1955 AHSME Problems/Problem 45
Problem 45
Given a geometric sequence with the first term and
and an arithmetic sequence with the first term
.
A third sequence
is formed by adding corresponding terms of the two given sequences.
The sum of the first ten terms of the third sequence is:
Solution
Let our geometric sequence be and let our arithmetic sequence be
. We know that
This implies that
, hence
and
. Solving this system yields
, so
or
. But since
,
and
.
So our two sequences are
and
, which means the third sequence will be
Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is
.