The sum to infinity of 1/7 + 2/72 + 1/73 + 2/74 +……………. is

1.	The sum to infinity of 1/7 + 2/72 + 1/73 + 2/74 +……………. isA) 5/48B) 7/24C) 3/16D) 1/5
Answer» Correct option is (C) 3/16 \(\because\frac17,\frac1{7^3},........\) will form a G.P. of common ratio \(=\frac1{7^2}<1\) And \(\frac2{7^2},\frac2{7^4},........\) will form a G.P. of common ratio \(=\frac1{7^2}<1\) We know that if r < 1 then sum of infinity terms in a G.P. is given by \(S_\infty=\frac a{1-r}.\) Now, \(\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+ ........\) \(=(\frac17+\frac1{7^3}+.......)+(\frac2{7^2}+\frac2{7^4}+........)\) \(=\cfrac{\frac{1}{7}}{1-\frac{1}{7^2}}+\cfrac{\frac{2}{7^2}}{1-\frac{1}{7^2}}\) \(=\cfrac{\frac{1}{7}}{\frac{49-1}{49}}+\cfrac{\frac{2}{49}}{\frac{49-1}{49}}\) \(=\frac{1}{7}\times\frac{49}{48}+\frac{2}{49}\times\frac{49}{48}\) \(=\frac{7}{48}+\frac{2}{48}\) \(=\frac{9}{48}=\frac{3}{16}\) Hence, \(\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+.......\) \(=\frac{3}{16}\) Correct option is C) 3/16

The sum to infinity of 1/7 + 2/72 + 1/73 + 2/74 +……………. isA) 5/48B) 7/24C) 3/16D) 1/5

Answer»

Correct option is (C) 3/16

\(\because\frac17,\frac1{7^3},........\) will form a G.P. of common ratio \(=\frac1{7^2}<1\)

And \(\frac2{7^2},\frac2{7^4},........\) will form a G.P. of common ratio \(=\frac1{7^2}<1\)

We know that if r < 1 then sum of infinity terms in a G.P. is given by \(S_\infty=\frac a{1-r}.\)

Now, \(\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+ ........\)

\(=(\frac17+\frac1{7^3}+.......)+(\frac2{7^2}+\frac2{7^4}+........)\)

\(=\cfrac{\frac{1}{7}}{1-\frac{1}{7^2}}+\cfrac{\frac{2}{7^2}}{1-\frac{1}{7^2}}\)

\(=\cfrac{\frac{1}{7}}{\frac{49-1}{49}}+\cfrac{\frac{2}{49}}{\frac{49-1}{49}}\)

\(=\frac{1}{7}\times\frac{49}{48}+\frac{2}{49}\times\frac{49}{48}\)

\(=\frac{7}{48}+\frac{2}{48}\)

\(=\frac{9}{48}=\frac{3}{16}\)

Hence, \(\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+.......\) \(=\frac{3}{16}\)

Correct option is C) 3/16