Question 18
The sum of first five multiples of3 is :

A) 45
B) 55
C) 65
D) 75

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Solution

The first five multiples of 3 are 3, 6, 9, 12 and 15.
Here, first term, a = 3,
Common difference, d = 3
Number of terms, n = 5
$∴ S_{5} = \frac{5}{2} [2 a + (5 - 1) d] [∵ S_{n} = \frac{n}{2} {2 a + (n - 1) d}]$
$= \frac{5}{2} [2 \times 3 + 4 \times 3]$
$= \frac{5}{2} (6 + 12) = 5 \times 9 = 45$

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Similar questions

In the given figure, AB || CD. If $∠ C A B = 180^{\circ}$ and $∠ E F C = 25^{\circ}$ then $∠ C E F = ?$

$(a) 65^{\circ}$

$(b) 55^{\circ}$

$(c) 45^{\circ}$

$(d) 75^{\circ}$

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Weight $(i n k g)$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$	$75 - 80$
Number of candidates	$13$	$18$	$45$	$16$	$06$	$02$

:

Weight	$45 - 55$	$55 - 65$	$65 - 75$	$75 - 85$	$85 - 95$
Number of persons	$1$	$2$	$3$	$10$	$7$

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