S Chand CLASS 10 Chapter 9 Arithmetic and Geometric Progression Exercise 9 C

  Exercise 9 C

Question 1 

Ans: (i) $27,9,3,1, \ldots$

$\begin{aligned}\therefore & a=27 \\r &=\frac{9}{27}=\frac{1}{3}, \frac{3}{9} \\&=\frac{1}{3}, .....\end{aligned}$

Hence, it is a G.P. and $r=\frac{1}{3}$

(ii) $-1,2,4,8, \ldots .$

$\therefore a=-1$,

$\quad b=\frac{2}{-1}=-2, \frac{4}{2} .$

$\quad=2, \frac{8}{4}=2 .$

So, it is not a G.P 

(iii) 

$\begin{aligned} & 2, \frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \ldots \\ \therefore & a=2, \\ &=\frac{1}{2} \div 2 \\ &=\frac{1}{2} \times \frac{1}{2} \\ &=\frac{1}{4} . \\ \frac{1}{8} &=\frac{1}{2} . \end{aligned}$

$=\frac{1}{4} .$

$\frac{1}{32}=\frac{1}{8} .$

$=\frac{1}{32} \times 8 .$

$=\frac{1}{4}$

∴ IT is GP and  $r=\frac{1}{4}$

$(i v)-12,-6,0,6, \ldots . .$

$\begin{aligned}&\therefore a=-13 \\&r=\frac{-6}{-12}=\frac{1}{2} \\&=\frac{0}{-6} \\&=0\end{aligned}$

Hence,  it is not G.P.

Question 2

Ans:  (i) 2,6....

$\left(r=\frac{6}{2}=3\right) .$

2,6,18,54,162

(ii) $\frac{1}{16},-\frac{1}{8}, \ldots$

$\left(r=-\frac{1}{8} \div \frac{1}{10}=-\frac{1}{8} \times 16=-2 .\right)$

$\frac{1}{16},-\frac{1}{8}, \frac{1}{4},-\frac{1}{2}, 1$

(iii)0.3, 0.06......

$r=\frac{0.01}{0.3}$= $\frac{1}{5}=0-2$

$0.3,0.03,0.012,0.0024,0.00048$

Question 3

Ans: 6th term of the G.P 2, 10 , 50......

∴ a = 2 

r=  $\frac{10}{2}=5$

$\begin{aligned} So  T_{6} &=9 x^{h-1} \\ &=2 \times 5^{6-1} \\ &=2 \times 5^{5} \\ &=2 \times 3125 \\ &=6250 \end{aligned}$

(ii)  11th term of the G.P. $4,12,3, \ldots .$

$\begin{aligned} \therefore & a=4 \\ & r=\frac{12}{4}=3 . \end{aligned}$

$\begin{aligned} T_{11} &=ar^{h-1} \\ &=4 \times(3)^{11-1} \\ &=4 \times 3^{10} \\ &=4 \times 59,049 \\ &=236196 \end{aligned}$

Question 4

Ans:

 (i) $\begin{aligned} T_{n} &=4 \cdot 3^{h-1} \\ \therefore T_{1} &=4 \cdot 3^{1-1} \\ &=4 \cdot 3^{0} \\ &=4 \times 1=4 \\ T_{2} &=4 \cdot 3^{2-1} \\ &=4 \cdot 3^{1} \\ &=4 \times 3 \\ &=12 \\ T_{3} &=4 \cdot 3^{3-1} \\ &=4 \cdot 3^{2} \\ &=4 \times 9 \\ &=36 \end{aligned}$

$\begin{aligned} T_{4} &=4.3^{4-1} \\ &=4.3^{3} \\ &=4 \times 27 \\ &=108 \\ T_{5} &=4.3^{5-1} \\ &=4.3^{4} \\ &=4 \times 81 \\ &=324 \end{aligned}$

Hence, the terms are 4, 12, 36 , 108 , 324.

(ii) 

$\begin{aligned} T_{h} &=\frac{5^{h-1}}{2^{h+1}} \\ T_{1} &=\frac{5^{1-1}}{2^{n+1}} \\ &=\frac{5^{0}}{2^{2}} \\ &=\frac{1}{4} \\ T_{2} &=\frac{5^{2-1}}{2^{2+1}} \\ &=\frac{5^{1}}{2^{3}} \\ &=\frac{5}{8} \\ T_{3} &=\frac{5^{3-1}}{2^{3+1}} \\ &=\frac{5^{2}}{2^{4}} \\ &=\frac{25}{16} \end{aligned}$

$\begin{aligned} T_{4} &=\frac{5^{4-1}}{5^{9+1}} \\ &=\frac{5^{3}}{2^{5}} \\ &=\frac{125}{2^{5}} \\ &=\frac{125}{32} \\ T_{5} &=\frac{5^{5-1}}{2^{5+1}} \\ &=\frac{5^{4}}{2^{6}} \\ &=\frac{625}{64} . \end{aligned}$

Hence, the 5 terms are  $\frac{1}{4}, \frac{5}{8}, \frac{25}{16} ,\frac{125}{32}, \frac{625}{64}$

Question 5

Ans: (i) 12 , -36 .... sixth term

$\begin{aligned} \therefore \quad a &=12, \\ r &=\frac{-36}{12} \\ r &=-3 . \end{aligned}$

$\begin{aligned} \therefore T_{6} &=arn^{-1}  \\ &=12 \times(-3)^{6-1} \\ &=12 \times(-3)^{5} \\ &=12 \times(-243) \\ &=-2916 \end{aligned}$

(ii)  $3,-\frac{1}{3}, \ldots, 8$ th term.

$\begin{aligned} \therefore \quad q &=3, \\ r &=-\frac{1}{3} \div 3 . \\ &=-\frac{1}{3} \times \frac{1}{3} \\ &=-\frac{1}{9} \end{aligned}$

So , $\begin{aligned} T_{8} &=arn^{-1} \\ &=3 \cdot\left(-\frac{1}{9}\right)^{81} \\ &=3\left(-\frac{1}{9}\right)^{7} \end{aligned}$

(iii)

 $\begin{aligned} & b^{2} c^{3}, b^{3} c^{2}, \ldots, & 5th term \\ & \therefore a=b^{2} c^{3} \end{aligned}$

$r=\frac{b^{2} c^{2}}{b^{2} c^{8}}$

$r=\frac{b}{c}$

So $T_{5}= arn^{-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{5-1}$

$=b^{2} c^{3} \times\left(\frac{b}{c}\right)^{4} .$

$=b^{2} c^{3} \times \frac{b^{4}}{c^{4}}$

$=b^{2} c^{2} \times b^{4} \times c^{-4}$

$=b^{2+4} \times c^{3}-4$

$=b^{6} \times c^{-1}$

$\frac{b^{6}}{c}$

Question 6

Ans: G.P. is $27,-18,12,-8, \ldots$ is $\frac{1024}{2187}$ (Given)

$\begin{aligned} \therefore a &=27, \\ r &=\frac{-18}{27} \\ r &=\frac{-2}{3} . \end{aligned}$

Let  $\frac{1024}{2187}$ be the nth term 

$\therefore a_{n}=a r n^{-1}$

$\frac{1024}{2187}=27\left(-\frac{2}{3}\right)^{n-1}$

$\frac{1024}{2187 \times 27}=\left(-\frac{2}{3}\right)^{n-1}$

$\frac{2^{10}}{3^{7} \times 3^{3}}=\left(-\frac{2}{3}\right)^{n-1}$

$\frac{2^{10}}{3^{10}}=\left(\frac{-2}{3}\right)^{n-1}$

$\left(\frac{-2}{3}\right)^{10}=\left(\frac{-2}{3}\right)^{n-1}$

On comparing, 

$10=n-1$

$10+1=n$

$11=n$

n=11

Hence, it is 11th term.

 

Question 7

 

Ans: In a G.P 

$T_{4}=54$

$T_{7}=1458$

Let a be the first term and r be the common ratio

$So T_{4}=a r^{n-1}$

$54=ar^{4-1}$

$54=ar^{3}$

$a r^{3}=5 4$.........(i)

Dividing ,eqn (i) and (ii)

$r^{3}=\frac{1458}{54}$

$r^{3}=27$

$r=\sqrt[3]{27}$

$r=3$

Put the value of r = 3 in eq (i)

$a r^{3}=54$

$a \times(3)^{3}=54$

$a \times 27=54$

$a=\frac{54}{27} .$

$a=2$

$\therefore a=2, r=3 .$

Hence , G.P will be 2, 6 ,18 ,54....

 

Question 8

Ans: In a G.P. $3,3 \sqrt{3}, 9, \ldots$

Last term (l) = 2187

∴ a = 3

r =  $\frac{3 \sqrt{3}}{3}$

$T_{n}=l=arn^{-1}$

$2187=3(\sqrt{3})^{n-1}$

$\frac{2187}{3}=(\sqrt{3})^{n-1}$

$729=(\sqrt{3})^{n-1}$

$3^{6}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{6 \times 2}=(\sqrt{3})^{n-1}$

$(\sqrt{3})^{12}=(\sqrt{3})^{n-1} .$

On comparing,

$12=n-1$

$12+1=n$

$13=n$

$n=13$

Hence, it is 13th term

Question 9

Ans: Given , 

1, x, y , z 16 are in G.P 

So, first term (a) = 1, 

Common ratio(r) = $\frac{x}{1}$

$T_{5}=16$

$\begin{aligned} \therefore T_{5} &=a r(h-1) . \\ 16 &=9 r(5-1) \\ 16 &=9 r^{4} \\ 16 &=1 \times r^{4} \\(2)^{4} &=r^{4} \end{aligned}$

On comparing, 

r= 3

 So, the common ratio is 2 

Then, 

$\begin{aligned} r=\frac{x}{1} &=\frac{2}{1} \\ x &=2 \end{aligned}$

Also , the common ratio

$\frac{16}{z}=2$

$\begin{aligned}&z=\frac{16}{2} \\&z=8\end{aligned}$

Also, the common ratio 

$\begin{aligned} \frac{2}{y} &=2 \\ \frac{y}{y} &=2 \\ y &=\frac{8}{2} \\ y &=4 . \end{aligned}$

According to the question 

$\therefore x+y+z$

$=2+4+8$

=14 

Question 10

 

Ans:  In a G.P 

$T_{3}=18$,

$T_{7}=3 \frac{5}{9}=\frac{32}{9}$

Let a be the first term and r be the common ratio 

∴ $T_{n}=a r^{n-1}$

$T_{3}=a r^{3-1}$

$18=a r^{2}$

$a r^{2}=18$ ........(i)

$T_{7}=a r^{7-1}$

$\frac{32}{9}=a r^{6}$

$a r^{6}=\frac{32}{9}$...........(ii)

On dividing , 

$\frac{ar^{6}}{ar^{2}}=\frac{32}{9\times 18}$

$r^{4}=\frac{16}{81}$

$r^{4}=\left(\frac{2}{3}\right)^{4}$

On comparing, 

$\begin{aligned} & r=\frac{2}{3} \\ \therefore \quad & a r^{2}=18 \end{aligned}$

$a \times\left(\frac{2}{3}\right)^{2}=18$

$a \times \frac{4}{9}=18$

$a=\frac{81}{2}$

Then , 

$T_{10}=ar^{9}$

$=\frac{81}{2} \times\left(\frac{2}{3}\right)^{9}$

$=\frac{81}{2} \times \frac{2^{9}}{3^{9}}$

$\begin{aligned} &=\frac{3^{4} \times 2^{9}}{2 \times 3^{9}} \\=& \frac{2^{9-1}}{3^{9-4}} \\=& \frac{2^{8}}{3^{5}} \\=& \frac{256}{243} \end{aligned}$

Question 11

 

Ans: Given, 

In a G.P 

$T_{5}=P$,

$T_{8}=Q$,

$T_{11}=S$

Let a be the first term and r be the common ratio 

$\begin{aligned} \therefore \quad T_{5} &=ar^{5-1} \\ &=ar^{4} \\ &=p \end{aligned}$

$\begin{aligned} T_{8} &=a_{r} 8-1 \\ &=a^{7} \\ &=a \end{aligned}$

$\begin{aligned} T_{11} &=a r^{11-1} \\ &=a r^{10} \end{aligned}$

=5

$\begin{aligned} Q^{2} &=\left(ar7^{2}\right)^{2} \\ &=a^{2} r^{14} . \end{aligned}$

and p $\times s$ = $a r^{4} \times a r^{16} .$

$=a^{2} r^{4}$

Hence , prove, 

$Q^{2}=P \times S$