Algebra> find the sum to n trms of the A.G.P 3+5/4...

find the sum to n trms of the A.G.P3+5/4+7/42+9/43+...

krishanu dev , 10 Years ago

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2 Answers

Aziz Alam

Last Activity: 10 Years ago

the sum of n terms of A.G.P.

$? [S_n = \sum_{k = 1}^n \left[a + (k - 1) d\right] r^{k - 1} = \frac{a - [a+(n - 1)d] r^n}{1 - r}+\frac{dr(1 - r^{n - 1})}{(1 - r)^2}.]$ $? [S_n = \sum_{k = 1}^n \left[a + (k - 1) d\right] r^{k - 1} = \frac{a - [a+(n - 1)d] r^n}{1 - r}+\frac{dr(1 - r^{n - 1})}{(1 - r)^2}.]$

To derive it call the sum of series be S_n. Then multiply it by “r” of the G.P. and then subtract S_n with rS_n

Ravi

Last Activity: 10 Years ago

Take the LCM of the denominator and keep it outside of the series. Now from the nth term of the series, simplify and get the 2 GP’s in addition. Solve for the sum of the GP to get the final answer.
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