Asked by Anonymous
Find the 7th term of a Geometric progressions, if the first and the 5th term are 16 and A respectively
Answers
Answered by
oobleck
a = 16
r^4 = A
a_7 = a_5 * r^2 = 16√A
r^4 = A
a_7 = a_5 * r^2 = 16√A
Answered by
mathhelper
a=16
ar^4 = A
16r^4 = A
r^2 = √A/4
term(7) = a r^6
= a(r^4)(r^2)
= 16 (A/16)(√A/4)
= A^(3/2) / 4
testing:
suppose r = 3, then my first 7 terms are:
16, 48, 144, 432, 1296, 3888, 11644
According to my answer, term(7) = 1296^(3/2) / 4 = 46656/4 = 11664
ar^4 = A
16r^4 = A
r^2 = √A/4
term(7) = a r^6
= a(r^4)(r^2)
= 16 (A/16)(√A/4)
= A^(3/2) / 4
testing:
suppose r = 3, then my first 7 terms are:
16, 48, 144, 432, 1296, 3888, 11644
According to my answer, term(7) = 1296^(3/2) / 4 = 46656/4 = 11664
Answered by
oobleck
dang! Go with ol' reliable mathhelper once again.
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