Question

Find the sum of GP with steps and i will mark it brainliest..please do it fast.​

Answer 1

✦ A.P. Series:-

If in an A.P. series "a" be the first term and "d" be the common difference then ,

(1) The n'th term is given by the formula .

$\sf\longrightarrow\boxed{ a_n=a+(n-1)d }$

(2)Sum of n number of terms ,

$\sf\longrightarrow\boxed{ S_n=\frac{n[2a+(n-1)d]}{2} }$

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✦ G.P. Series:-

If in an G.P. series "a" be the first term and "r" be the common ratio then ,

(1) The n'th term is given by the formula .

$\sf\longrightarrow\boxed{ a_n=at{}^{n-1} }$

(2)Sum of n number of terms ,

$\sf\longrightarrow\boxed{ S_n=\frac{n[r{}^{n}-1]}{r-1}}$

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❏ Question:-

Find the sum of GP,

$\sf\longrightarrow \frac{x+y}{x-y}+1+\frac{x-y}{x+y}+........+upto\: n\: terms$

❏ Solution:-

•first term(a)$\sf\longrightarrow\frac{x+y}{x-y}$

• common difference (d)=$\sf\longrightarrow\frac{1}{\frac{x+y}{x-y}}=\frac{x-y}{x+y}$

$\bf\therefore$ Sum of n terms is ,

$\bf\sf\longrightarrow S_n=\large{\frac{a(r^{n}-1)}{r-1}}$

$\sf\longrightarrow\bf S_n=\frac{\frac{x+y}{x-y}[(\frac{x-y}{x+y})^{n}-1]}{\frac{x-y}{x+y}-1}$

$\bf\sf\longrightarrow\bf S_n=\frac{\frac{x+y}{x-y}[\frac{x-y}{x+y})^{n}-1]}{\frac{\cancel x-y-\cancel x-y}{x+y}}$

$\bf\sf\longrightarrow\bf S_n=\frac{\frac{x+y}{x-y}[(\frac{x-y}{x+y})^{n}-1]}{\frac{-2y}{x+y}}$

$\bf\sf\longrightarrow\bf S_n=\frac{(x+y)(x+y)}{-2y(x-y)}\times[(\frac{x-y}{x+y})^{n}-1]$

$\sf\bf \longrightarrow\bf \boxed{\red{S_n=\frac{(x+y)^{2}}{2y(x-y)}\times[1-(\frac{x-y}{x+y})^{n}]}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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