Find the sum of the first 40 positive integer | vedclass

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(A) The positive integers that are divisible by $6$ are $6, 12, 18, 24, \ldots$This sequence forms an Arithmetic Progression $(A.P.)$ where the first

Vedclass · Accepted Answer

$4920$

Vedclass · Answer

(A) The positive integers that are divisible by $6$ are $6, 12, 18, 24, \ldots$This sequence forms an Arithmetic Progression $(A.P.)$ where the first term $a = 6$ and the common difference $d = 6$.We need to find the sum of the first $n = 40$ 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 $n = 40$,$a = 6$,and $d = 6$:$S_{40} = \frac{40}{2}[2(6) + (40 - 1)6]$$S_{40} = 20[12 + (39)(6)]$$S_{40} = 20[12 + 234]$$S_{40} = 20 	imes 246$$S_{40} = 4920$.

$S.No.$	$a$	$d$	$n$	$a_{n}$
$(i)$	$7$	$3$	$8$	$...$
$(ii)$	$-18$	$...$	$10$	$0$
$(iii)$	$...$	$-3$	$18$	$-5$
$(iv)$	$-18.9$	$2.5$	$...$	$3.6$
$(v)$	$3.5$	$0$	$105$	$...$

Find the sum of the first $40$ positive integers divisible by $6$.

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