Exercise 2.4 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.4$
Question $1 .$
Find the next three terms of the following sequence.
(i) $8,24,72, \ldots \ldots$
(ii) $5,1,-3, \ldots \ldots$
(iii) $\frac{1}{4}, \frac{2}{9}, \frac{3}{16}$
Solution:
(i) $8,24,72 \ldots$
In an arithmetic sequence $a=8$,
$\begin{aligned}
&\mathrm{d}=\mathrm{t}_{1}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2} \\
&=24-8 \quad 72-24 \\
&=16 \neq 48
\end{aligned}$
So, it is not an arithmetic sequence. In a geometric sequence,
$\begin{aligned}
&\mathrm{r}=\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}} \\
&\Rightarrow \frac{24}{8}=\frac{72}{24} \\
&\Rightarrow 3=3
\end{aligned}$
$\therefore$ It is a geometric sequence
$\therefore$ The $n^{\text {th }}$ term of a G.P is $t_{n}=a r^{n-1}$
$\begin{aligned}
&\therefore \mathrm{t}_{4}=8 \times 3^{4-1} \\
&=8 \times 3^{3} \\
&=8 \times 27 \\
&=216 \\
&\mathrm{t}_{5}=8 \times 3^{5-1} \\
&=8 \times 3^{4} \\
&=8 \times 81 \\
&=648
\end{aligned}$

$\begin{aligned}
&\mathrm{t}_{6}=8 \times 3^{6-1} \\
&=8 \times 3^{5} \\
&=8 \times 243 \\
&=1944
\end{aligned}$
The next 3 terms are $8,24,72,216,648,1944$.
(ii) $5,1,-3, \ldots$
$\begin{aligned}
&\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2} \\
&\Rightarrow 1-5=-3-1 \\
&-4=-4 \therefore \text { It is an A.P. } \\
&\mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \\
&\mathrm{t}_{4}=5+3 \times-4 \\
&=5-12 \\
&=-7 \\
&15-\mathrm{a}+4 \mathrm{~d} \\
&=5+4 \times-4 \\
&=5-16 \\
&=-11 \\
&\mathrm{t}_{6}=\mathrm{a}+5 \mathrm{~d} \\
&=5+5 \times-4 \\
&=5-20 \\
&=-15
\end{aligned}$
$\therefore$ The next three terms are $5,1,-3$, 그, $\underline{-11}, \underline{-15}$
(iii) $\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \ldots \ldots \ldots .$
Here $\mathrm{a}_{\mathrm{n}}=$ Numerators are natural numbers and denominators are squares of the next numbers $\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \frac{4}{25}, \frac{5}{36}, \frac{6}{49} \ldots \ldots \ldots \ldots$

Question 2.
Find the first four terms of the sequences whose $\mathrm{n}^{\text {th }}$ terms are given by
(i) $a_{n}=n^{3}-2$
Answer:
$\begin{aligned}
&a_{n}=n^{3}-2 \\
&a_{1}=1^{3}-2=1-2=-1 \\
&a_{2}=2^{3}-2=8-2=6 \\
&a_{3}=3^{3}-2=27-2=25
\end{aligned}$
$\mathrm{a}_{4}=4^{3}-2=64-2=62$
The four terms are $-1,6,25$ and 62
(ii) $\mathrm{a}_{\mathrm{n}}=(-1)^{\mathrm{n}+1} \mathrm{n}(\mathrm{n}+1)$
Answer:
$\begin{aligned}
&\mathrm{a}_{\mathrm{n}}=(-1)^{\mathrm{n}+1} \mathrm{n}(\mathrm{n}+1) \\
&\mathrm{a}_{1}=(-1)^{2}(1)(2)=1 \times 1 \times 2=2 \\
&\mathrm{a}_{2}=(-1)^{3}(2)(3)=-1 \times 2 \times 3=-6 \\
&\mathrm{a}_{3}=(-1)^{4}(3)(4)=1 \times 3 \times 4=12 \\
&\mathrm{a}_{4}=(-1)^{5}(4)(5)=-1 \times 4 \times 5=-20
\end{aligned}$
The four terms are $2,-6,12$ and $-20$

(iii) $a_{n}=2 n^{2}-6$
Answer:
$\begin{aligned}
&a_{n}=2 n^{2}-6 \\
&a_{1}=2(1)^{2}-6=2-6=-4 \\
&a_{2}=2(2)^{2}-6=8-6=2 \\
&a_{3}=2(3)^{2}-6=18-6=12 \\
&a_{4}=2(4)^{2}-6=32-6=26
\end{aligned}$
The four terms are $-4,2,12,26$
 

Question $3 .$
Find the $n^{\text {th }}$ term of the following sequences
(i) $2,5,10,17, \ldots \ldots \ldots$
(ii) $0, \frac{1}{2}, \frac{2}{3}, \ldots .$
(iii) $3,8,13,18, \ldots \ldots .$
Solution:
(i) $2,5,10,17$
$=1^{2}+1,2^{2}+1,3^{2}+1,4^{2}+1 \ldots \ldots \ldots .$
$\therefore n^{\text {th }}$ term is $n^{2}+1$
(ii) $0, \frac{1}{2}, \frac{2}{3}, \ldots \ldots \ldots \ldots$
$=\frac{1-1}{1}, \frac{2-1}{2}, \frac{3-1}{3} \ldots .$
$\Rightarrow \frac{n-1}{n}$
$\therefore$ nth term is $\frac{n-1}{n}$
(iii) $3,8,13,18$

$a=3$ $d=5$ $t_{n}=a+(n-1) d$ $=3+(n-1) 5$ $=3+5 n-5$ $=5 n-2$ $\therefore n^{\text {th }}$ term is $5 n-2$
Question $4 .$
Find the indicated terms of the sequences whose $n^{\text {th }}$ terms are given by
(i) $\mathrm{a}_{\mathrm{n}}=\frac{5 n}{n+2} ; \mathrm{a}_{6}$ and $\mathrm{a}_{13}$
(ii) $a_{n}=-\left(n^{2}-4\right)$; $a_{4}$ and $a_{11}$
Solution:
(i)
$\begin{aligned}
a_{n} &=\frac{5 n}{n+2} \\
a_{13} &=\frac{5 \times 13}{13+2}=\frac{65^{13}}{15^{3}}=\frac{13}{3} \\
a_{6} &=\frac{5 \times 6}{6+2}=\frac{30^{15}}{8^{4}} \\
&=\frac{15}{4}
\end{aligned}$
(ii)
$\begin{aligned}
a_{n} &=-\left(n^{2}-4\right) ; a_{4} \text { and } a_{11} \\
a_{4} &=-\left(4^{2}-4\right) \\
&=-(16-4)=-12 \\
a_{11} &=-\left(11^{2}-4\right) \\
&=-(121-4)=-117
\end{aligned}$

Question $5 .$
Find $a_{8}$ and $a_{15}$ whose $n^{\text {th }}$ term is
$a_{n}=\left\{\begin{array}{l}\frac{n^{2}-1}{n+3} ; n \text { is even, } n \in \mathrm{N} \\ \frac{n^{2}}{2 n+1} ; n \text { is odd, } n \in \mathrm{N}\end{array}\right.$
Solution:
$\begin{aligned}
&a_{n}=\left\{\begin{array}{l}
\frac{n^{2}-1}{n+3}, n \text { is even } \\
\frac{n^{2}}{2 n+1}, n \text { is odd } \\
a_{8}=\frac{n^{2}-1}{n+3}=\frac{8^{2}-1}{8+3}=\frac{64-1}{11}=\frac{63}{11} \\
a_{15}=\frac{n^{2}}{2 n+1}=\frac{15^{2}}{2 \times 15+1}=\frac{225}{30+1}=\frac{225}{31}
\end{array}\right.
\end{aligned}$

Question $6 .$
If $a_{1}=1, a_{2}=1$ and $a_{n}=2 a_{n-1}+a_{n-2} n \geq 3, n \in N$. Then find the first six terms of the sequence.
Answer:
$a_{1}=a_{2}=1$
$\mathrm{a}_{\mathrm{n}}=2 \mathrm{a}_{\mathrm{nt}-1}+\mathrm{a}_{\mathrm{n}-2}$
$a_{3}=2 a_{3-1}+a_{3-2}=2 a_{2}+a_{1}$
$=2(1)+1=3$
$\mathrm{a}_{4}=2 \mathrm{a}_{4-1}+\mathrm{a}_{4-2}$
$=2 \mathrm{a}_{3}+\mathrm{a}_{2}$
$=2(3)+1=6+1=7$
$\mathrm{a}_{5}=2 \mathrm{a}_{5-1}+\mathrm{a}_{5-2}$
$=2 \mathrm{a}_{4}+\mathrm{a}_{3}$
$=2(7)+3=17$
$a_{6}=2 a_{6-1}+a_{6-2}$
$=2 \mathrm{a}_{5}+\mathrm{a}_{4}$
$=2(17)+7$
$=34+7=41$
The sequence is $1,1,3,7,17,41, \ldots$