Question 11 & 12 Exercise 4.5

Solutions of Question 11 & 12 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.

If $p^{t h}, q^{t h}$ and $r^{t h}$ terms of G.P be $a, b, c$ respectively. Prove that $a^{q-r} b^{r-p} c^{p-q}=1$.

The general term of G.P $a_n=a_1 r^{n-1}$.
Therefore, $a_p=a_1 r^{p-1}=a \quad a_q=a_1 r^{q-1}=b$ and $a_r=a_1 r^{r-1}$. Then
$$\begin{align}a^{q-r}&=(a_1 r^{p-1})^{q-r} . \\ b^{r-p}&=(a_1 r^{q-1})^{r-p}, \text { and } \\ c^{p-q}&=(a_1 r^{r-1})^{p-q}.\end{align}$$
Multiplying the above three equations
$$\begin{align}a^{q-r} b^{r-p} c^{p-q}& =(a_1 r^{p-1})^{q-r} \cdot(a_1 r^{q-1} r^{r-p} \cdot(a_1 r^{r-1})^{p-q}. \\ & =a_1^{q-r+r-p+p-q}\times r^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \\ & =a_1^0 \cdot r^{p q-p r-q+r+q r-p q r+p+p r-q r-p+q} \\ & =1 \cdot r^0=1 . \text { Thus } \\ a^{q-r} b^{r-p} c^{p-q}&=1.\end{align}$$

Find an infinite geometric series whose sum is 6 and such that each term is four times the sum of all the terms that follow it.

Let considering the geometric series with common ratio $r$ and first term $a_1$
$$a_1+a_1 r+a_1 r^2+a_1 r^3+....(i)$$
Then $S_{\infty}=\dfrac{a_1}{1-r}$, but we are given $S_{\infty}=6$. Therefore,
$$\dfrac{a_1}{1-r}=6 \text { or } 6 a_1=1-r ...(ii)$$
Also we are given that each term is four times the sum of all the terms following it, therefore
$$\begin{align}a_1&=4(a_1 r+a_1 r^2+a_1 r^3+\cdots) \\ \Rightarrow a_1&=4 a_1(r+r ^2+r^3+\ldots) \\ \Rightarrow \dfrac{1}{4}&=\dfrac{r}{1-r} \quad \because r+r^2+r^3+\ldots=\dfrac{r}{1-r} \\ \Rightarrow 4 r&=1-r\\ \Rightarrow 5 r&=1\\ r&=\dfrac{1}{5}\end{align}$$
putting in (ii), we get
$$\begin{align}6 a_1&=1-\dfrac{1}{5}=\dfrac{4}{5}\\ \Rightarrow a_1&=\dfrac{4}{5 \times 6}=\dfrac{2}{15} \text {. }\end{align}$$
Thus the infinite gcometric serics is:
$$\begin{align}a_1+a_1 r+a_1 r^2+a_1 r^3+\ldots & =\dfrac{2}{15}+\dfrac{2}{15}(\dfrac{1}{5})+\dfrac{2}{15}(\dfrac{1}{5})^2+\ldots \\ & =\dfrac{2}{15}+\dfrac{2}{45}+\dfrac{2}{375}+\ldots\end{align}$$

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