The 3rd and 7th term of a Geometric Progression (GP) are 81 and 16. Find the 5th term.
-
A.
(frac{4}{729}) -
B.
(frac{81}{16}) -
C.
27 -
D.
36
Correct Answer: Option D
Explanation
The nth term of a GP is given by: (T_{n} = ar^{n-1}).
(T_{3} = ar^{3-1} = ar^{2} = 81)…….(1)
(T_{7} = ar^{7-1} = ar^{6} = 16 ) …… (2)
Dividing (2) by (1), we have (r^{4} = frac{16}{81} = (frac{2}{3})^{4} implies r = frac{2}{3})
Putting (r = frac{2}{3}) in equation (1), we have (81 = a times (frac{2}{3})^{2} = a times frac{4}{9} implies a = frac{729}{4})
(T_{5} = ar^{5-1} = ar^{4} = frac{729}{4} times (frac{2}{3})^{4})
= (frac{729}{4} times frac{16}{81} = 36)