Browse Questions

# Find the sum of even numbers between 50 and 100

(a) 1800 (b) 2800 (c) 5800 (d) 5000

Toolbox:
• Sum of an AP is given by $S_{n} = \large \frac {n}{2} [2a + (n-1)d]$
• nth term of an AP is given by $T_{n} = \large a + (n-1)d$
Sum of an AP is given by $S_{n} = \large \frac {n}{2} [2a + (n-1)d]$
Here $a = 52, \quad d = 2, \quad T_{n} = 98$
$T_{n} = \large a + (n-1)d$
$\therefore 98 = 52 + (n - 1)2 \implies 42 = (n - 1)2 \implies n = 24$

$S_{n} = \large \frac {n}{2} [2a + (n-1)d]$

$S_{n} = \large \frac {24}{2} [2 X 52 + (24 - 1) X 2]$

$S_{n} = 1800$