Exercise 5.3 - Chapter 5 Binomial Theorem Sequences and Series 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 5.3
Question 1.
Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
Solution:
Let ' $a$ ' be the first term and $d$ be the common difference of A.P.
Let the A.P. be $a, a+d, a+2 d, \ldots .$. we are given $\mathrm{S}_{10}=52$ and $\mathrm{S}_{15}=77$.
(i.e.) $\frac{10}{2}[2 a+(10-1) d]=52$ and $\frac{15}{2}[2 a+(15-1) d]=77$
(i.e) $2 a+9 d=52 \times \frac{2}{10}=\frac{104}{10}=\frac{52}{5}$ and $2 a+14 d=\frac{77 \times 2}{15}=\frac{154}{15}$
$2 a+9 d=\frac{52}{5}$ $2 a+14 d=\frac{154}{15}$
(1) $-(2) \Rightarrow-5 d=\frac{52}{5}-\frac{154}{15}=\frac{156-154}{15}=\frac{2}{15}$ $d=\frac{-2}{15 \times 5}=-2 / 75$
Substituting $d=\frac{-2}{75}$ in (1) we get $2 a+9\left(\frac{-2}{75}\right)=\frac{52}{5}$
$
2 a=\frac{52}{5}+\frac{18}{75}=\frac{52}{5}+\frac{6}{25}=\frac{260+6}{25}=\frac{266}{25}
$
$
\Rightarrow a=\frac{266}{25 \times 2}=\frac{133}{25}
$

Now $a=\frac{133}{25}$ and $d=\frac{-2}{75}$
$
\begin{aligned}
\therefore S_{20} & =\frac{20}{2}[2 a+(20-1) d]=10[2 a+19 d]=10\left[2\left(\frac{133}{25}\right)+19\left(\frac{-2}{75}\right)\right] \\
& =10\left[\frac{266}{25}-\frac{38}{75}\right]=\frac{10}{75}[798-38]=\frac{10}{75}(760)=\frac{7600}{75}=\frac{304}{3}
\end{aligned}
$
Question 2.
Find the sum up to the $17^{\text {th }}$ term of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots$
Solution:
$
\begin{aligned}
& t_1=\frac{1^3}{1} ; t_2=\frac{1^3+2^3}{1+3} ; t_3=\frac{1^3+2^3+3^3}{1+3+5} \ldots \\
& \therefore t_n=\frac{1^3+2^3+3^3+\ldots n^3}{1+3+5+\ldots n \text { terms }}=\frac{\sum n^3}{n^2}\left(\frac{2 n-1+1}{2}\right)^2 \\
& \quad=\frac{n^2(n+1)^2}{4\left(n^2\right)}=\frac{(n+1)^2}{4}=\frac{n^2+2 n+1}{4} \\
& \therefore S_n=\sum t_n=\frac{1}{4} \sum n^2+2 n+1=\frac{1}{4} \sum n^2+\sum 2 n+\sum 1 \\
& =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{2(n)(n+1)}{2}+n\right]
\end{aligned}
$
To find $S_{17}$ put $n=17$

$
\begin{aligned}
S_{17} & =\frac{1}{4}\left[\frac{17 \times 18 \times 35}{6}+\frac{2(17)(18)}{2}+17\right] \\
& =\frac{1}{4}[17 \times 105+17 \times 18+17] \\
& =\frac{17(105+18+1)}{4}=17 \times 31=527
\end{aligned}
$
Question 3.
Compute the sum of first $n$ terms of the following series:
(i) $8+88+888+8888+\ldots \ldots$
(i) $6+66+666+6666+$
Solution:

(i) $S_n=8+88+888+\ldots n$ terms $=8[1+11+111+\ldots \ldots . n$ terms $]$
$
\begin{aligned}
& =\frac{8}{9}[9+99+999+\ldots . . n \text { terms }] \\
& =\frac{8}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\ldots . n \text { terms }\right] \\
& =\frac{8}{9}\left[10+10^2+\ldots . m \text { terms }-(1+1+1 \ldots . . n \text { terms })\right] \\
& =\frac{8}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]=\frac{8}{9}\left[\frac{10\left(10^n-1\right)}{9}-n\right] \\
& =\frac{80}{81}\left(10^n-1\right)-\frac{8 n}{9}
\end{aligned}
$
(ii) $S_n=6+66+666+\ldots . . n$ terms $=6[1+11+111+\ldots . . n$ terms $]$
$
=\frac{6}{9}[9+99+999+\ldots . . . n \text { terms }]
$
$
=\frac{6}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\ldots . . . n \text { terms }\right]
$
$=\frac{6}{9}\left[10+10^2+10^3+\ldots . . m\right.$ terms $-(1+1+1 \ldots . . n$ terms $\left.)\right]$
$
=\frac{6}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]=\frac{6}{9}\left(\frac{10^n-1}{9}-n\right)
$
$
=\frac{6}{9}\left[\frac{10\left(10^n-1\right)-9 n}{9}\right]=\frac{6}{81}\left[10\left(10^n-1\right)-9 n\right]
$
Question 4.
Compute the sum of first $n$ terms of $1+(1+4)+\left(1+4+4^2\right)+\left(1+4+4^2+4^3\right)+$

Solution:
$t_1=1, t_2=1+4, t_3=1+4+4^2, t_n=1+4+4^2+\ldots . n$ terms which is a G.P
$
\therefore t_n=\frac{1\left(4^n-1\right)}{4-1}=\frac{4^n-1}{3}
$
So $S_n=\sum t_n=\frac{\sum 4^n-1}{\sum 3}=\frac{\sum 4^n-\sum 1}{\sum 3}$
$\sum 4^n=4^1+4^2+\ldots \ldots n$ terms $=\frac{4\left(4^n-1\right)}{4-1}=\frac{4\left(4^n-1\right)}{3}$
So $\frac{\sum 4^n-\sum 1}{\sum 3}=\frac{\frac{4\left(4^n-1\right)}{3}-n}{3}=\frac{4\left(4^n-1\right)-3 n}{9}=\frac{4}{9}\left(4^n-1\right)-\frac{n}{3}$
Question 5.
Find the general term and sum to $\mathrm{n}$ terms of the sequence $1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \ldots \ldots \ldots$
Solution:
$t_1=1, t_2=\frac{4}{3}, t_3=\frac{7}{9}, t_4=\frac{10}{27}$
Numerator $1,4,7,10, \ldots \ldots$ (A.P $a=1, d=4-1=3$ )
So $t_n=a+(n-1) d=1+(n-1) 3=1+3 n-3=3 n-2$
Denominator $1,3,9,27$, which is a G.P. $a=1, r=3$
So $t_n=a r^{n-1}=1\left(3^{n-1}\right)=3^{n-1}$
$
\therefore t_n=\frac{3 n-2}{3^{n-1}}
$
It is an arithmetico Geometric series. Here the $\mathrm{n}^{\text {th }}$ term is $\mathrm{t}_{\mathrm{n}}=[\mathrm{a}+(\mathrm{n}-1) d] \mathrm{r}^{\mathrm{n}-1}$ where $\mathrm{a}=1, \mathrm{~d}=3$ and $\mathrm{r}$ $=1 / 3$

Now the sum to $\mathrm{n}$ terms is
$
\begin{aligned}
S_n & =\frac{a-[a+(n-1) d] r^n}{1-r}+d r\left(\frac{1-r^{n-1}}{(1-r)^2}\right) \\
& =\frac{1-[1+(n-1) 3] \frac{1}{3^n}}{1-\frac{1}{3}}+3 \times \frac{1}{3}\left(\frac{1-\frac{1}{3^{n-1}}}{(1-1 / 3)^2}\right) \\
& =\frac{1-(3 n-2) \frac{1}{3^n}}{2 / 3}+\frac{3^{n-1}-1}{3^{n-1}(2 / 3)^2}=\frac{3^n-(3 n-2)}{2\left(3^{n-1}\right)}+\frac{3^{n-1}-1}{4\left(3^{n-3}\right)}
\end{aligned}
$
Question 6.
Find the value of $\boldsymbol{n}$, if the sum to $n$ terms of the series $\sqrt{3}+\sqrt{75}+\sqrt{243}+\ldots$. is $435 \sqrt{3}$.
Solution:
$
t_1=\sqrt{3}, t_2=\sqrt{75}=\sqrt{25 \times 3}=5 \sqrt{3}, t_3=\sqrt{243}=\sqrt{81 \times 3}=9 \sqrt{3}
$
Here $t_1=\sqrt{3}, t_2=5 \sqrt{3}, t_3=9 \sqrt{3}$
(i.e) $a=\sqrt{3}, d=5 \sqrt{3}-\sqrt{3}=4 \sqrt{3}$
$
S_n=\frac{n}{2}[2 a+(n-1) d]=435 \sqrt{3} \text { (given) }
$
$
\begin{aligned}
& \Rightarrow \frac{n}{2}[2 \sqrt{3}+(n-1) 4 \sqrt{3}]=435 \sqrt{3} \\
& \Rightarrow \frac{n \times \sqrt{3}}{2}[2+4 n-4]=435 \times \sqrt{3} \\
& \Rightarrow n[4 n-2]=870
\end{aligned}
$
$
\begin{aligned}
& 4 n^2-2 n-870=0 \\
& (\div b y 2) 2 n^2-n-435=0 \\
& 2 n^2-30 n+29 n-435=0 \Rightarrow 2 n(n-15)+29(n-15)=0
\end{aligned}
$
$(2 n+29)(n-15)=0 \Rightarrow n=\frac{-29}{2}$ or 15

$n=\frac{-29}{2}$ not possible, So $n=15$
Question 7.
Show that the sum of $(m+n)^{\text {th }}$ and $(m-n)^{\text {th }}$ term of an A.P. is equal to twice the $m^{\text {th }}$ term.
Solution:
Let the A.P. be $a, a+d, a+2 d, \ldots \ldots$.
$
\begin{aligned}
& \mathrm{t}_{\mathrm{m}+\mathrm{n}}=\mathrm{a}+(\mathrm{m}+\mathrm{n}-1) \mathrm{d} \\
& \mathrm{t}_{\mathrm{m}-\mathrm{n}}=\mathrm{a}+(\mathrm{m}-\mathrm{n}-1) \mathrm{d} \\
& \mathrm{t}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d} \\
& 2 \mathrm{t}_{\mathrm{m}}=2[\mathrm{a}+(\mathrm{m}-1) \mathrm{d}] \\
& \text { To prove } \mathrm{t}_{\mathrm{m}+\mathrm{n}}+\mathrm{t}_{\mathrm{m}-\mathrm{n}}=2 \mathrm{t}_{\mathrm{m}} \\
& \text { LHS } \mathrm{t}_{\mathrm{m}+\mathrm{n}}+\mathrm{t}_{\mathrm{m}-\mathrm{n}}=\mathrm{a}+(\mathrm{m}+\mathrm{n}-1) \mathrm{d}+\mathrm{a}+(\mathrm{m}-\mathrm{n}-1) \mathrm{d} \\
& =2 a+d[m+h-1+m-h-1] \\
& =2 \mathrm{a}+\mathrm{d}[2 \mathrm{~m}-2] \\
& =2[\mathrm{a}+(\mathrm{m}-1) \mathrm{d})=2 \mathrm{t}_{\mathrm{m}}=\text { RHS }
\end{aligned}
$
Question 8.
A man repays an amount of ₹ 3250 by paying ₹ 20 in the first month and then increases the payment by ₹ 15 per month. How long will it take him to clear the amount?
Solution:
$\mathrm{a}=20, \mathrm{~d}=15, \mathrm{~S}_{\mathrm{n}}=3250$ to find $\mathrm{n}$.
$
\begin{aligned}
& \text { Now } S_n=\frac{n}{2}[2 a+(n-1) d]=3250 \\
& \Rightarrow \frac{n}{2}[40+(n-1) 15]=3250 \\
& n[40+15 n-15]=3250 \times 2 \\
& n[25+15 n]=6500
\end{aligned}
$

$
\begin{aligned}
& 5 n[5+3 n]=6500 \\
& \Rightarrow n(5+3 n)=\frac{6500}{5}=1300 \\
& 3 n^2+5 n-1300=0 \\
& 3 n^2-60 n+65 n-1300=0 \\
& 3 n(n-20)+65(n-20)=0 \\
& (3 n+65)(n-20)=0 \\
& n=-65 / 3 \text { or } 20 \\
& n=-65 / 3 \text { is not possible } \\
& \therefore n=20
\end{aligned}
$
So he will take 20 months to clear the amount.
Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball 24 meters away from the
starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Solution:
$\mathrm{t}_1=24 \times 2=48, \mathrm{t}_2=48+8=56$ or $(24+4) 2, \mathrm{t}_3=(28+4) 2=64$ which is an A.P.
Here $\mathrm{a}=48$,
$
\mathrm{d}=56-48=8
$
$
\therefore S_{20}=\frac{20}{2}[2 a+(20-1) d]=10[2(48)+19(8)]=10(96+152)=10(248)=2480 \mathrm{~m}
$
The contestant has to run $2480 \mathrm{~m}$ to bring all the balls.

Question 10 .
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $\mathrm{n}^{\text {th }}$ hour?
Solution:
Number of bacteria at the beginning $=30$
Number of bacteria after 1 hour $=30 \times 2=60$
Number of bacteria after 2 hours $=30 \times 2^2=120$
Number of bacteria after 4 hours $=30 \times 24=30 \times 16=480$
$\therefore$ Number of bacteria after $\mathrm{n}^{\text {th }}$ hour $=30 \times 2^{\mathrm{n}}$
Question 11.
What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of $10 \%$ compounded annually?

Solution:
$
\mathrm{P}=₹ 500, \text { C. } . \mathrm{I}=10 \%, \text { Amount after } 1 \text { year }=\mathrm{P}\left(1+\frac{r}{100}\right)=500+\left(1+\frac{10}{100}\right)=500\left(\frac{11}{10}\right)
$
Amount after 2 years $=500\left(\frac{11}{10}\right)^2$
Now $a=500\left(\frac{11}{10}\right), r=\frac{500\left(\frac{11}{10}\right)^2}{500\left(\frac{11}{10}\right)}=\frac{11}{10}$
$
t_n=a r^{n-1}, t_{10}=500\left(\frac{11}{10}\right)\left(\frac{11}{10}\right)^9=500\left(\frac{11}{10}\right)^{10}=500(1.1)^{10}=500(2.5937)=1296.87
$
Aliter:
Amount after 1 year $=500\left(1+\frac{10}{100}\right)=500\left(\frac{11}{10}\right)$
Amount after 2 years $=500\left(\frac{11}{10}\right)\left(\frac{11}{10}\right)=500\left(\frac{11}{10}\right)^2$
So amount after 10 years $=500\left(\frac{11}{10}\right)^{10}=1296.87$

Question 12 .
In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over $1,50,000$ units?
Solution:
$
\begin{aligned}
& \mathrm{a}=5, \mathrm{r}=2, \mathrm{t}_{\mathrm{n}}>150000 \text {, To find ' } \mathrm{n} \text { ' } \\
& t_n=a r^{n-1}=150000(\text { i.e }) 5(2)^{n-1}=150000 \\
& \Rightarrow 2^{n-1}=\frac{150000}{5}=30000 \\
& 2^{15}=32768 \text { and } 2^{14}=16384
\end{aligned}
$
We are given $2^{n-1}>30000 \Rightarrow 2^{15}=32768>30000 \Rightarrow n-1=14 \Rightarrow n=15$
On the $15^{\text {th }}$ day it will grow over 150000 units.