Question

Please anyone solve this problem.​

Answer 1

Question :–

▪︎ In A.P. $\: { \bold{x \th \: \: term \: \: T_{x} = \dfrac{1}{y} }} \:$ and $\: { \bold{y \th \: \: term \: \: T_{y} = \dfrac{1}{x} }} \:$ then Prove that –

$\: { \bold{ S_{xy} = \dfrac{1}{2}(xy + 1) }} \:$

ANSWER :–

• We know that nth term of A.P. is –

$\\ \longrightarrow \large { \red{ \boxed { \bold{ t_{n} = a + (n - 1)d }}}} \:$

• Here –

$\\ { \bold{ \: \: \: \: \: \: \: . \: \: \: a = first \: \: term}} \:$

$\\ { \bold{ \: \: \: \: \: \: \: . \: \: \: d = common \: \: difference}} \:$

$\\ { \bold{ \: \: \: \: \: \: \: . \: \: \: t_{n} = n \th \: \: term}} \:$

$\\ { \bold{ \: \: \: \: \: \: \: . \: \: \: n = total \: \: number \: \: of \: \: term}} \:$

• So that –

$\\ \implies { \bold{x \th \: \: term \: \: ( t_{x}) = a + (x - 1)d}} \:$

$\\ \implies { \bold{ \dfrac{1}{y} = a + (x - 1)d \: \: \: \: - - - - eq.(1)}} \:$

$\\ \implies { \bold{y \th \: \: term \: \: ( t_{y}) = a + (y - 1)d}} \:$

$\\ \implies { \bold{ \dfrac{1}{x} = a + (y - 1)d \: \: \: \: - - - - eq.(2)}} \:$

• Now subtract eq.(1) by eq.(2) –

$\\ \implies { \bold{ \dfrac{1}{y} - \dfrac{1}{x} = (a - a) + [(x - 1) - (y - 1)]d}} \:$

$\\ \implies { \bold{ \dfrac{1}{y} - \dfrac{1}{x} = (x - y)d}} \:$

$\\ \implies { \bold{ \dfrac{ \cancel{ (x - y) }}{xy} = \cancel{(x - y)}d}} \:$

$\\ \implies { \pink{ \bold{d = \dfrac{1}{xy} }}} \:$

• Put the value of 'd' in eq.(1) –

$\\ \implies { \bold{ \dfrac{1}{y} = a + (x - 1)( \dfrac{1}{xy} )}} \:$

$\\ \implies { \bold{ \cancel \dfrac{1}{y} = a + \cancel \dfrac{1}{y} - \dfrac{1}{xy} }} \:$

$\\ \implies { \bold{ 0 = a - \dfrac{1}{xy} }} \:$

$\\ \implies { \pink{ \bold{ a = \dfrac{1}{xy} }}}\:$

• We know that sum of n terms of A.P. is –

$\\ \longrightarrow \large { \red{ \boxed { \bold{ S_{n} = \dfrac{n}{2} [ 2a + (n - 1)d ] }}}} \:$

• Now put the values –

$\\ \implies { \bold{ S_{xy} = \dfrac{xy}{2} [ 2( \dfrac{1}{xy}) + (xy - 1)( \dfrac{1}{xy} ) ] }} \:$

$\\ \implies { \bold{ S_{xy} = \dfrac{ \cancel{xy}}{2} \times \dfrac{1}{ \cancel{xy}} [ 2 + (xy - 1) ] }} \:$

$\\ \implies { \bold{ S_{xy} = \dfrac{ 1}{2} (2 + xy - 1) }} \:$

$\\ \implies { \pink{ \boxed{ \bold{ S_{xy} = \dfrac{ 1}{2} ( xy + 1) }}}} \:$

$\\ \longrightarrow \large { \boxed{ \boxed{ \bold{Hence \: \: proved}}}} \:$