Question

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

Answer 1

Question:-

• Find the sum of the first 40 positive integers divisible by 6

Answer:-

• 4920

Formula Used:-

• Sn=n/2×[2a+(n-1)d]

Solution:-

The positive integers divisible by 6 are 6,12,18,.......

This is an example of A.P. (Arithmetic progression).

Sum of n terms of an A.P. is given by,

Sn=n/2×[2a+(n-1)d]

where,

a = first term = 6

d=common difference = 6

Thus,

Sn = 40/2×{2(6)+(39)6}

= 20{12+234}

= 20×246

= 4920

Hence, the Answer is 4920

Answer 2

Question:-

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

Answer:-

4920

Formula Used:-

Sn=n/2×[2a+(n-1)d]

Solution:-

The positive integers divisible by 6 are 6,12,18,.......

This is an example of A.P. (Arithmetic progression).

Sum of n terms of an A.P. is given by,

Sn=n/2×[2a+(n-1)d]

where,

a = first term = 6

d=common difference = 6

Thus,

Sn = 40/2×{2(6)+(39)6}

= 20{12+234}

= 20×246

= 4920

Hence, the Answer is 4920