What is the sum of the geometric series in which a1 = 3, r = 2, and an = 192?
Hint: Sn=a1 (1-rn)/ 1-r , r ≠ 1, where a1 is the first term and r is the common ratio.
A. Sn = −381 B. Sn = −274 C. Sn = 381 D. Sn = 274
We need to solve this, first
192 = 3(2)n-1 divide both sides by 3
64 = 2n-1 and 26 = 64, so n = 7
So we have the sum of 7 terms
Sn = 3 [ (1 - 27) / (1 - 2) ] = 3 [ (1 - 128) / (-1) ] = 3* 127 = 381
