Question 6 & 7 Exercise 4.4

Solutions of Question 6 & 7 of Exercise 4.4 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 $a_{10}=l, a_{13}=m$ and $a_{16}=n;\quad$ show that $\ln =m^2$

We know that $a_n=a_1 r^{n-1}$ therefore
$$\begin{align}a_{10}&=a_1 r^9=l \\ a_{13}&=a_1 r^{12}=m\\ \text{and} \quad a_{16}&=a_1 e^{\mathbf{A 5}}=n\end{align}$$ Multiplying (i) and (iii), we get
$$\begin{align}a_{10} \cdot a_{16}&=\ln =(a_1 r^9)(a_1 r^{15})\\ \Rightarrow \quad \ln& =a_1^2 r^{9+15^{\circ}} \\ \Rightarrow \quad \ln &=a_1^2 r^{24} \\ \Rightarrow \quad \ln &=(a_1 r^{12})^2 \\ \Rightarrow \quad \ln &=m^2 \because m=a_1 r^{12} \text { by (ii) }\end{align}$$ Hence showed that $\ln =m^2$.

Show that the reciprocal of the terms of a geometric sequence also form a geometric sequence.

Let we are considering the standard geometric sequence with common ratio $r$ that is
$$a_1, a_1 r, a_1 r^2, a_1 r^3, \ldots, a_1 r^{n-1},$$ Taking reciprocal of the terms, we get $\dfrac{1}{a_1}, \dfrac{1}{a_1 r}, \dfrac{1}{a_1 r^2}, \ldots, \dfrac{1}{a_1 r^{n-1}}$,
General term of the (ii) sequence is:
$$\begin{align}a_n&=\dfrac{1}{a_1 r^{n-1}}\\ a_{n+1}&=\dfrac{1}{a_1 r^n}\\ \text{Now we check}\\ \dfrac{a_{n+1}}{a_n}&=\dfrac{\dfrac{1}{a_1 r^n}}{\dfrac{1}{a_1 r^{n-1}}}\\ \Rightarrow \dfrac{a_{n+1}}{a_n}&=\dfrac{1}{a_1 r^n} \cdot a_1 r^{n-1}\\ \Rightarrow \dfrac{a_{n+1}}{a_n}&=\dfrac{r^n}{r^n} \dfrac{1}{r}=\dfrac{1}{r}\end{align}$$ We see that $\dfrac{a_{n+1}}{a_n}$ is independent of $n$,
which implies that sequence in (ii) is also geometric sequence with common ratio $\dfrac{1}{r}$.
Thus the reciprocal of geometric sequence is also a geometric sequence.

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