Help pls (1 Viewer)

Epicman69 · Nov 18, 2020

pls help with this question

Trebla · Nov 18, 2020

For b), find the sum of n terms of a GP with first term a and common ratio r. Do the same but with first term a and common ratio r². Take the ratio of the sums and factor the differences of two squares and it should simplify to the result.

Epicman69 · Nov 18, 2020

Trebla said:
For b), find the sum of n terms of a GP with first term a and common ratio r. Do the same but with first term a and common ratio r². Take the ratio of the sums and factor the differences of two squares and it should simplify to the result.

I did that but I got r^n + 1 : 1

Trebla · Nov 18, 2020

Epicman69 said:
I did that but I got r^n + 1 : 1

Double check your denominator terms

CM_Tutor · Nov 19, 2020

Epicman69 said:
View attachment 29541

pls help with this question

For (b): Let the common first term of the series be a.

$S_n = a + ar + ar^2 + ... + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$

$\Sigma_n = a + ar^2 + ar^4 + ... + ar^{2n-1} = \frac{a(r^{2n} - 1)}{r^2 - 1}$

$\begin{align*} \frac{\Sigma_n}{S_n} &= \frac{a(r^{2n} - 1)}{r^2 - 1} \div \frac{a(r^n - 1)}{r - 1} \\ &= \frac{a\left[(r^n)^2 - 1\right]}{r^2 - 1} \times \frac{r - 1}{a(r^n - 1)} \\ &= \frac{a(r^n + 1)(r^n - 1)(r - 1)}{a(r + 1)(r - 1)(r^n - 1)} \\ &= \frac{r^n + 1}{r + 1} \\ \Sigma_n : S_n &= (r^n + 1) : (r + 1) \end{align*}$

CM_Tutor · Nov 19, 2020

Epicman69 said:
View attachment 29541

pls help with this question

For part (c): Let the GP have a first term a.

$S_n = T_1 + T_2 + T_3 + ... + T_n = a + ar + ar^2 + ... ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$

$\begin{align*} R_n &= T_{n+1} + T_{n+2} + T_{n+3} + ... + T_{2n} \\ &= (T_1 + T_2 + T_3 + ... + T_n) + T_{n+1} + T_{n+2} + T_{n+3} + ... + T_{2n} - (T_1 + T_2 + T_3 + ... + T_n) \\ &= S_{2n} - S_n \\ &= \frac{a(r^{2n} - 1)}{r - 1} - \frac{a(r^n - 1)}{r - 1} \\ &= \frac{a[(r^n)^2 - 1] - a(r^n - 1)}{r - 1} \\ &= \frac{a(r^n - 1)(r^n + 1 - 1)}{r - 1} \\ &= \frac{ar^n(r^n - 1)}{r - 1} \\ \\ \frac{R_n}{S_n} &= \frac{ar^n(r^n - 1)}{r - 1} \div \frac{a(r^n - 1)}{r - 1} \\ &= \frac{ar^n(r^n - 1)}{r - 1} \times \frac{r - 1}{a(r^n - 1)} \\ &= \frac{ar^n(r^n - 1)(r - 1)}{a(r^n - 1)(r - 1)} \\ &= r^n \\ R_n : S_n &= r^n : 1 \\ \\ \text{If $R_8 : S_8 = 1 : 81$, then } \frac{r^8}{1} &= \frac{1}{81} \\ &= \frac{1}{3^4} \\ \implies r &= \pm \frac{1}{\sqrt{3}} \end{align*}$

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