Question

Consider the multiples of 7m between 100and500 What are the first and last number
How many terms are there in the sequence

Answer 1

Answer:

the multiples of 7 starts from 105 and the last no. is 497

by the simple formula of

n th term when the numbers are in AP

Tn = a+( n - 1 )d

let last term is 497

then

497 = 105 +( n - 1 )7

392/7 = n - 1

then the value of n is 57

therefore the no. of terms in the sequence is 57