Question -
Answer -
Let n be the number of terms which are divisible by 3 and d are of two digit numbers
Let a be the first term and d be the common difference, then
a = 12, d = 3, last term = 99
an = a + (n – 1) d
99 = 12 + (n – 1) x 3
=> 99 = 12 + 3n – 3
=> 3n = 99 – 9
=> n = 30
Number of terms = 30