Question

The sum of 100 terms of an A.P whose all the terms are natures numbers , lies in open interval (10000,10100) and its 50 term is 100 , then its first term is

• A. 49
• B. 50
• C. 51
• D. 2

Answer 1

Answer: (C)

51

$\frac{100}{2}(2a+99d)\in (10000,10100)$

where $a=$ first term and $d=$ common difference

$a_{50}=a+49d=100$

$\Rightarrow 2a+98d=200$

$\Rightarrow 2a+99d=200+d$

substitute this in above eq., $\frac{100}{2}(200+d)\in (10000,10100)$

200+d $\in (\frac{10000}{50},\frac{10100}{50})$

200+d $\in (200,202)$

$\Rightarrow d\in (0,2)$

$\Rightarrow d=1$ ( $d$ will be natural number as all terms are natural)

Substitute $d$ in $a+49d=100$ numbers

$\Rightarrow a=100-(49\times 1)$

$\Rightarrow a=51$

$\therefore$ option $C$ is correct.