Question

3. Finbd sum of first 40 positive integers divisible by 6.​

Answer 1

Answer:

Here, we have to find sum of series,

6,12,18,24...234,240

Here, first term, a=6

Last term, l=240

Number of terms, n=40

So, sum of series,S40=n2(a+l)=402(6+240)

S40=20∗246=4920

Step-by-step explanation:

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Answer 2

Answer:

Here, we have to find sum of series,

6,12,18,24...234,240

Here, first term, a=6

Last term, l=240

Number of terms, n=40

So, sum of series,S40=n2(a+l)=402(6+240)

S40=20∗246=4920

 

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