Find the sum of all 3 digit numbers which leave a remainder of 6 when divided by 10.
Answers
Answer:
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 a
n
=998
Now, let us find the number of terms
We know that the nth term of AP(Arithmetic Progression) is given by
a
n
=a+(n−1)d
Now, substituting the values, we get:
a
n
=a+(n−1)d
⇒998=101+(n−1)3
⇒(n−1)3=998−101
⇒(n−1)3=897
⇒n−1=
3
897
⇒n−1=299
⇒n=299+1
⇒n=300
We also know that the sum to nth term in an AP is given by,
S=
2
n
(1st term + last term)
Therefore, we have:
S=
2
300
(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.
Step-by-step explanation: