SOLUTION: Please help me solve this question the fifth, ninth and sixteenth term of a linear sequence(a.p) are consecutive terms of an exponential sequence(g.p) (i) find the common differ

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Question 1184673: Please help me solve this question
the fifth, ninth and sixteenth term of a linear sequence(a.p) are consecutive terms of an exponential sequence(g.p)
(i) find the common difference of the linear sequence in terms of the first terms
(ii) show that the twenty-first,thirty-seventh and sixty-fifth terms of the linear sequence are consecutive terms of an exponential sequence whose common ratio is 7/4
Answer by greenestamps(13200) About Me  (Show Source):
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(i) 5th term of the ap: a+4d
9th term: a+8d
16th term: a+15d

In a gp, the square of any term is equal to the product of the terms before and after it:

%28a%2B8d%29%5E2=%28a%2B4d%29%28a%2B15d%29
a%5E2%2B16ad%2B64d%5E2=a%5E2%2B19ad%2B60d%5E2
4d%5E2-3ad=0
d%284d-3a%29=0

The problem is of little interest if d=0, so

4d-3a=0
4d=3a
d=%283%2F4%29a

(i) ANSWER: d = (3/4)a

(ii) 21st term: a+20d = a+15a = 16a
37th term: a+36d = a+27a = 28a
65th term: a+64d = a+48a = 49a

28a/16a = 7/4; 49a/28a = 7/4; this sequence is geometric with common ratio 7/4