Question

what is first term and common difference of two digit number divisible by 3​

Answer 1

Answer:

We know, first two digit number divisible by 3 is 12 and last two digit number divisible by 3 is 99. Thus, we get

12,15,18,...,99 which is an AP

Here, a=12,d=3

Let there be n terms. Then,

a

n

=99

a+(n−1)d=99

12+(n−1)3=99

n=29+1=30

Therefore, two digit numbers divisible by 3 are 30.

Answer 2

Answer:First two digit number divisible by 3 = 12

Last two digit number divisible by 3 = 99

An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

A.P = 12,15,18,…,99

here

First term (a) = 12

Common difference (d) = 3

Let us consider there are n numbers then

an = 99

a + (n – 1)d = 99

12 + (n – 1)3 = 99

12 + 3n – 3 = 99

n = 29+1

n = 30

∴ Two digit numbers divisible by 3 = 30.