Question

Find sum of all 3 digit numbers which leave the remainder 1 when divided by 4.​

Answer 1

The lowest 3 digit number is 101 and the highest number is 998.

Therefore, the AP is 101,104,107 \: \: \: \: \: \: ,998 where, the first term is a=101, the common difference is d=104−101=3 and the nth term is an=998

Now, let us find the number of terms 

 We know that the nth term of AP(Arithmetic Progression) is given by

an=a+(n−1)d

Now, substituting the values, we get:

an=a+(n−1)d⇒998=101+(n−1)3⇒(n−1)3=998−101⇒(n−1)3=897⇒n−1=3897⇒n−1=299⇒n=299+1⇒n=300

 We also know that the sum to nth term in an AP is given by,

S=2n(1st term + last term) 

Therefore, we have:

S=2300(101+998)=150×1099=164850

Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.

Answer 2

Answer:

sum = 123525

Step-by-step explanation:

the series is in AP