The third of geometric progression (G.P) is 10 and the sixth term is 80. Find the common ratio.
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Correct Answer: Option A
Explanation:
\(T_{n} = ar^{n - 1}\) ( Geometric Progression)
\(T_{3} = ar^{3 - 1} = ar^{2} = 10 .... (1)\)
\(T_{6} = ar^{6 - 1} = ar^{5} = 80 .....(2)\)
Divide (2) by (1)
\(r^{5 - 2} = r^{3} = 8 \)
\(r = \sqrt[3]{8} = 2\)
\(T_{n} = ar^{n - 1}\) ( Geometric Progression)
\(T_{3} = ar^{3 - 1} = ar^{2} = 10 .... (1)\)
\(T_{6} = ar^{6 - 1} = ar^{5} = 80 .....(2)\)
Divide (2) by (1)
\(r^{5 - 2} = r^{3} = 8 \)
\(r = \sqrt[3]{8} = 2\)