Question

1) Find the sum of all natural number between 1 to 140 which is divisible by 4?
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Answer 1

Answer:

the sum of all natural numbers between 1 to 140 which is divisible by 4 = 2520

Answer 2

Answer:

2520

Step-by-step explanation:

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140.

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP.

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n =140⇒a+(n−1)d

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n =140⇒a+(n−1)d140=4+(n−1)×4

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n =140⇒a+(n−1)d140=4+(n−1)×4∴n=35

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n =140⇒a+(n−1)d140=4+(n−1)×4∴n=35∴ required sum =Sn = n/2[a+l]

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n =140⇒a+(n−1)d140=4+(n−1)×4∴n=35∴ required sum =Sn = n/2[a+l] =35/2[4+140]

Clearly, the numbers from 1 to 140 which are divisible by 4 are 4,8,12,...140. This is an AP with first term a=4, common difference d=4 and last term l=140.Let there be n terms in this AP. Then,a n =140⇒a+(n−1)d140=4+(n−1)×4∴n=35∴ required sum =Sn = n/2[a+l] =35/2[4+140] =2520