Question

How many two_digit numbers are divisible by 3 ?

Answer 1

a=12

d=3

n=?

l=99

a+(n-1)d= l

12+(n-1)3=99

n-1= 99-12/3

n-1=29

n=30

therefore, 30 two-digit numbers are divisible by 3

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Answer 2

$\huge{\underline{\underline{\bf{Solution}}}}$

$\rule{200}{2}$

$\tt Given \begin{cases} \sf{A.P \: : 12, 15, 18 ....... 99} \\ \sf{First \: term (a) = 12} \\ \sf{Common \: Difference(d) = 3} \\ \sf{Last \: term (A_n) = 99} \\ \sf{Number \: of \: terms (n) = ?} \end{cases}$

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{To \: Find :}}}}$

We have to find number of terms (n)

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{Explanation :}}}}$

We know the formula to find the value of n.

$\large{\star{\boxed{\sf{A_n = a + (n - 1)d}}}}$

____________________[Put Values]

$\sf{99 = 12 + (n - 1)3} \\ \\ \sf{\mapsto 99 = 12 + 3n - 3} \\ \\ \sf{\mapsto 99 = 9 + 3n} \\ \\ \sf{\mapsto 99 - 9 = 3n} \\ \\ \sf{\mapsto 90 = 3n} \\ \\ \sf{\mapsto \frac{\cancel{90}}{\cancel{3}} = n} \\ \\ \sf{\mapsto n = 30}$

$\large{\star{\boxed{\sf{n = 30}}}}$

$\therefore$ Number of two digit numbers divisible by 3 is 30.