Exercise 5.2-Additional Questions - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions Solved
Question 1.
Find the first five terms of the sequence given by $\left\{\begin{array}{c}a_1=1 \\ a_n=a_{n-1}+2, n \geq 2\end{array}\right.$.
Solution:
Here $a_1=1$
Substituting $\mathrm{n}=2$, we obtain $\mathrm{a}_2=\mathrm{a}_1+2=1+2=3$
Substituting $\mathrm{n}=3,4$ and 5 , we obtain respectively
$
\begin{aligned}
& a_3=a_2+2=3+2=5, a_4=a_3+2=5+2=7 \\
& a_5=a_4+2=7+2=9
\end{aligned}
$
Thus, the first five terms are $1,3,5,7$ and 9 .
Question 2.
Find the $18^{\text {th }}$ and $25^{\text {th }}$ term of the sequence defined by
$
a_n= \begin{cases}n(n+2), & \text { if } n \text { is even natural number } \\ \frac{4 n}{n^2+1}, & \text { if } n \text { is odd natural number }\end{cases}
$
Solution:
When $\mathrm{n}=18$ (even)
$
\mathrm{a}_{\mathrm{n}}=\mathrm{n}(\mathrm{n}+2)=18(18+2)=18(20)=360
$

When $\mathrm{n}=25$ (odd)
$
a_n=\frac{4 n}{n^2+1}=\frac{4(25)}{(25)^2+1}=\frac{100}{625+1}=\frac{100}{626}=\frac{50}{313}
$
Question 3.
Write the first six terms of the sequences given by
(i) $a_1=a_2=1=a_{n-1}+a_{n-2}(n \geq 3)$
(ii) $\mathrm{a}_1=4, \mathrm{a}_{\mathrm{n}+1}=2 n \mathrm{a}_{\mathrm{n}}$
Solution:
(i) Here $\mathrm{a}_1=\mathrm{a}_2=1=\mathrm{a}_{\mathrm{n}-1}+\mathrm{a}_{\mathrm{n}-2}(\mathrm{n} \geq 3)$
Putting $n=3, \mathrm{a}_3=\mathrm{a}_2+\mathrm{a}_1=1+1=2$
Putting $n=4, a_4=a_3+a_2=2+1=3$
Putting $n=5, a_5=a_4+a_2=3+2=5$
Putting $n=6, a_6=a_5+a_4=5+3=8$
$\therefore$ First six terms of the sequence are $1,1,2,3,5,8$
(ii) Here $a_1=4$ and $a_{n+1}=2 n a_n$
Putting $\mathrm{n}=1, \mathrm{a}_2=2 \times 1 \times \mathrm{a}_1=2 \times 1 \times 4=8$
Putting $\mathrm{n}=2, \mathrm{a}_3=2 \times 2 \times \mathrm{a}_2=4 \times 8=32$
Putting $n=3, a_4=8 \times 192=1536$
Putting $\mathrm{n}=4, \mathrm{a}_5=2 \times 4 \times \mathrm{a}_4=8 \times 192=1536$
Putting $\mathrm{n}=5, \mathrm{a}_6=2 \times 5 \times \mathrm{a}_5=10 \times 1536=15360$
Question 4.
An A.P. consists of 21 terms. The sum of the three terms in the middle is 129 and of the last three is 237. Find the series.
Solution:
Let $\mathrm{a}_1$ be the first term and $\mathrm{d}$, be the common difference. Here $\mathrm{n}=21$.
$\therefore$ The three middle terms are $\mathrm{a}_{10}, \mathrm{a}_{11}, \mathrm{a}_{12}$
Now, $\mathrm{a}_{10}+\mathrm{a}_{11}+\mathrm{a}_{12}=129[$ Given $]$
$
\begin{aligned}
& \therefore\left(a_1+9 d\right)+\left(a_1+10 d\right)+\left(a_1+11 d\right)=129 \\
& \Rightarrow 3 a_1+30 d=129 \Rightarrow a_1+10 d=43 \ldots \ldots \text { (i) }
\end{aligned}
$
The last three terms are $a_{19}, a_{20}, a_{21}$
$
\begin{aligned}
& \mathrm{a}_{19}, \mathrm{a}_{20}, \mathrm{a}_{21}=237 \text { [Given] } \\
& \therefore\left(\mathrm{a}_1+18 \mathrm{~d}\right)+\left(\mathrm{a}_1+19 \mathrm{~d}\right)+\left(\mathrm{a}_1+20 \mathrm{~d}\right)=237 \\
& \text { (i.e.,) } 3 \mathrm{a}_1+57 \mathrm{~d}=237 \Rightarrow \mathrm{a}_1+19 \mathrm{~d}=79 \ldots \text { (2) }
\end{aligned}
$
Subtracting (i) from (ii), we get $9 \mathrm{~d}=36, \Rightarrow d=4$
$\therefore$ From (i), $\mathrm{a}_1+40=43 \Rightarrow \mathrm{a}_1=3$
Hence, the series is $3,7,11,15 \ldots \ldots$

Question 5.
Prove that the product of the $2^{\text {nd }}$ and $3^{\text {rd }}$ terms of an arithmetic progression exceeds the product of the

first and fourth by twice the square of the difference between the $1^{\text {st }}$ and $2^{\text {nd }}$.
Solution:
Let ' $a$ ' be the first term and 'd' be the common difference of A.P.
Then, $a_1=a, a_2=a+(2-1) d=a+d$
$a_3=a+(3-1) d=a+2 d, a_4=a+(4-1) d=a+3 d$
We have to show that $a_2 \cdot a_3-a_1 \cdot a_4=2\left(a_2-a_1\right)^2$
$
\begin{aligned}
& \text { LHS }=a_2 \cdot a_3-a_1 \cdot a_4=(a+d)(a+2 d)-a(a+3 d) \\
& =a_2+3 a d+2 d_2-a_2-3 a d=2 d_2 \\
& \text { RHS }=2\left(a_2-a_1\right)^2=2 d_2
\end{aligned}
$
Since LHS = RHS. Hence proved.
Question 6.
If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A.P. are $a, b, c$ respectively, prove that $a(q-r)+b(r-p)+c(p-q)=0$.
Solution:
Let $\mathrm{A}$ be the first term and $\mathrm{D}$ be the common difference of A.P.
$
\begin{aligned}
& \mathrm{a}_{\mathrm{p}}=\mathrm{a}, \therefore \mathrm{A}+(\mathrm{p}-1) \mathrm{D}=\mathrm{a} \ldots \ldots(1) \\
& \mathrm{a}_{\mathrm{q}}=\mathrm{b}, \therefore \mathrm{A}+(\mathrm{q}-1) \mathrm{D}=\mathrm{b} \ldots \ldots(2) \\
& \mathrm{a}_{\mathrm{r}}=\mathrm{c}, \therefore \mathrm{A}+(\mathrm{r}-1) \mathrm{D}=\mathrm{c} \ldots \ldots(3) \\
& \therefore \mathrm{a}(\mathrm{q}-\mathrm{r})+\mathrm{b}(\mathrm{r}-\mathrm{p})+\mathrm{c}(\mathrm{p}-\mathrm{q})=[\mathrm{A}+(\mathrm{p}-1) \mathrm{D}](\mathrm{q}-\mathrm{r})+[\mathrm{A}+(\mathrm{q}-1) \mathrm{D}] \\
& (\mathrm{r}-\mathrm{p})+[\mathrm{A}+(\mathrm{r}-1) \mathrm{D}](\mathrm{p}-\mathrm{q})[\mathrm{U} \operatorname{sing}(1),(2) \text { and (3) }] \\
& =(\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q}) \mathrm{A}+[(\mathrm{p}-1)(\mathrm{q}-\mathrm{r})+(\mathrm{q}-1)(\mathrm{r}-\mathrm{p})+(\mathrm{r}-1)(\mathrm{p}-\mathrm{q})] \mathrm{D} \\
& =(0) \mathrm{A}+(\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{pq}-\mathrm{r}+\mathrm{p}+\mathrm{pr}-\mathrm{p}-\mathrm{q}+\mathrm{q}) \mathrm{D} \\
& =(0) \mathrm{A}+(0) \mathrm{D}=0
\end{aligned}
$

Question 7.
If $a, b, c$ are in A.P. and $p$ is the A.M. between a and $b$ and $q$ is the A.M. between $b$ and $c$, show that $b$ is the A.M. between $\mathrm{p}$ and $\mathrm{q}$.
Solution:
a, b, c are in A.P.
$2 \mathrm{~b}=\mathrm{a}+\mathrm{c} \ldots \ldots(1)$
$\mathrm{p}$ is the A.M. between a and $\mathrm{b}$
$
p=\frac{a+b}{2}
$
$\mathrm{q}$ is the A.M. between $\mathrm{b}$ and $\mathrm{c}$
$
q=\frac{b+c}{2}
$
Adding (2) and (3), we get $p+q=\frac{a+b}{2}+\frac{b+c}{2}=\frac{a+c+2 b}{2}=\frac{2 b+2 b}{2}=2 b$ [Using 1] $2 b=p+q$ or $b=\frac{p+q}{2}$
Hence, $b$ is the A.M. between $p$ and $\mathrm{q}$

Question 8 .
If $x, y, z$ be respectively the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a G.P. show that $x^{q-r}, y^{r-p}, z^{p-q}=1$ Solution:
Let $A$ be the first term and $\mathrm{R}$ be the common ratio of $G . P$.
$
\begin{aligned}
& a_{\mathrm{p}}=\mathrm{x} \Rightarrow \mathrm{x}=A R^{\mathrm{p}-1} \\
& \mathrm{a}_{\mathrm{q}}=\mathrm{y} \Rightarrow \mathrm{x}=A R^{\mathrm{q}-1} \\
& \mathrm{a}_{\mathrm{r}}=\mathrm{z} \Rightarrow \mathrm{x}=A R^{\mathrm{r}-1}
\end{aligned}
$
Raising (1), (2), (3) to the powers $q-r, r-p, p-q$ respectively, we get
$
\begin{aligned}
& x^{q-r}=\mathrm{A}^{q-r}, \mathrm{R}^{(q-r)(p-1)} \\
& y^{r-p}=\mathrm{A}^{r-p}, \mathrm{R}^{(r-p)(q-1)} \\
& z^{p-q}=\mathrm{A}^{p-q}, \mathrm{R}^{(p-q)(r-1)}
\end{aligned}
$
Multiplying (4), (5) and (6), we get
$
x^{q-r}, y^{r-p}, z^{p-q}=\mathrm{A}^{q-r+r-p+p-q}, \mathrm{R}^{(q-r)(p-1)+(r-p)(q-1)+(p-q)(r-1)}=\mathrm{A}^0 \times \mathrm{R}^0=1 \times 1=1
$
Hence, $x^{q-r}, y^{r-p}, z^{p-q}=1$