Question

1. Find the sum of all 3 digit numbers between 200 and 400 which are divisible by 13. 2. Given the geometric progression 3, 6, 12, 24, ..., find the sixteenth term.​

Answer 1

Question 1

These numbers are
`208, 221, \cdots, 390`.

This arithmetic series has a first term of 208, a common difference of 13, and
`(390-208)/(13)+1=15` terms.

The sum is
`(15)/(2)[2(208)+(15-1)(13)]=\boxed{4485}`.

Question 2

The first term is 3 and the common ratio is 2.

So, the sixteenth term is
`3(2)^(16-1)=\boxed{98304}`.