Question 5 & 6 Exercise 4.5

Solutions of Question 5 & 6 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find $r$ such that $S_{10}=244 S_5$.

We know that $$S_n=\dfrac{a_1(r^n-1)}{r-1}$$
then
$$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$
Putting both the $S_{10}$ and $S_S$ in the given equation, we get
\begin{align}\dfrac{a_1(r^{10}-1)}{r-1}&=244 \dfrac{a_1(r^5-1)}{r-1} \\ \Rightarrow r^{10}-1&=244(r^5-1) \\ \Rightarrow r^{10}-244 r^5 \cdots 1+244&=0 \\ \Rightarrow r^{10}-244 r^5+243&=0 \\ \Rightarrow r^{10}-r^5-243 r^5+243&=0 \\ \Rightarrow r^5(r^5-1)-243(r^5-1)&=0 \\ \Rightarrow(r^5-1)(r^5-243)&=0 \\ \Rightarrow \text { Either } r^5-10&=0 \text { or } r^5-243=0 \\ \Rightarrow r^5&=1 \text { or } r^5=3^5 \\ \Rightarrow r&=1 \text { or } r=3 . \\ \text { But } r^5 \neq 1 \text {, thus } r=3.\end{align}

Prove that: $S_n(S_{3 n}-S_{2 n})=(S_n-S_{2 n})^2$

We know that:
$$S_n=\dfrac{a_1(r^n-1)}{r-1}....(i)$$
Replacing $n$ by $2 n$, and $3 n$ in the above, then we get
\begin{align}S_{2 n}&=\dfrac{a_1(r^{2 n}-1)}{r-1} \ldots . . . (ii)\\ \text { and } S_{3 n}&=\dfrac{a_1(r^{3 n}-1)}{r-1}...(iii)\end{align}
Puting (i),(ii) and (iii) in L.H.S of the given, we get \begin{align}S_n(S_{3 n}-S_{2 n})&=\dfrac{a_1(r^n-1)}{r-1}[\dfrac{a_1(r^{3 n}-1)}{r-1}-\dfrac{a_1(r^{2 n}-1)}{r-1}]\\ & =\dfrac{a_1^2(r^n-1)}{(r-1)^2}[r^{3 n-1}-1-(r^{2 n}-1)]\\ & =\dfrac{a_1^2(r^n-1)}{(r-1)^2}[r^{3 n}-1-r^{2 n}+1] \\ & =\dfrac{a_1^2(r^n-1)}{(r-1)^2}[r^{2 n}r^n-r^{2 n}]. \\ & =\dfrac{a_1^2 r^{2 n}(r^n-1)}{(r-1)^2}(r^n-1) \\ \Rightarrow S_n(S_{3 n}-S_{2 n})&=[\dfrac{a_1 r^n(r^n-1)}{r-1}]^2...(1)\end{align} Now taking R.H.S, then
\begin{align}(S_n-S_{2 n})^2&=[\dfrac{a_1(r^n-1)}{r-1}-\dfrac{a_1(r^{2 n}-1)}{r-1}]^2 \\ & =[\dfrac{a_1 r^n-a_1-a_1 r^{2 n}+a_1}{r-1}]^2 \\ & =[\dfrac{a_1(r^n-r^n r^n)}{r-1}]^2 . \\ & =[\dfrac{a_1 r^n(1-r^n)}{r-1}]^2 \\ \Rightarrow(S_n-S_{2 n})^2&=[\dfrac{a_1 r^n(r^n-1)}{r-1}]^2 \ldots .(2)\end{align} From (i) and (ii), we see that
$$S_n(S_{3 n}-S_{2 n})=(S_n-S_{2 n})^2.$$

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