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)