Question

$1+3+5+7+...+29+30+31+32+...+60=$

• A. 1611
• B. 1620
• C. 1609
• D. 1600

Answer 1

Answer: (B)

$1+3+5+7+...+29+30+31+32+...+60$
$=[1+3+5+7+...+29]+(30+31+...+60]$
$1^{st}$ A.P. $2^{nd}$ A.P.
$1^{st}$ A.P. has first term= 1,
common difference = 2 and last term = 29
$2^{nd}$ A.P. has first term = 30,
common difference = 1 and last term = 60
$1+3+...+60=$ Sum of $1^{st}$ A.P. + Sum of $2^{nd}$ A.P.
$=\frac{n_1}{2}(1+29)+\frac{n_2}{n}(30+60)$
$=\frac{n_1}{2}(30)+\frac{n_2}{n}(90)=15n_1+45n_2$
We know that $l=a+(n-1)d$
$29=1+(n_1-1)2\Rightarrow 2(n_1-1)=28$
$n_1-1=14\Rightarrow n_1=15$
and $60=30+(n_2-1)(1)$
$\Rightarrow n_2-1=30\Rightarrow n_2=31$
$\therefore 1+3+5+....+29+30+31+....60$
$=15\times 15+45\times 31=225+1395=1620$