Question

Find the sum of the 40 positive integers divisible by 6 ...​

Answer 1

$\large\mathtt \green {Answer }$

The first 40 positive integers divisible by 6 are 6, 12, 18 ..........240.

Clearly,

these numbers form and Ap with first term , a = 6 common difference, D = 6 and last term, l = 240 .

↪sum of the n term , Sn= n/2 [a+l]

↪sum of the Frist 40 term ,

$↪(s)40 = \frac{40}{2} (6 + 240)$

$↪20 \times 246 = 4920$

Hence,

• the sum of first positive integers divisible by 6 is 4920

Answer 2

the first 40 positive integers divisible by 6 are 6,12,18,....... upto

40 terms

the given series is in arthimetic progression with first term a=6 and common difference d=6

sum of n terms of an A.p is  

n/2×{2a+(n-1)d}

→required sum = 40/2×{2(6)+(39)6}

=20{12+234}

=20×246

=4920

I hope this will help u ;)