Mathematics
Answer
Three digit numbers divisible by 87 are,
174, 261, ………, 957.
The above series is an A.P. with a = 174 and d = 261 - 174 = 87 and last term = 957.
Let no. of terms in the A.P. be n.
∴ an = 957
⇒ a + (n - 1)d = 957
⇒ 174 + (n - 1)(87) = 957
⇒ 174 + 87n - 87 = 957
⇒ 87n + 87 = 957
⇒ 87n = 957 - 87
⇒ 87n = 870
⇒ n = 10.
Hence, there are 10 three digit numbers which are divisible by 87.
Related Questions
Find the value of p, if x, 2x + p and 3x + 6 are in A.P.
If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, which term of it is zero ?
For what value of n, the nth term of A.P. 63, 65, 67, ……. and nth term of A.P. 3, 10, 17, ….., are equal to each other?
Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.