Question

Proof of sum of infinite terms of AGP​

Answer 1

Answer:

We are now ready to state the sum of an infinite AGP, and will present the proof below: The sum of infinite terms of an AGP is given by S ∞ = a 1 − r + d r ( 1 − r ) 2 S_{\infty}=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2} S∞=1−ra+(1−r)2dr , where ∣ r ∣ < 1 |r|<1 ∣r∣<1.

Answer 2

For Any General Arithmetico Geometric Series:

$\sf ab,(a + d)(br),(a + 2d)(b {r}^{2} ),...,(a + (n - 1)d)(b {r}^{n - 1} )$

The Sum of first n Terms of this AGP is given by:

$\sf S_{n} = \frac{ab}{1 - r} + \frac{dbr(1 - {r}^{n - 1}) }{(1 - r) ^{2} } - \frac{(a + (n - 1)d)b {r}^{n} }{1 - r}$

Now, If we take the limit on both the sides as n approaches to infinity.

$\\ \sf\lim _{n\rightarrow + \infty }(S _{n}) = \small{\frac{ab}{1 - r} + \frac{dbr}{ {(1 - r)}^{2} } } \lim_{n \rightarrow + \infty }(1 - {r}^{n - 1} ) - \frac{b(a + (n - 1)d)}{ 1 - r} \lim_{n \rightarrow + \infty }( {r}^{n} )$

Now, if

$|r| < 1$

$\\ \implies \sf \boxed{\lim _{n\rightarrow + \infty }(S _{n}) = \frac{ab}{1 - r} + \frac{dbr}{ {(1 - r)}^{2} }}$