Question

find the sum of all natural numbers from 1 to 100 divisible by 2 and 5

Answer 1

Answer:

Step-by-step explanation:

Basically what we need to do is

Integers divisible by 2 + Integers divisible by 5 — Integers divisible by 10

(As some common integers are also included in it)

Sn=n[2a+(n+1)d]/2

Sn=n[A + An]/2

So,

50[2+100]/2 + 20[5+100]/2 + 10[10+100]/2

= 50*51 + 10*105 — 10*55

= 2550 + 1050 — 550

= 3050

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Answer 2

Answer:

Step-by-step explanation:

Integers between 1 and 100 divisible 2 and 5 can be expressed in the form of the AP given below.

10,20,30,40,...,100

Here,

First term(a)=10

Common difference(d)=20-10=10

Tn=a+(n-1)d=100

10+(n-1)10=100

10+10n-10=100

10n=100

n=100/10=10

The given AP has 10 terms.

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

S10=10/2[2*10+(10-1)10]

=5[20+9*10]

=5[20+90]

=5*110=550 ans