# 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