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 ter...
Question 1 Report
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.
Answer Details
(a) Common ratio. For a GP, \(T_3 = ar^2 = 360\) and \(T_6 = ar^5 = 1215\). Dividing:
\[\frac{ar^5}{ar^2} = r^3 = \frac{1215}{360} = \frac{27}{8}\Rightarrow r = \sqrt[3]{\tfrac{27}{8}} = \tfrac{3}{2}.\]
(b) First term.
\[ar^2 = 360 \Rightarrow a\left(\tfrac{3}{2}\right)^2 = 360 \Rightarrow a\times\tfrac{9}{4} = 360 \Rightarrow a = 160.\]
(c) Sum of the first four terms. The terms are \(160, 240, 360, 540\):
\[S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{160\left(\left(\tfrac{3}{2}\right)^4 - 1\right)}{\tfrac{3}{2} - 1} = \frac{160(5.0625 - 1)}{0.5} = 1300.\]
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