Question

Consider the sequence of three digit numbers which leave a remainder 1 on divisible by 3

a) What is its common difference ?

b) Which is the smallest number in this sequence ?

c)How many three digit numbers are there ,which leave a remainder 1 on divisible by 3?

d) What is the sum of such numbers​

Answer 1

a) Let the first term be a and the common difference be d.

So, the series will be a, a+d, a+2d

If we subtract 1 from each number in the sequence, it will be divisible by 3 and will look like( a-1,a+d-1,a+2d-1), now their common difference will be divisible by 3

Such as,

(a+d-1) -(a-1) = 3n

a+d-1-a+1=3n

d=3n

So,common difference will be 3n where n is a Natural number

b) Smallest 3 digit number divisible by 3 = 102, so the smallest number in this sequence is 102+1=103

c) The largest 3 digit number that leaves remainder 1 when divided by 3 = 997

We have smallest number as 103 and largest as 997,if we put all such numbers in sequence, we will get 3 as a common difference, for example,

103,106,109.......997

So, by the following formula:

$\boxed{T_n=a+(n-1)d}$

$997= 103+(n-1)3 \\\\\sf 997= 103+3n-3 \\\\\sf 997-100=3n \\\\\sf 3n=897 \\\\\sf n=299$

SO, THERE ARE 299 SUCH NUMBERS