The third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the
(a) common ratio;
(b) first term ;
(c) sum of the first four terms.
(a) common ratio;
(b) first term ;
(c) sum of the first four terms.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
\(T_{n} = ar^{n - 1}\) (terms of a geometric progression)
\(T_{3} = ar^{3 - 1} = ar^{2} = 360 ... (1)\)
\(T_{6} = ar^{6 - 1} = ar^{5} = 1215 ... (2)\)
Divide (2) by (1).
\(\frac{ar^{5}}{ar^{2}} = \frac{1215}{360}\)
\(r^{3} = \frac{27}{8}\)
\(\therefore r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}\)
(b) Putting r in (1), we have
\(ar^{2} = 360\)
\(a \times (\frac{3}{2})^{2} = 360 \implies \frac{9a}{4} = 360\)
\(a = \frac{360 \times 4}{9} = 160\)
(c) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) (sum of terms of a G.P)
\(S_{4} = \frac{160((\frac{3}{2})^{4} - 1)}{\frac{3}{2} - 1}\)
= \(\frac{160(\frac{81}{16} - 1)}{\frac{1}{2}}\)
= \(\frac{160 \times \frac{65}{16}}{\frac{1}{2}}\)
= \(1300\)
\(T_{n} = ar^{n - 1}\) (terms of a geometric progression)
\(T_{3} = ar^{3 - 1} = ar^{2} = 360 ... (1)\)
\(T_{6} = ar^{6 - 1} = ar^{5} = 1215 ... (2)\)
Divide (2) by (1).
\(\frac{ar^{5}}{ar^{2}} = \frac{1215}{360}\)
\(r^{3} = \frac{27}{8}\)
\(\therefore r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}\)
(b) Putting r in (1), we have
\(ar^{2} = 360\)
\(a \times (\frac{3}{2})^{2} = 360 \implies \frac{9a}{4} = 360\)
\(a = \frac{360 \times 4}{9} = 160\)
(c) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) (sum of terms of a G.P)
\(S_{4} = \frac{160((\frac{3}{2})^{4} - 1)}{\frac{3}{2} - 1}\)
= \(\frac{160(\frac{81}{16} - 1)}{\frac{1}{2}}\)
= \(\frac{160 \times \frac{65}{16}}{\frac{1}{2}}\)
= \(1300\)