Question 10 and 11, Exercise 4.4
Solutions of Question 10 and 11 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Question 10
Find the next two terms of the geometric sequence: $$20,30,45 \ldots$$
Solution.
Given sequence is geometric with \(a_1=20\) and \(r=\frac{30}{20}=\frac{3}{2}\).
General term of the geometric series is given as
$$a_{n}=a_{1} r^{n-1}.$$
Thus
\begin{align*}
& a_{4}=a_{1} r^3=(20)\left(\frac{3}{2}\right)^3=20 \times \frac{27}{8} = \frac{540}{8} = 67.5 \\
& a_{5}=a_{1} r^4=(20)\left(\frac{3}{2}\right)^4=20 \times \frac{81}{16} = \frac{1620}{16} = 101.25
\end{align*}
Hence $a_4=67.5$, $a_5=101.25$.
Question 11
Find the next two terms of the geometric sequence: $$729,243,81,\ldots$$
Solution.
Given sequence is geometric with \(a_1=729\) and \(r=\frac{243}{729}=\frac{1}{3}\).
General term of the geometric series is given as
$a_{n}=a_{1} r^{n-1}.$
Thus
\begin{align*}
& a_{4}=a_{1} r^3=(729)\left(\frac{1}{3}\right)^3=729 \times \frac{1}{27} = 27 \\
& a_{5}=a_{1} r^4=(729)\left(\frac{1}{3}\right)^4=729 \times \frac{1}{81} = 9
\end{align*}
Hence $a_4=27$, $a_5=9$.
Go to