====== 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.

=====Question 5=====
Find $r$ such that $S_{10}=244 S_5$.
====Solution====
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}


=====Question 6=====
Prove that: $S_n(S_{3 n}-S_{2 n})=(S_n-S_{2 n})^2$
====Solution====
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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