Asked by john
Find the 10th term in an AP where the first term is 3 and whose 1st, 4th and 13th terms form a GP.
Answers
Answered by
Steve
a = 3
Since the GP has a common ratio
(a+3d)/(a) = (a+12d)/(a+3d)
(3+3d)/3 = (3+12d)/(3+3d)
1+d = (1+4d)/(1+d)
d = 2
a+9d = 3+9*2 = 21
Since the GP has a common ratio
(a+3d)/(a) = (a+12d)/(a+3d)
(3+3d)/3 = (3+12d)/(3+3d)
1+d = (1+4d)/(1+d)
d = 2
a+9d = 3+9*2 = 21
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