The sum of first five multiples of $3$ is

45

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55

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65

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75

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Solution

The correct option is C $45$
Sum of an AP is: $\frac{n}{2} \times (2 a + (n - 1) d)$
$a = 3, d = 3, n = 5$
Then,
Sum = $\frac{5}{2} \times (2 \times 3 + (5 - 1) \times (3))$
= $\frac{5}{2} \times (6 + 12)$
= $45$

Suggest Corrections

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

Class interval:	25−35	35−45	45−55	55−65	65−75
Frequency:	6	10	8	12	4