Question 1187314: The first term of an arithmetic progression is 12 and the sum of the first 16 terms is 282.
a) Find the common difference of this progression.
The first, fifth and nth term of this arithmetic progression are the first, second and third
term respectively of a geometric progression.
b) Find the common ratio of the geometric progression and the value of n.
Answer by greenestamps(13198)   (Show Source):
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The first 16 terms are
a, a+d, a+2d, ..., a+14d, a+14d
Combining like terms, the sum of the first 16 terms is
16a + (1+2+3+...+14+15)d = 16a+120d
The first term is 12; the sum of the first 16 terms is 282:
16(12)+120d = 282
192+120d = 282
120d = 90
d = 3/4
ANSWER a): the common difference is 3/4
1st term: 12
5th term: 12+4(3/4) = 12+3 = 15
The 1st and 5th terms of the arithmetic progression are the 1st and second terms of a geometric progression.
common ratio: 15/12 = 5/4
3rd term in the geometric progression: 15(5/4) = 75/4 = 12+27/4 = 12+9d, which is the 10th term of the arithmetic progression.
The 3rd term of the geometric progression is the 10th term of the arithmetic progression.
ANSWER b): common ratio 3/4; n=10
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