Question

Solve the question in attachment

Class 11th
Maths { HP , GP }

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Answer 1

Step-by-step explanation:

explanation attached.

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Answer 2

Question :

If $\sf x = a + \frac{a}{r} + \frac{a}{r {}^{2} } + ... \infty$,$\sf y = b - \frac{b}{r} + \frac{b}{r {}^{2} } + ... \infty$ and $\sf c = c + \frac{c}{r {}^{2} } + \frac{c}{r {}^{4} } + ... \infty$

Prove that $\sf \dfrac{xy}{z} = \dfrac{ab}{c}$

Formula used:

1) Sum of n terms of a GP

The sum of first n terms of an GP is given by

$\sf\:S_{n}=a(\dfrac{1-r{}^{n}}{1-r})$,when r<1

or $\sf\:S_{n}=a(\dfrac{r{}^{n}-1}{r-1})$,when r>1

2) For infinite GP series

$\sf\:S_{n}=(\dfrac{a}{1-r})$

Solution :

$\sf x = a + \frac{a}{r} + \frac{a}{r { }^{2} } + ... \infty$

Here first term =a

common ratio = $\sf\:=\dfrac{1}{r}$

Therefore

$\sf x = s _{n} = \dfrac{a}{1 - \frac{1}{r} } = \dfrac{ar}{r - 1} ..(1)$

$\sf y = 1 - \frac{b}{r} + \frac{b}{r {}^{2} } + .... \infty$

Here first term =b

common ratio = $\sf\:=\dfrac{-1}{r}$

Therefore,

$\sf \: y = s _{n} = \dfrac{b}{1 - ( \frac{ - 1}{r} )}$

$\sf = \dfrac{br}{r + 1} ...(2)$

$\sf z = c + \frac{c}{r {}^{2} } + \frac{c}{r {}^{4} } + ... \infty$

Here ,first term = c

Common ratio =$\sf\:=\dfrac{1}{r{}^{2}}$

$\sf z = s_{n} = \dfrac{c}{1 - \frac{1}{r {}^{2} } }$

$= \sf \dfrac{cr {}^{2} }{r {}^{2} - 1} ...(3)$

________________________

Now LHS

$= \sf \dfrac{xy}{z}$

put the values of (1),(2)&(3)

$\sf = \dfrac{ \frac{ar}{r - 1} \times \frac{br}{r + 1} }{ \frac{cr}{r {}^{2} - 1 } }$

$\sf = \dfrac{ar}{r - 1} \times \dfrac{br}{r + 1} \times \dfrac{r {}^{2} - 1 }{cr {}^{2} }$

$\sf = \dfrac{ab \times \cancel{r {}^{2} }}{ \cancel{r {}^{2} - 1} } \times \dfrac{ \cancel{r {}^{2} - 1}}{c \times \cancel{r {}^{2}} }$

$\sf = \dfrac{ab}{c}$

= RHS

⇒LHS = RHS

Hence proved !