1955 AHSME Problems/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:
Let our geometric sequence be and let our arithmetic sequence be . We know that This implies that , hence and . We can rewrite as , plugging this into , we get , we can simplify this to get , 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 .