Mathematics
Answer
Two digits no. divisible by 3 are,
12, 15, 18 ………, 99.
The above series is an A.P. with, a = 12 and d = 15 - 12 = 3 and last term = 99.
Let total no. of terms be n,
∴ an = a + (n - 1)d
⇒ 99 = 12 + (n - 1)3
⇒ 99 = 12 + 3n - 3
⇒ 99 = 9 + 3n
⇒ 99 - 9 = 3n
⇒ 90 = 3n
⇒ n = 30.
Hence, two-digit numbers divisible by 3 are 30.
Related Questions
The nth term of the A.P. 6, 11, 16, 21, ……. is 106, the the value of n - 4 is :
17
15
16
20
In an A.P. ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Which term of A.P. 5, 15, 25, ……. will be 130 more than its 31st term ?
Find the value of p, if x, 2x + p and 3x + 6 are in A.P.