Question

Find the sum of those integers between 120 and 480 which are multiples of 3

or 12.
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Answer 1

The sum is 35700.

Step-by-step explanation:

Sum of the integers between 120 and 480 which are multiples of 3 or 12.

123 + 126 + 129 + 132 + ...........+ 477.

Here 1st term = a = 123

Common difference = d = 126 - 123 = 129 - 126 = 3

nth term = 477

=> a + (n - 1)d = 477

=> 123 + (n - 1) x 3 = 477

=> (n - 1) x 3 = 477 - 123 = 354

=> (n - 1) = $\frac{354}{3}$

=> n - 1 = 118

=> n = 118 + 1 = 119

Sum of the integers between 120 and 480 which are multiples of 3 or 12

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

$=\frac{119}{2}[2\times 123 + (119 - 1) \times 3]\\\\= \frac{119}{2} [246 + 118\times 3]\\\\= \frac{119}{2}[246 + 354]\\\\= \frac{119}{2}\times 600\\\\= 119\times 300\\\\= 35700$

Hence, the sum is 35700.