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`        find the sum of n terms in the arhtematic geometric series 4*7+7*7^2+10*7^3+13*7^4+.......`
one year ago

```							To get the sum of the first N  terms of an AGP, we need to find the value ofS= 4.7+7.7^2+10.7^3.........+{4.7+(n-1)3}7^(n-1)Now let`s multiply S by `r=7`, then we get7S or (Sr) = 4.7^2+7.7^3+10.7^4....Subtracting 7S from S , we getS = 4.7+7.7^2+10.7^3.......+{4.7+(n-1)3}7^(n-1)7S =      4.7^2+7.7^3+.......+{4.7+(n-1)3}7^n-6S = 4.7+ 3×7^2 + 3×7^3 +....+3.7^(n-1) - {4.7+(n-1)3}7^nIf we were to exclude the first term and the last term, then the rest is a sum of geometric progression with first term 3.7 and common ratio 7 for (N-1) terms. Thus-6S  = 1st Term + (sum of GP ) + Last Term-6S = 4.7 + [3.7 {1-7^(n-1)}]/(1-7) - {4.7+(n-1)3}7^nS= [{4.7 - [ 4.7+(n-1)3}7^n]}/-6] + { [3.7 {1-7^(n-1)}]/36}Now... For n=1.   S = 28.For n=2.   S = 371...
```
one year ago
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