If x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞, y = b −

1.	If x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞, y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ and z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ \(\frac{x.y}{z}\) = \(\frac{ab}{c}\)
Answer» Given series, x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞ is in G.P with common ratio \(\frac{1}{r}\). and series y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ is in G.P with common ratio − \(\frac{1}{r}\). and series z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ is in G.P with common ratio − \(\frac{1}{r^2}\) Sum of infinite terms of series x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞ is x = \(\frac{a}{1-\frac{1}{r}}\) Sum of infinite term of series y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ is y = \(\frac{b}{1-\big(\frac{1}{r}\big)}\) = \(\frac{b}{1+\frac{1}{r}}\) Sum of infinite terms of series z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ is z = \(\frac{c}{1-\frac{1}{r^2}}\) Now, LHS = \(\frac{xy}{z}\) = \(\frac{\bigg\{\frac{a}{1 − \frac{1}{ r}} \bigg\} \bigg\{\frac{b}{1 + \frac{1}{ r}}\bigg\}}{\frac{c}{1-\frac{1}{r^2}}}\) = \(\frac{\frac{ab}{\big(1-\frac{1}{r^2}\big)}}{\frac{c}{\big(1-\frac{1}{r^2}\big)}}\) = \(\frac{ab}{c}\) = R.H.S

If x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞, y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ and z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ \(\frac{x.y}{z}\) = \(\frac{ab}{c}\)

Answer»

Given series,

x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞ is in G.P with common ratio \(\frac{1}{r}\).

and series y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ is in G.P with common ratio − \(\frac{1}{r}\).

and series z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ is in G.P with common ratio − \(\frac{1}{r^2}\)

Sum of infinite terms of series

x = a + \(\frac{a}{r}\) + \(\frac{a}{r^2}\) + ⋯ + ∞ is x = \(\frac{a}{1-\frac{1}{r}}\)

Sum of infinite term of series

y = b − \(\frac{b}{r}\) + \(\frac{b}{r^2}\) − ⋯ + ∞ is y = \(\frac{b}{1-\big(\frac{1}{r}\big)}\) = \(\frac{b}{1+\frac{1}{r}}\)

Sum of infinite terms of series

z = c + \(\frac{c}{r^2}\) + \(\frac{c}{r^4}\) + ⋯ + ∞ is z = \(\frac{c}{1-\frac{1}{r^2}}\)

Now, LHS = \(\frac{xy}{z}\)

= \(\frac{\bigg\{\frac{a}{1 − \frac{1}{ r}} \bigg\} \bigg\{\frac{b}{1 + \frac{1}{ r}}\bigg\}}{\frac{c}{1-\frac{1}{r^2}}}\)

= \(\frac{\frac{ab}{\big(1-\frac{1}{r^2}\big)}}{\frac{c}{\big(1-\frac{1}{r^2}\big)}}\)

= \(\frac{ab}{c}\) = R.H.S