The sum to n terms of the series 1+ 4/5 + 7/5^2 + 10/5^3 + ..... is

Question

The sum to n terms of the series 1+ \(\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}\)+ ..... is

(a) \(\bigg(\frac{25}{14}-\frac{7n+10}{14\times5^{n-1}}\bigg)\)

(b) \(\bigg(\frac{17}{12}-\frac{4n+7}{12\times5^{n-1}}\bigg)\)

(c) \(\bigg(\frac{35}{16}-\frac{12n+7}{16\times5^{n-1}}\bigg)\)

(d) \(\bigg(\frac{15}{11}-\frac{10n+2}{11\times5^{n-1}}\bigg)\)

Answer 1

(c) \(\bigg(\frac{35}{16}-\frac{12n+7}{16\times5^{n-1}}\bigg)\)

 1+ \(\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}\)+ ..... is an A.G.P. with

A.P. : 1 + 4 + 7 + 10 + ...... 

G.P. : 1 + \(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}\) + ........

nth term of A.P. = 1 + 3(n – 1) = 3n – 2

nth term of G.P. = 1 x \(\big(\frac{1}{5}\big)^{n-1}\) = \(\frac{1}{5^{n-1}}\)

∴ nth term of the A.G.P. = \(\frac{3n-2}{5^{n-1}}\)

(n – 1)th term of the A.G.P. = \(\frac{3(n-1)-2}{5^{n-2}}\) = \(\frac{3n-5}{5^{n-2}}\)