Question

find the sum of the first 40 positive integers divisible by 5​

Answer 1

Hope my answer helps you :)

Solution:-

• First 40 positive integers divisible by 5

• Hence, the first multiple is 5 and the 40th multiple is 200.

• And, these terms will form an A.P. with the common difference of 5.

Here, First term  = 5

Number of terms  = 40

Common difference  = 5

So, the sum of 40 terms:-

=> 40  = 40/2 [2(5) + (40 − 1)5]

=> 20[10 + (39)5]  

=> 20(10 + 195)

=> 20(205) = 4100

Hence, the sum of first 40 multiples of 5 is 4100.

Thank you!

Answer 2

Answer:

The sum of the first 40 positive integers divisible by 5​ is  4100.

Step-by-step explanation:

The first 40 positive integers that are divisible by 5 are 5, 10, 15, 20,....

First number (a) = 5

and Common difference (d) = 5.

n = 40

We need to find $S_{40}$

​Sn = n/2[2a+(n−1)d]

= 40/2 [2(5) + (40 - 1)5]

= 20 [10 + (39)5]

= 20 [10 + 195]

= 20[205]

=4100

Sn = 4100