Exercise 2.7 - 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 $2.7$
Question $1 .$
Which of the following sequences are in G.P?
(i) $3,9,27,81$,
(ii) $4,44,444,4444$,
(iii) $0.5,0.05,0.005, \ldots \ldots$....
(iv) $\frac{1}{3}, \frac{1}{1}, 12 . \ldots .5$,
(iv) $\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \ldots \ldots .$ (v) $1,-5,25,-125$,
(v) $1,-5,25,-125, \ldots \ldots$
(vi) $120,60,30,18$, (vii) $16,4,1, \frac{1}{4}, \ldots .$
Solution:
(i) $3,9,27,81$
$\mathrm{r}=$ Common ratio

$r=\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}} \text { in G.P }$
Here $\frac{t_{2}}{t_{1}}=\frac{9}{3}=3$
$\frac{t_{3}}{t_{2}}=\frac{27}{9}=3$
$\therefore$ It is a G.P.
(ii) $4,44,444,4444, \ldots . .$
$r=\frac{t_{2}}{t_{1}}=\frac{44}{4}=11$
$\begin{aligned}
r &=\frac{t_{3}}{t_{2}}=\frac{444}{44}=\frac{111}{11} \\
11 & \neq \frac{111}{11}
\end{aligned}$
$\therefore$ It is not a G.P.
(iii) $0.5,0.05,0.005, \ldots$
$\begin{aligned}
r=\frac{t_{2}}{t_{1}}=& \frac{0.05}{0.5}=\frac{0.05 \times 100}{0.5 \times 100} \\
&=\frac{5}{50}=\frac{1}{10}
\end{aligned}$

$\therefore \text { It is a G.P. }$

(iv) $\frac{1}{3}, \frac{1}{6}, \frac{1}{12} \cdots$
$r=\frac{t_{2}}{t_{1}}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{6} \times \frac{\not 3}{1}=\frac{1}{2}$
$\begin{gathered}
r=\frac{t_{3}}{t_{2}}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{12} \times \frac{6}{1}=\frac{1}{2} \\
r=\frac{1}{2}=\frac{1}{2}
\end{gathered}$
$\therefore$ It is a G.P.
(v) $1,-5,25,-125$
$r=\frac{t_{2}}{t_{1}}=\frac{-5}{1}=-5$
$r=\frac{t_{3}}{t_{2}}=\frac{25}{-5}=-5$
$-5=-5$
$\therefore$ It is a G.P

(vi) $120,60,30,18, \ldots$
$r=\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}$
Here $r$ is not equal i.e $\frac{60}{120}=\frac{30}{60} \neq \frac{18}{30}$
$\therefore$ It is not a G.P
(vii) $16,4,1, \frac{1}{4}, \ldots$
$r=\frac{t_{2}}{t_{1}}=\frac{4}{16}=\frac{1}{4}$
$\begin{aligned}
&r=\frac{t_{3}}{t_{2}}=\frac{1}{4} \\
&r=\frac{1}{4}=\frac{1}{4}
\end{aligned}$
$\therefore$ It is a G.P
 

Question $2 .$
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) $a=6, r=3$
(ii) $\mathrm{a}=\sqrt{2}, \mathrm{r}=\sqrt{2}$
(iii) $\mathrm{a}=1000, \mathrm{r}=\frac{2}{5}$
Solution:
(i) $a=6, r=3$
$\mathrm{t}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$
$\mathrm{t}_{1}=\mathrm{ar}^{1-1}=\mathrm{ar}{ }^{0}=\mathrm{a}=6$
$t_{2}=a r^{2-1}=a r^{1}=6 \times 3=18$
$t_{3}-a r^{3-1}-a r^{2}-6 \times 3^{2}-54$
$\therefore$ The 3 terms are $6,18,54, \ldots .$

(ii) $a=\sqrt{2}, r=\sqrt{2}$
$\begin{aligned}
t_{n} &=a r^{n-1} \\
t_{1}=a r^{1-1}=a r^{0} &=\sqrt{2} \times 1=\sqrt{2} \\
t_{2}=a r^{2-1}=a r^{1} &=\sqrt{2} \times \sqrt{2}=2 \\
t_{3}=a r^{3-1}=a r^{2} &=\sqrt{2} \times(\sqrt{2})^{2} \\
&=\sqrt{2} \times 2=2 \sqrt{2}
\end{aligned}$
$\therefore$ The 3 terms are $\sqrt{2}, 2,2 \sqrt{2}, \ldots$
(iii) $a=1000, r=\frac{2}{5}$
$\begin{aligned}
t_{n} &=a r^{n-1} \\
t_{1}=a r^{1-1}=a r^{0} &=1000 \times 1=1000 \\
t_{2}=a r^{2-1}=a r &=1000^{200} \times \frac{2}{8}=400 \\
t_{3}=a r^{3-1}=a r^{2} &=1000\left(\frac{2}{5}\right)^{2} \\
&=1000^{40} \times \frac{4}{25} \\
&=160
\end{aligned}$
The 3 terms are $1000,400,160, \ldots \ldots \ldots \ldots . .$

Question 3.
In a G.P. $729,243,81, \ldots$ find t 7 .
Solution:
$\text { G.P }=729,243,81 \ldots . .$

$\mathrm{t}_{7}=$ ?
$\begin{aligned}
t_{n}=a^{n-1}, \text { here } a &=729, r=\frac{t_{2}}{t_{1}} \\
\therefore &=\frac{243}{729}=\frac{1}{3} \\
\therefore t_{7}=729\left(\frac{1}{3}\right)^{7-1} &=729 \times\left(\frac{1}{3}\right)^{6} \\
&=729 \times \frac{1}{729} \\
&=1
\end{aligned}$

Question $4 .$
Find $x$ so that $x+6, x+12$ and $x+15$ are consecutive terms of a Geometric Progression.
Answer:
$\frac{t_{2}}{t_{1}}=\frac{x+12}{x+6}, \frac{t_{3}}{t_{2}}=\frac{x+15}{x+12}$
Since it is a G.P.
$\begin{aligned}
&\frac{x+12}{x+6}=\frac{x+15}{x+12} \\
&(x+12)^{2}=(x+6)(x+15) \\
&x^{2}+24 x+144=x^{2}+21 x+90 \\
&3 x=-54 \Rightarrow x=\frac{-54}{3}=-18
\end{aligned}$

Question $5 .$
Find the number of terms in the following G.P.
(i) $4,8,16, \ldots, 8192$
(ii) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \ldots \frac{1}{2187}$
Solution:

(1) $4,8,16, \ldots \ldots .8192$

$a=4$
$r=\frac{8}{4}=2$
$t_{n}=8192$
$t_{n}=a r^{n-1}$
$8192=4 \times(2)^{n-1}$
$4 \times 2^{n-1}=8 \times 92^{2048}$
$2^{n-1}=2048$
$\dot{2}^{n-1}=2^{11}$
$n=11$
$\therefore$ No. of terms $=12-11+1=12$
$\therefore$ No. of terms $=12$
(ii) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots, \frac{1}{2187}$
Here $a=\frac{1}{3}, r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{g_{3}} \times \frac{\not \beta}{1}=\frac{1}{3}$
$t_{n}=a r^{n-1}$
$\frac{1}{3} \times\left(\frac{1}{3}\right)^{n-1}=\frac{1}{2187}$

$\left(\frac{1}{3}\right)^{n-1}=\frac{1}{2187} \times \not 3$ $=\frac{1}{729}$ $\left(\frac{1}{3}\right)^{n-1}=\frac{1}{3^{6}}=\left(\frac{1}{3}\right)^{6}$ $n-1=6$ $\therefore$ No. of terms $=7$

Question 6.

In a G.P. the $9^{\text {th }}$ term is 32805 and $6^{\text {th }}$ term is $1215 .$ Find the 12 th term. Solution: In a G.P $t_{n}=$ ar ${ }^{n-1}$ $t_{9}=32805$ $t_{6}=1215$ $t_{12}=$ ? Let $t_{9}=$ ar $8=32805 \ldots \ldots . .(1)$
In a G.P. the $9^{\text {th }}$ term is 32805 and $6^{\text {th }}$ term is 1215 . Find the 12 th term.
Solution:
In a G.P
$t_{n}=a r^{n-1}$
$t_{9}=32805$
$\mathrm{t}_{6}=1215$
$t_{12}=$ ?
Let
$t_{9}=a r^{8}=32805$

$\begin{aligned}
\mathrm{t}_{6}=\mathrm{ar}^{5}=1215 \ldots \ldots &(2) \\
\frac{(1)}{(2)} &=\frac{a r^{8}}{a r^{5}}=\frac{32805}{1215} \\
r^{8-5} &=27 \\
r^{3} &=3^{3} \\
r &=3 \\
\Rightarrow \quad \text { Substitute } r &=3 \text { in }(2) \\
a \times 3^{5} &=1215 \\
a &=5 \\
t_{12} &=a r 11 \\
\therefore \quad &=5 \times 3^{11} \\
&=5 \times 177147 \\
&=885735
\end{aligned}$

Question $7 .$
Find the 10 th term of a G.P. whose $8^{\text {th }}$ term is 768 and the common ratio is $2 .$
Answer:
Here $r=2, t_{8}=768$
$\mathrm{t}_{8}=768\left(\mathrm{t}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}\right)$
a. $r^{8-1}=768$
$\left.a r^{7}=768 \ldots . .1\right)$
$10^{\text {th }}$ term of a G.P. $=$ a.r $10^{-1}$
$=\mathrm{ar}^{9}$ $=\left(\mathrm{ar}^{7}\right) \times\left(\mathrm{r}^{2}\right)$ $=768 \times 2^{2}($ from $\mathrm{l})$ $=768 \times 4=3072$ $\therefore 10^{\text {th }}$ term of a G.P. $=3072$

Question $8 .$
If a, b, c are in A.P. then show that $3^{a}, j^{b}, j^{c}$ are in $(i . P$.
Solution
If $\mathbf{a}, \mathbf{b}, \mathrm{c}$ are in A.P
$\begin{aligned}
&\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2} \\
&\mathrm{~b}-\mathrm{a}=\mathrm{c}-\mathrm{b} \\
&2 \mathrm{~b}=\mathrm{c}+\mathrm{a}
\end{aligned}$
To prove that $3^{\mathrm{a}}, 3^{\mathrm{b}}, 3^{\mathrm{c}}$ are in G.P
$\Rightarrow 3^{2 b}=3^{c+a}+a$ [Raising the power both sides]
$\Rightarrow 3^{\mathrm{b}} \cdot 3^{\mathrm{b}}=3^{\mathrm{c}} \cdot 3^{\mathrm{a}}$

$\Rightarrow \frac{3^{b}}{3^{a}}=\frac{3^{\mathrm{c}}}{3^{b}}$ $\Rightarrow \frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{1}}$ $\Rightarrow$ Common ratio is same for $3^{\mathrm{a}}, 3^{\mathrm{b}}, 3^{\mathrm{c}}$ $\Rightarrow 3^{\mathrm{a}}, 3^{\mathrm{b}}, 3^{\mathrm{c}}$ forms a G.P $\therefore$ Hence it is proved.
 

Question $9 .$
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is $\frac{57}{2}$. Find the three terms.
Solution:
Let the three consecutive terms in a G.P are $\frac{a}{r}$, a, ar.
Their Product $=\frac{a}{r} \times a \times a r=27$
$\begin{aligned}
&a^{3}=27=3^{3} \\
&a=3
\end{aligned}$
Sum of the product of terms taken two at a time is $\frac{57}{2}$

$\begin{aligned}
\frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r} &=\frac{57}{2} \\
\frac{a^{2}}{r}+a^{2} r+a^{2} &=\frac{57}{2} \\
3^{2}\left(\frac{1}{r}+r+1\right) &=\frac{57}{2} \\
\frac{1+r^{2}+r}{6_{3}} \frac{-y^{3}}{6_{2}} \\
18+18 r^{2}+18 r &=\frac{57}{2} \times \frac{1}{9}=\frac{57}{18} \\
18 r^{2}+18 r-57 r+18 &=0 \\
18 r^{2}-39 r+18 &=0 \div 3
\end{aligned}$

$\begin{aligned}
\Rightarrow \quad 6 r^{2}-13 r+6 &=0 \\
\left(r-\frac{2}{3}\right)\left(r-\frac{3}{2}\right) &=0 \\
r &=\frac{2}{3}, \frac{3}{2} \\
\text { If } a=3, r &=\frac{2}{3}
\end{aligned}$
$\therefore$ The three numbers are $\frac{3}{2}, 3,3 \times \frac{2}{3}$
(or) $3 \times \frac{2}{3}, 3, \not 3 \times \frac{2}{\not}$
$\frac{9}{2}, 3,2$
If $a=3, r=\frac{3}{2}$, the three numbers are
$\begin{aligned}
\frac{a}{r}, a, a r &=\frac{3}{\frac{3}{2}}, 3,3 \times \frac{3}{2} \\
&=\frac{6}{3}, 3, \frac{9}{2} \\
&=2.3 . \frac{9}{2}
\end{aligned}$

Question $10 .$
A man joined a company as Assistant Manager. The company gave him a starting salary of $\square 60,000$ and agreed to increase his salary $5 \%$ annually. What will be his salary after 5 years?
Solution:
Starting salary $=\square 60,000$
Increase per year $=5 \%$
$\therefore$ At the end of 1 year the increase
$=60,0,00 \times \frac{5}{100}$
$\square 3000$

$\therefore$ At the end of first year his salary
$=\square 60,000+3000$
I year salary $=\square 63,000$
II Year increase $=63000 \times \frac{5}{100}$
At the end of II year, salary
$=63000+3150$
$=\square 66150$
III Year increase $=66150 \times \frac{5}{100}$
$=3307.50$
At the end of III year, salary $=66150+3307.50$
$=\square 69457.50$
IV year increase $=69457.50 \times \frac{5}{100}$
$=\square 3472.87$

Question $11 .$
Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: 20,000 to start with followed by a guaranteed annual increase of $3 \%$ for the first 5 years. Offer B: $\square 22,000$ to start with followed by a guaranteed annual increase of $3 \%$ for the first 5 years.
What is his salary in the 4th year with respect to the offers $A$ and $B$ ?
Solution:
Offer A
Starting salary $\square 20,000$
Annual increase 6\%
i.e.
$\begin{aligned}
&₹ 20,000 \times \frac{6}{100} \\
&=₹ 1200
\end{aligned}$
At the end of I year salary $=20000+1200$
$\begin{aligned}
&=₹ 21200 \\
\text { II year increase } &=21200 \times \frac{6}{100}
\end{aligned}$

$=₹ 1272$
At the end of II year salary
$=21200+1272=22472$
III year increase $=22472 \times \frac{6}{100}=1348.32$
At the end of
III year, salary $=22472+1348=23820$
$\therefore$ IV year salary $=\square 23820$
Offer B
Starting salary $=\square 22,000$

Annual increase $=3 \%=\frac{3}{100}$
I year, increase $=22000 \times \frac{3}{100}=₹ 660$
At the end of
I year, salary $=22000+660$
$=₹ 22660$
II year increase $=2266 \phi \times \frac{3}{10 \varnothing}$
$=₹ 679.8$
At the end of
II year, salary $=₹ 23339.80$
III year increase $=23339.8 \times \frac{3}{100}$
$=₹ 700$
At the end of
III year, salary $=₹ 24039.80$
$\therefore$ IV year salary $=₹ 24040$
Salary as per Option $A=\square 23820$
Salary as per Option B $=\square 24040$
$\therefore$ Option B is better.
 

Question $12 .$
If $a, b, c$ are three consecutive terms of an A.P. and $x, y, z$ are three consecutive terms of a G.P. then prove that $x^{b-c} \times y^{c-a} \times z^{a-b}=1$.
Solution:
a, b, c are three consecutive terms of an AP.
$\therefore$ Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be $\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}$ respectively
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three consecutive terms of a GP.
$\therefore$ Assume $\mathrm{x}, \mathrm{y}, \mathrm{z}$ as $\mathrm{x}, \mathrm{x} . \mathrm{r}, \mathrm{x}^{2} \mathrm{r}^{2}$ respectively ........ (2)
PT : $x^{b-c}, y^{c-a}, z^{a-b}=1$

Substituting (1) and (2) in LHS, we get $\mathrm{I} \cdot \mathrm{HS}=\mathrm{x}^{\mathrm{a}+\mathrm{d}-\mathrm{a}-2 \mathrm{~d}} \times(\mathrm{xr})^{\mathrm{a}+2 \mathrm{~d}-\mathrm{a}} \times\left(\mathrm{xr}^{2}\right)^{\mathrm{a}-\mathrm{a}-\mathrm{d}}$
$\begin{aligned}
&=(x)^{-d} \cdot(x r)^{2 d}\left(x r^{2}\right)^{-d} \\
&=\frac{1}{x^{d}} \times x^{2 d} \cdot r^{2 d} \times \frac{1}{x^{d} r^{2 d}}=1=\text { RHS }
\end{aligned}$