Question

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

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

Answer 1

Answer: (B)

1620

Answer 2

$1 + 3 + 5 + 7 + ... + 29 + 30 + 31 + 32 + ... + 60$
$= [1 + 3 + 5 + 7 + ... + 29] + (30 + 31 + ... + 60]$
$1^{st}$ 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.
$= \frac{n_1}{2} (1+29) + \frac{n_2}{n} (30+60)$
$= \frac{n_1}{2} (30) + \frac{n_2}{n} (90) = 15 n_1 +45 n_2$
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 $60 =30 + (n_2 - 1) (1)$
$\Rightarrow\:\:\: n_2 -1 = 30 \Rightarrow n_2 = 31$
$\therefore\:\:\:\: 1 + 3 + 5 + .... + 29 + 30 + 31 + .... 60$
$= 15 \times 15 + 45 \times 31 = 225 + 1395 = 1620$