Find the sum of the first 15 multiples of 8. | vedclass

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(B) The multiples of $8$ are $8, 16, 24, 32, \ldots$These numbers form an Arithmetic Progression $(A.P.)$ where the first term $a = 8$ and the common

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$960$

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(B) The multiples of $8$ are $8, 16, 24, 32, \ldots$These numbers form an Arithmetic Progression $(A.P.)$ where the first term $a = 8$ and the common difference $d = 8$.We need to find the sum of the first $n = 15$ terms.The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.Substituting the values: $S_{15} = \frac{15}{2}[2(8) + (15 - 1)8]$.$S_{15} = \frac{15}{2}[16 + 14(8)]$.$S_{15} = \frac{15}{2}[16 + 112]$.$S_{15} = \frac{15}{2}(128)$.$S_{15} = 15 	imes 64 = 960$.

Find the sum of the first $15$ multiples of $8.$

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