Question

Find the sum of first 40 positive integers divisible by 6 also find the sum of first 20
positive integers divisible by 5 or 6.

Answer 1

Hi there!
Here's the answer:

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¶ Sum of terms in A.P = $\frac{n}{2}(a+l)$

where a- first term ;
l- last term ;
n- No. of terms


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¶¶¶ SOLUTION :

(a)
The No.s divisible by 6 are :

6, 12, ………… , 6n

Here, the No.s are in A.P

where,
first term = 6
last term = 6n

Given,
n = 40

=> last term = 6n = 6×40 = 240.

Sum of terms = $\frac{40}{2}(6+240)$

= 20 × 246

= 4920

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(b)

The No.s divisible by 5 and 6 are :

30, 60, ………… , 30n

Here, the No.s are in A.P

where,
first term = 30
last term = 30n

Given,
n = 20

=> last term = 30n = 30×20 = 600.

Sum of terms = $\frac{20}{2}(30+600)$

= 10 × 630

= 6300

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Answer 2

Step-by-step explanation:

positive integers divisible by 6 are

6,12,18,24...........

since difference is same,it is an AP

we need to find the sum of 40 integers

we can use formula

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

here n=40,a=6&d=12-6=6

putting values in formula

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

=40/2(2×6+(40-1)×6)

=20(12+39×6)

=20(12+234)

=20×246=4920

therefore sum of the first 40 positive integers which are divisible by 6 is 4920