Question

find the sum of first 40 terms of positive integers divisible by 6

Answer 1

a=6. ,d=6, n=40
S40 =40/2[2(6)+39(6)]
=20[12+234]
=20 × 246
=4920

Answer 2

Answer:

4,920

Step-by-step explanation:

here 40 positive integers are from numbering 1-40

here the first term is 6

diffrence is of 6

.:. here

a=t1=6

tn=36

d=6

Sn=?

Sn=n/2 {2a+(n-1)d}

S40=40/2{2*6+(40-1)6}

       =20{12+39*6}

       =20*246

       =4,920

That is sum of first 40 integers divisible by 6 is 4,920