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.
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) \(S_{4} = \frac{160((\frac{3}{2})^{4} - 1)}{\frac{3}{2} - 1}\) = \(\frac{160(\frac{81}{16} - 1)}{\frac{1}{2}}\) = \(\frac{160 \times \frac{65}{16}}{\frac{1}{2}}\) = \(1300\)