Question

How many two-digit numbers are divisible by 3 in AP?

Answer 1

Lowest two digit number divisible by 3 is 12

Highest two digit number divisible by 3 is 99

First term, a=12

Common difference, d=3

nth term=99

So, a + (n-1)d = 99 [by the formula]

12 + (n-1)3 = 99

(n-1)3 = 87

(n-1) = 29

n = 29+1

n= 30

So number of two digit numbers divisible by 3 is 30

Answer 2

Answer:

The numbers divisible by 3 are:

3,6,9,12,15........

the lowest no. divisible by 3 is 12 and

the highest no. divisible by 3 is 99

so the difference is 3 which is 12-15

last term, $a_{n}$=99, first term a=12, d=3

$a_{n}$=a+(n-1) x d

99=12+(n - 1) x 3

99=12+3n-3

99=9+3n

3n=99-9

3n=90

n=$\frac{90}{3}$

n=30

Therefore total nos divisible by 3 is 30

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