ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Exercise 9.4

 Exercise 9.4

Question 1

Can 0 be a term of a geometric progression?

Sol :

No, 0 is not a term of geometric progression

Question 2

(i) Find the next term of the list of numbers $\frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots$

(ii) Find the next term of the list of numbers  $\frac{3}{16},-\frac{3}{8}, \frac{3}{4},-\frac{3}{2}, \ldots$

(iii) Find the 15th term of the series $\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{3}}+\cdots$

(iv) Find the nth term of the list of numbers $\frac{1}{\sqrt{2}},-2,4 \sqrt{2},-16, \ldots$

(v) Find the 10th and nth terms of the list of numbers 5, 25, 125, …

(vi) Find the 6th and the nth terms of the list of numbers $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots$

(vii) Find the 6th term from the end of the list of numbers 3, – 6, 12, – 24, …, 12288.

Sol :

Given :

$\Rightarrow \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots$

Here, $a=\frac{1}{6}, r=\frac{1}{3} \div \frac{1}{6}=\frac{1}{3} \times \frac{6}{1}=2$

$\therefore$ Next term $=\frac{2}{3} \times 2=\frac{4}{3}$

(ii) $\frac{3}{16},-\frac{3}{8}, \frac{3}{4},-\frac{3}{2}, \ldots$

Here, $a=\frac{3}{16}, r=\frac{-3}{8} \div \frac{3}{16}=\frac{-3}{8} \times \frac{16}{3}=-2$

$\therefore$ Next term $=\frac{-3}{2} \times(-2)=3$

(iii) $\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{3}}+\ldots$

Here, $a=\sqrt{3}, r=\frac{1}{\sqrt{3}} \div \sqrt{3}=\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}$

$=\frac{1}{3}$

$\therefore a_{15}=a r^{n-1}=\sqrt{3}\left(\frac{1}{3}\right)^{15-1}$

$=\sqrt{3} \times\left(\frac{1}{3}\right)^{14}=\sqrt{3} \times \frac{1}{3^{14}}$

(iv) $\frac{1}{\sqrt{2}},-2,4 \sqrt{2},-16, \ldots$

Here, $a=\frac{1}{\sqrt{2}}, r=-2 \div \frac{1}{\sqrt{2}}=-2 \times \sqrt{2}=-2 \sqrt{2}$

$a_{n}=a r^{n-1}=\frac{1}{\sqrt{2}} \times(-2 \sqrt{2})^{n-1}$

$=\frac{1}{\sqrt{2}} \times(-1)^{n-1} \times\left[(\sqrt{2})^{2} \times \sqrt{2}\right]^{n-1}$

$=(-1)^{n-1} \times \frac{1}{\sqrt{2}} \times\left[(\sqrt{2})^{3}\right]^{n-1}$

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

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

$=(-1)^{n-1}(\sqrt{2})^{3 n-4}$

$=(-1)^{n-1} \times 2^{\frac{3 n-4}{2}}$

(v) 5,25,125,...

Here, a=5, $r=25 \div 5=5$

$a_{10}=a r^{n-1}=5 \times(5)^{10-1}$

$=5 \times 5^{9}=5^{9+1}=5^{10}$

$a_{n}=a r^{n-1}=5 \times 5^{n-1}=5^{n-1+1}=5^{n}$

(vi) $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots$

Here, $a=\frac{3}{2}, r=\frac{3}{4} \div \frac{3}{2}=\frac{3}{4} \times \frac{2}{3}=\frac{1}{2}$

$\therefore a_{n}=a r^{n-1}=\frac{3}{2} \times \frac{1}{2}^{n-1}$

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

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

$a_{6}=\frac{3}{2^{n}}=\frac{3}{2^{6}}=\frac{3}{64}$

(vii) 3,-6,12,-24,..., 12288

6th term from the end of the list

Here, $a=3, r=-6 \div 3=-2, l=12288$

Now. 6th term from the end

$=l \times\left(\frac{1}{r}\right)^{n-1}=12288 \times\left(\frac{1}{-2}\right)^{6-1}$

$=12288 \times \frac{1}{(-2)^{5}}=\frac{12288}{-32}=-384$

Question 3

Which term of the G.P.

(i) 2, 2√2, 4, … is 128?

(ii) $1, \frac{1}{3}, \frac{1}{9}, \ldots, 28 \text{is} \frac{1}{243} ?$

(iii) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683} ?$

Sol :

Given

(i) 2, 2√2, 4, … is 128?

Here $a=2, r=\frac{2 \sqrt{2}}{2}=\sqrt{2}, l=128$

Let 128 be the n th term, then

$a_{n}=128=a r^{n-1}$

$\Rightarrow 128=2(\sqrt{2})^{n-1} \Rightarrow 2(\sqrt{2})^{n-1}=2^{7}$

$(\sqrt{2})^{n-1}=2^{7-1}=2^{6}$

$(\sqrt{2})^{n-1}=(\sqrt{2})^{12}$

Comparing, we get n-1=12 

$\Rightarrow n=12+1=13$

$\therefore 128$ is the 13 th term

(ii) $1, \frac{1}{3}, \frac{1}{9}, \ldots$ is $\frac{1}{243}$

Here, $a=1, r=\frac{1}{3} \div 1=\frac{1}{3}, l=\frac{1}{243}$

Let $\frac{1}{243}$ is the n th term, then

$a_{n}=\frac{1}{243}=a r^{n-1}=1 \times\left(\frac{1}{3}\right)^{n-1}$

$\Rightarrow\left(\frac{1}{3}\right)^{n-1}=\left(\frac{1}{3}\right)^{5}$

Comparing, we get

n-1=5 

$\Rightarrow n=5+1=6$

$\therefore \frac{1}{243}$ is the 6 th term

(iii) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$

Here, $a=\frac{1}{3}, r=\frac{1}{9} \div \frac{1}{3}$

$=\frac{1}{9} \times \frac{3}{1}=\frac{1}{3}, l=\frac{1}{19683}$

Let $\frac{1}{19683}$ is the $n$ th term, then

$a_{n}=\frac{1}{19683}=a r^{h-1}=\frac{1}{3}\left(\frac{1}{3}\right)^{n-1}$

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

$\Rightarrow \left(\frac{1}{3}\right)^{n}=\left(\frac{1}{3}\right)^{9}$

Comparing, we get

$\therefore n=9$

$\begin{array}{l|l} 3 & 19683 \\ \hline 3 & 6561 \\ \hline 3 & 2187 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Hence, $\frac{1}{19683}$ is the 9 th term

Question 4

Which term of the G.P. 3, – 3√3, 9, – 9√3, … is 729 ?

Sol :

G.P. 3, -3√3, 9, – 9√3, … is 729 ?

Here $a=3, r=\frac{-3 \sqrt{3}}{3}=\sqrt{-3}, l=729$

Let 729 is the $n$ th term, then

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

$\Rightarrow \frac{729}{3}=(-\sqrt{3})^{n-1} \Rightarrow 243=(-\sqrt{3})^{n-1}$

$\Rightarrow(-\sqrt{3})^{10}=(-\sqrt{3})^{n-1}$

Comparing, we get

$n-1=10 \Rightarrow n=10+1=11$

$\therefore 729$ is the 11 th term

Question 5

Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.

Sol :

In a G.P.

$a_8$ = 192 and r = 2

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

$a_{8}=a r^{n-1} \Rightarrow 192=a(2)^{n-1}=a 2^{8-1}=a 2^{7}$

$\Rightarrow a=\frac{192}{2^{7}}=\frac{192}{128}=\frac{3}{2}$

$\therefore a_{12}=a(r)^{n-1}$

$\Rightarrow a_{12}=\frac{3}{2}(2)^{12-1}=\frac{3}{2} \times 2^{11}$

$=\frac{3}{2} \times 2048=3072$

$\therefore \mathrm{a}_{12}=3072$

Question 6

In a GP., the third term is 24 and 6th term is 192. Find the 10th term

Sol :

In a GP.

$a_{3}=24$ and $a_{6}=192, a_{10}=?$

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

$a_{6}=a r^{6-1}=a r^{5}=192 \quad\left\{\because a_{n}=a r^{n-1}\right\}$

$a_{3}=a r^{3-1}=a r^{2}=24$

Dividing, we get

$\frac{a r^{5}}{a r^{2}}=\frac{192}{24} \Rightarrow r^{3}=8=(2)^{3}$

$\therefore r=2$

$\quad$ Now, $a r^{2}=24 \Rightarrow a \times 2^{2}=24$

$\Rightarrow a=\frac{24}{2^{2}}=\frac{24}{4}=6$

$\therefore \quad a=6$

Now, $a_{10}=a r^{10-1}=a r^{9}$

$=6 \times(2)^{9}=6 \times 512=3072$

Question 7

Find the number of terms of a G.P. whose first term is $\frac{3}{4}$, common ratio is 2 and the last term is 384.

Sol :

First term of a G.P. (a) $=\frac{3}{4}$

and common ratio (r) = 2

Last term = 384

Let number of terms is n

then $a_{n}=a r^{n-1}$

$\Rightarrow 384=\frac{3}{4}(2)^{n-1}$

$\Rightarrow 2^{n-1}=\frac{384 \times 4}{3}=512=2^{9}$

$\therefore n-1=9$

$\Rightarrow n=9+1=10$

$\therefore$ Number of terms in G.P. =10

Question 8

Find the value of x such that

(i) $-\frac{2}{7}, x ,-\frac{7}{2}$  are three consecutive terms of a G.P.

(ii) x + 9, x – 6 and 4 are three consecutive terms of a G.P.

(iii) x, x + 3, x + 9 are first three terms of a G.P. Sol. Find the value of x

Sol :

Find the value of x

(i) $-\frac{2}{7}, x_{3}-\frac{7}{2}$  are three consecutive terms of a G.P.

$\therefore x^{2}=\frac{-2}{7} \times \frac{-7}{2}=1=(\pm 1)^{2}$

$\therefore x=+1$

(ii) x+9, x-6 and 4 are three consecutive terms of a G.P., then

$(x-6)^{2}=(x+9) \times 4$

$\Rightarrow x^{2}-12 x+36=4 x+36$

$\Rightarrow x^{2}-12 x-4 x+36-36=0$

$\Rightarrow x^{2}-16 x=0 \Rightarrow x(x-16)=0$

Either x-16=0, then x=16

or x=0

$\therefore x=0,16$

(iii) x, x+3, x+9 are first three terms of a G.P.

$\therefore(x+3)^{2}=x(x+9)$

$\Rightarrow x^{2}+6 x+9=x^{2}+9 x$

9=9x-6x=3x

$\therefore x=\frac{9}{3}=3$

Question 9

If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.

Sol :

In a G.P.

$a_{4}=x, a_{7}=y, a_{10}=z$

Let a be the first term and r be the common

ratio, therefore 

$a_{4}=a r^{n-1}=a r^{4-1}=a r^{3}=x$

Similarly, 

$a_{7}=a r^{6}=y$

$a_{10}=a r^{9}=z$

If x, y and z are in G.P., then

$y^{2}=x z$

Now, $x z=a r^{3} \times a r^{9}=a^{2} r^{3+9}=a^{2} r^{12}$

$y^{2}=\left(a r^{6}\right)^{2}=a^{2} \cdot r^{12}$

$\because$ L.H.S. $=$ R.H.S.

$\therefore x, y$ and $z$ are in G.P.

Question 10

The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q² = ps.

Sol :

In a G.P.

$a_{5}=p, a_{8}=q$ and $a_{11}=s$

To show that $q^{2}=p x$

Let a be the first term and r be the common

$a_{5}=a r^{n-1}=a r^{5-1}=a r^{4}=p$

Similarly, $a_{8}=a r^{7}=q$ and

$a_{11}=a r^{10}=s$

$q^{2}=\left(a r^{7}\right)^{2}=a r^{14}$

and $p x=a r^{4} \times a r^{10}=a^{2} r^{4+10}=a^{2} r^{14}$

Hence, $q^{2}=p s$

Question 11

If a, b, c are in G.P., then show that a², b², c² are also in G.P.

Sol :

a, b, c are in G.P.

Show that a², b², c² are also in G.P

∵ a, b, c are in G.P., then

b² = ac …(i)

a², b², c² will be in G.P.

if (b²)² = a² x c²

⇒ (ac)² = a²c² [From (i)]

⇒ a²c² = a²c² which is true.

Hence proved.

Question 12

If a, b, c are in A.P., then show that $3^{a}, 3^{b}, 3^{c}$ are in G.P.

Sol :

a, b and c are in A.P.

Then, 2b = a + c

Now, $3^{a}, 3^{b}, 3^{c}$ will be in G.P.

if $\left(3^{b}\right)^{2}=3^{a} \cdot 3^{c}$

if $3^{2 b}=3^{a+c}$

Comparing, we get

if 2b = a + c

Which are in A.P. is given

Question 13

If a, b, c are in A.P., then show that $10^{a x+10}, 10^{b x+10}, 10^{c x+10}, x \neq 0$, are in G.P.

Sol :

a, b, c are in A.P.

To show that are in G.P. $10^{a x+10}, 10^{b x+10}, 10^{c x+10}, x \neq 0$

∵ a, b, c are in A.P.

$\therefore 2 b=a+c$...(i)

Now 

$\left(10^{\alpha x+10}\right),\left(10^{b x+10}\right),\left(10^{c x+10}\right)$ will be in

G.P. if $\left(10^{b x+10}\right)^{2}=\left(10^{a x+10}\right) \times\left(10^{c x+10}\right)$

if $10^{2 b x+20}=10^{a x+10+c x+10}$

if $10^{2 b x+20}=10^{a x+c x+20}$

Comparing, if 2bx+20=ax+cx+20

if 2bx=ax+cx

if 2b=a+c

Which is given

Question 14

If $a, a^{2}+2$ and $a^{3}+10$ are in G.P., then find the values(s) of a.

Sol :

$a, a^{2}+2$ and $a^{3}+10$ are in G.P.

$\because\left(a^{2}+2\right)^{2}=a\left(a^{3}+10\right)$

$\Rightarrow a^{4}+4 a^{2}+4=a^{4}+10 a$

$\Rightarrow 4 a^{2}-10 a+4=0$

$\Rightarrow 2 a^{2}-5 a+2=0$

$\Rightarrow 2 a^{2}-a-4 a+2=0$

⇒a(2 a-1)-2(2 a-1)=0

⇒(2 a-1)(a-2)=0

Either 2a-1=0, then 2a=1 $\Rightarrow a=\frac{1}{2}$

or a-2=0, then a=2

Hence a=2 or $\frac{1}{2}$

Question 15

If k, 2k + 2, 3k + 3, … are in G.P., then find the common ratio of the G.P.

Sol :

k, 2k + 2, 3k + 3, … are in G.P.

then $(2 k+2)^{2}=k(3 k+3)$

$\Rightarrow 4 k^{2}+8 k+4=3 k^{2}+3 k$

$\Rightarrow 4 k^{2}+8 k+4-3 k^{2}-3 k=0$

$\Rightarrow k^{2}+5 k+4=0$

$\Rightarrow k^{2}+k+4 k+4=0$

⇒k(k+1)+4(k+1)=0

⇒(k+1)(k+4)=0

Hence , k=-1 or -4

Question 16

The first and the second terms of a GP. are $x^{4}$ and $x^{m}$ . If its 8th term is $x^{52}$, then find the value of m.

Sol :

In a G.P.,

First term $\left(a_{1}\right)=x^{-4}$ …(i)

Second term $\left(a_{2}\right)=x^{m}$

Eighth term $\left(a_{8}\right)=x^{52}$

$r=\frac{a_{2}}{a_{1}}=\frac{x^{m}}{x^{-4}}=x^{m+4}$...(ii)

Now, $a_{8}=a r^{n-1}=a r^{8-1}=a r^{7}$

$x^{52}=x^{-4} \times r^{7}=x^{-4} \times x^{7(m+4)}$

$x^{52}=x^{-4+7 m+28}=x^{7 m+24}$

Comparing 

$52=7 m+24 \Rightarrow 7 m=52-24=28$

$\Rightarrow m=\frac{28}{7}=4$

Hence m=4

Question 17

Find the geometric progression whose 4th term is 54 and the 7th Term is 1458.

Sol :

In a G.P.,

4 th term $\left(a_{4}\right)=54$

and 7 th term $\left(a_{7}\right)=1458$ Let $a$ be the first term and r be the common ratio, then $a r^{3}=54$ and $a r^{6}=1458$

Dividing

$\frac{a r^{6}}{a r^{3}}=\frac{1458}{54}$

$\Rightarrow r^{3}=27=\left(3^{3}\right)$

$\therefore r=3$

Now, $a r^{3}=54$

$\Rightarrow a \times 27=54 \Rightarrow a=\frac{54}{27}=2$

Hence G.P. is

2,6,18,54, ....

Question 18

The fourth term of a GP. is the square of its second term and the first term is – 3. Determine its seventh term.

Sol :

In a GP.

$a_{n}$ is square of $a_{2}$ i.e. 

$a n=\left(a_{2}\right)^{2}$ $a_{1}=-3$

Let $a$ be the first term and $r$ be the common ratio, then $\quad a_{4}=a r^{n-1}=a r^{4-1}=a r^{3}$

and $a_{2}=a r^{2}-1=a r$

$\therefore a r^{3}=(a r)^{2}$

$\Rightarrow a r^{3}=a^{2} r^{2}$

$\frac{r^{3}}{r^{2}}=\frac{a^{2}}{a}=r=a=-3 \quad\left(\because a_{1}=-3\right)$

Now, $a_{7}=a r^{7-1}=a r^{6}=(-3)(-3)^{6}=(-3)^{7}$

=-2187

Question 19

The sum of first three terms of a G.P. is $\frac{39}{10}$  and their product is 1. Find the common ratio and the terms.

Sol :

Sum of first three terms of G.P. $=\frac{39}{10}$ and their product =1

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

Let $\frac{a}{r}, a, a r$ be the three terms of $\mathrm{G}$. ., then

$\frac{a}{r}+a+a r=\frac{39}{10}$ and $\frac{a}{r} \times a \times a r=1$

$\Rightarrow a\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$ and $a^{3}=1 \Rightarrow a=1$

Substitute the value of $a$

$\therefore 1\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$

$\frac{1+r+r^{2}}{r}=\frac{39}{10}$

$10+10 r+10 r^{2}=39 r$

$\Rightarrow 10 r^{2}+10 r-39 r+10=0$

$\Rightarrow 10 r^{2}-29 r+10=0$

$\left\{\begin{array}{c}\because 10 \times 10=100 \\ \therefore 100=-25 \times(-4) \\ -29=-25-4\end{array}\right\}$

$\Rightarrow 10 r^{2}-25 r-4 r+10=0$

$\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$

$\Rightarrow(2 r-5)(5 r-2)=0$

Either 2r-5=0, then $r=\frac{5}{2}$

or 5r-2=0, then $r=\frac{2}{5}$

Hence $r=\frac{5}{2}$ or $\frac{2}{5}$

and terms will be if $r=\frac{5}{2}$

$1, \frac{5}{2}, \frac{25}{4}, \frac{125}{8}, \ldots$

if $r=\frac{2}{5},$ then terins will be

$1, \frac{2}{5}, \frac{4}{25}, \frac{8}{125}, \ldots$

Question 20

Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.

Sol :

Given: Three numbers are in A.P. and their sum = 15

Let a – d, a, a + d be the three number in A.P.

$\therefore a-d+a+a+d=15 \Rightarrow 3 a=15$

$\Leftrightarrow a=\frac{15}{3}=5$

By adding 1,4,19 in then,

We get

a-d+1, a+4, a+d+19

These are in G.P. $\therefore b^{2}=a \cdot c$

$\therefore(a+4)^{2}=(a-d+1)(a+d+19)$

$\Rightarrow a^{2}+8 a+16=a^{2}+a d+19 a-a d-d^{2}-19 d+a+d+19$

$\Rightarrow a^{2}+8 a+16=a^{2}-d^{2}-18 d+20 a+19$

$\Rightarrow 8 a+16=20 a-18 d-d^{2}+19$

$\Rightarrow 8 a+16-20 a+18 d+d^{2}-19=0$

$\Rightarrow d^{2}+18 d-12 a-3=0$

$\Rightarrow d^{2}+18 d-12 \times 5-3=0$

$\Rightarrow d^{2}+18 d-60-3=0$

$\Rightarrow d^{2}+18 d-63=0$

$\left\{\begin{aligned} \because-63 &=+21 \times-3 \\+& 18=+21-3 \end{aligned}\right\}$

$\Rightarrow d^{2}+21 d-3 d-63=0$

$\Rightarrow d(d+21)-3(d+21)=0$

$\Rightarrow(d+21)(d-3)=0$

Either d+21=0, then d=-21

or d-3=0, then d=+3

If d=3 and a=5, then G.P.

5-3,5,5+3 i.e. 2,5,8

If d=-21, then

5+21,5,5-21

$\Rightarrow 26,5,-16$

Question 21

Three numbers form an increasing G.P. If the middle term is doubled, then the new numbers are in A.P. Find the common ratio of the G.P.

Sol :

Three numbers form an increasing G.P.

Let $\frac{a}{r}$ ,a,ar be three numbers in G.P.

Double the middle term, we get

$\frac{a}{r}, 2 a, a r$ will be in A.P.

If $2(2 a)=\frac{a}{r}+a r$

If $4 a=a\left(\frac{1}{r}+r\right)$

If $4=\frac{1}{r}+r \Rightarrow 4 r=1+r^{2}$

$\Rightarrow r^{2}-4 r+1=0$

$\Rightarrow r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-4) \pm \sqrt{(-4)^{2}-4 \times 1 \times 1}}{2 \times 1}$

$=\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm \sqrt{12}}{2}$

$=\frac{4 \pm 2 \sqrt{3}}{2}=2 \pm \sqrt{3}$

$\therefore r=2 \pm \sqrt{3}$

$\because$ The numbers are increasing.

$\therefore r=2+\sqrt{3}$

Question 22

Three numbers whose sum is 70 are in GP. If each of the extremes is multiplied by 4 and the mean by 5, the numbers will be in A.P. Find the numbers.

Sol :

Three numbers are in G.P.

Let numbers be

$\frac{r}{a}, a, a r$

$\therefore \frac{a}{r}+a+a r=70$

$\Rightarrow a\left(\frac{1}{r}+1+r\right)=70 \ldots(i)$

By multiplying the extremes by 4 and mean by $5,$ then

$\frac{a}{r} \times 4, a \times 5, a r \times 4$

$\frac{4 a}{r}, 5 a, 4 a r$

But these are in A.P.

$\therefore 2(5 a)=\frac{4 a}{r}+4 a r$

$10 a=4 a\left(\frac{1}{r}+r\right)$

$5=2\left(\frac{1}{r}+r\right)$

$\Rightarrow 5 r=2+2 r^{2}$

$\Rightarrow 2 r^{2}-5 r+2=0$

$\Rightarrow 2 r^{2}-r-4 r+2=0$

$\left\{\begin{array}{l}\because 2 \times 2=4 \\ \therefore 4=-1 \times(-4) \\ -5=-1-4\end{array}\right\}$

$\Rightarrow r(2 r-1)-2(2 r-1)=0$

$\Rightarrow(2 r-1)(r-2)=0$

Either 2r-1=0, then $r=\frac{1}{2}$

or r-2=0, then r=2

From (i)

$a\left(\frac{1}{2}+1+2\right)=70$

$\frac{7}{2} a=70$

$\Rightarrow a=70 \times \frac{2}{7}=20$

$\therefore$ Numbers are (if $r=2)$

$\frac{20}{2}, 20,20 \times 2 \Rightarrow 10,20,40$

If $r=\frac{1}{2},$ then

$\frac{20}{\frac{1}{2}}, 20,20 \times \frac{1}{2}$

$\Rightarrow \frac{20 \times 2}{1}, 20,20 \times \frac{1}{2}$

$\Rightarrow 40,20,10$

Question 23

There are four numbers such that first three of them form an A.P. and the last three form a GP. The sum of the first and third number is 2 and that of second and fourth is 26. What are these numbers?

Sol :

There are 4 numbers, such that

First 3 numbers are in A.P. and

last 3 numbers are in GP.

Sum of first and third numbers = 2

and sum of 2nd and 4th = 26

Let a, b, c and d are numbers such that

a, b, c are in A.P.

$\therefore 2 b=a+c$...(i)

and b, c, d are in G.P. $\therefore c^{2}=b d$...(ii)

Also. a+c=2 

$\Rightarrow 2 b=2 \Rightarrow b=\frac{2}{2}=1$...(iii)

and b+d=26 $\Rightarrow 1+d=26$

$\Rightarrow d=26-1=25$...(iv)

Now, $c^{2}=b d=1 \times 25=25=(5)^{2}$

$\therefore c=5$

$a=2 b-c=2 \times 1-5=2-5=-3$

$\therefore$ Numbers are -3,1,5,25

Question 24

(i) If a, b, c are in A.P. as well in G.P., prove that a = b = c.

(ii) If a, b, c are in A.P as well as in G.P., then find the value of $a^{b-c}+b^{c-a}+c^{a-b}$

Sol :

(i) a, b, c are in A.P. as well as in GP.

To prove: a = b = c

a, b, c are in A.P.

$\therefore 2 b=a+c \Rightarrow b=\frac{a+c}{2}$...(i)

$\therefore a, b, c$ are in G.P. 

$\therefore b^{2}=a c$...(ii)

$\Rightarrow\left(\frac{a+c}{2}\right)^{2}=a c$

$\Rightarrow \frac{(a+c)^{2}}{4}=a c$

$\Rightarrow(a+c)^{2}=4 a c$

$\Rightarrow(a+c)^{2}-4 a c=0$

$\Rightarrow(a-c)^{2}=0$

$\Rightarrow a-c=0$

$\Rightarrow a=c$...(iii)

From $(i), 2 b=a+c=a+a=2 a$

$\therefore b=a$...(iv)

From (iii) and (iv)

Hence , a=b=c

(ii) a, b, c are in A.P. as well as in G.P. 

$\therefore 2 b=a+c$

and $b^{2}=a c$

and a=b=c [proved in (i)]

Now, $a^{b-c}+b^{c-a}+c^{a-b}$

Since, a=b=c

$\therefore b-c=0, c-a=0$ and $a-b=0$

$\therefore a^{0}+b^{0}+c^{0}=1+1+1$ $\left(\because x^{0}=1\right)$

=3

Question 25

The terms of a G.P. with first term a and common ratio r are squared. Prove that resulting numbers form a G.P. Find its first term, common ratio and the nth term.

Sol :

In a G.P.,

The first term = a

and common ratio = r

GP. is a, ar, ar²

Squaring we get

$a^{2}, a^{2} r^{2}, a^{2} r^{4}$ are in G.P.

if $b^{2}=4 a c$

$\Rightarrow\left(a^{2} r^{2}\right)^{2}=a^{2} \times a^{2} r^{4} \Rightarrow a^{4} r^{4}=a^{4} r^{4}$

Which is true The first term is $a^{2}$ and common ratio is $r^{2}$ 

The n th term will be $a_{n}=a r^{n-1}=a^{2}\left(r^{n-1}\right)^{2}=a^{2} r^{2 n-2}$

Question 26

Show that the products of the corresponding terms of two G.P.’s a, ar, ar², …, a r^{n-1} \text { and } A, A R, A R^{2}, \ldots, A R^{n-1} form a G.P. and find the common ratio.

Sol :

It has to be proved that the sequence

$\mathrm{aA}, \operatorname{arAR}, \mathrm{ar}^{2} \mathrm{AR}^{2}, \ldots, \mathrm{ar}^{n-1} \mathrm{AR}^{n-1}$ and forms a G.P.

Hence, $\frac{\text { Second term }}{\text { First term }}=\frac{a r \mathrm{AR}}{a \mathrm{~A}}=r \mathrm{R}$

and $\frac{\text { Third term }}{\text { Second term }}=\frac{a r^{2} A R^{2}}{a r A R}=r R$

Thus, the above sequence forms a G.P. and the common ratio is rR

Question 27

(i) If a, b, c are in G.P. show that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are also in G.P.

(ii) If K is any positive real number and $K^{a}, K^{b} K^{c}$ are three consecutive terms of a G.P., prove that a, b, c are three consecutive terms of an A.P.

(iii) If p, q, r are in A.P., show that pth, qth and rth terms of any G.P. are themselves in GP.

Sol :

(i) a, b, c are in G.P.

$\therefore b^{2}=a c$

$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ will be in G.P.

if $\left(\frac{1}{b}\right)^{2}=\frac{1}{a} \times \frac{1}{c} \Rightarrow \frac{1}{b^{2}}=\frac{1}{a c}$

$\Rightarrow a c=b^{2}$ (By cross multiplication)

which is given

Hence proved

(ii) $\mathrm{K}$ is any positive number $\mathrm{K}^{a}, \mathrm{~K}^{b}, \mathrm{~K}^{c}$ are in G.P.

then $\left(\mathrm{K}^{b}\right)^{2}=\mathrm{K}^{a} \times \mathrm{K}^{c}$

$\Rightarrow \mathrm{K}^{2 b}=\mathrm{K}^{a+c}$

$\Rightarrow 2 b=a+c$

Hence, a, b, c are in A.P.

(iii) $p, q, r$ are in A.P. $\therefore 2 q=p+r$

p th term in $\mathrm{G.P.}=\mathrm{AR}^{p-1}$

q th term $=\mathrm{AR}^{q-1}$

rth term $=\mathrm{AR}^{r-1}$

These will be in G.P. if $\left(\mathrm{AR}^{q-1}\right)^{2}=\mathrm{AR}^{p-1} \times \mathrm{AR}^{r-1}$

if $\mathrm{A}^{2} \mathrm{R}^{2 q-2}=\mathrm{A}^{2} \mathrm{R}^{p-1+r-1}$

if $A^{2} R^{2 q-2}=A^{2} R^{p+r-2}$

if $R^{2 q-2}=R^{p+r-2}$

Comparing, we get $2 q-2=p+r^{\bullet}-2$

$\Rightarrow 2 q=p+r$

$\Rightarrow p, q, r$ are in A.P. which is given

Hence proved

Question 28

If a, b, c are in GP., prove that the following are also in G.P.

(i) $a^{3}, b^{3}, c^{3}$

(ii) $a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$

Sol :

(i) a, b, c are in G.P.

$\therefore b^{2}=a c$

$a^{3}, b^{3}, c^{3}$ are in G.P.

if $\left(b^{3}\right)^{2}=a^{3} \times c^{3}$

if $\left(b^{2}\right)^{3}=(a \times c)^{3}$

if $b^{2}=a c$

which is given

Hence proved

(ii) $a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$ will be in G.P. $a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$ are in G.P.

If $\frac{a b+b c}{a^{2}+b^{2}}=\frac{b^{2}+c^{2}}{a b+b c}$

i.e., if $\frac{a(a r)+\operatorname{ar}\left(a r^{2}\right)}{a^{2}+(a r)^{2}}=\frac{(a r)^{2}+\left(a r^{2}\right)^{2}}{a(a r)+a r\left(a r^{2}\right)}$

$\Rightarrow$ If $\frac{a^{2} r\left(1+r^{2}\right)}{a^{2}\left(1+r^{2}\right)}=\frac{a^{2} r^{2}\left(1+r^{2}\right)}{a^{2} r\left(1+r^{2}\right)}$

$\Rightarrow$ If r=r which is true. $\therefore a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$ are in G.P.

Question 29

If a, b, c, d are in G.P., show that

(i) $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P.

(ii) $(b-c)^{2}+(c-a)^{2}+(d-b)^{2}=(a-d)^{2}$

Sol :

a, b, c, d are in G.P.

Let r be the common ratio, then a = a

$b=a r, c=a r^{2}, d=a r^{3}$

(i) $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P

$\therefore a^{2}+b^{2}=a^{2}+a^{2} r^{2}=a^{2}\left(1+r^{2}\right)$

$b^{2}+c^{2}=a^{2} r^{2}+a^{2} r^{4}=a^{2} r^{2}\left(1+r^{2}\right)$

$c^{2}+d^{2}=a^{2} r^{4}+a^{2} r^{6}=a^{2} r^{4}\left(1+r^{2}\right)$

$a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ will be in G.P.

if $\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$

Now, $\left(b^{2}+c^{2}\right)^{2}=\left[a^{2} r^{2}\left(1+r^{2}\right)\right]^{2}$

$=a^{4} r^{4}\left(1+r^{2}\right)^{2}$...(i)

$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left[a^{2}\left(1+r^{2}\right)\right]\left[a^{2} r^{4}\left(1+r^{2}\right)\right]^{2}$

$=a^{4} r^{4}\left(1+r^{2}\right)^{2}$...(ii)

From (i) and (ii)

$\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$

Hence, $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P.

(ii) Show that

$(b-c)^{2}+(c-a)^{2}+(d-b)^{2}=(a-d)^{2}$

L.H.S. $=(b-c)^{2}+(c-a)^{2}+(d-b)^{2}$

$=\left(a r-a r^{2}\right)^{2}+\left(a r^{2}-a\right)^{2}+\left(a r^{3}-a r\right)^{2}$

$=a^{2} r^{2}(1-r)^{2}+a^{2}\left(r^{2}-1\right)^{2}+a^{2} r^{2}\left(r^{2}-1\right)^{2}$

$=a^{2}\left[r^{2}\left(1-r^{2}-2 a r\right)+r^{4}-2 r^{2}+1+r^{2}\left(r^{4}-\right.\right.\left.\left.2 r^{2}+1\right)\right]$

$=a^{2}\left[r^{2}-r^{4}-2 a r^{3}+r^{4}-2 r^{2}+1+r^{6}-2 r^{4}\right.\left.\left.+r^{2}\right)\right]$

$=a^{2}\left(r^{6}-2 r^{3}+1\right)$

R.H.S. $=(a-d)^{2}=\left(a-a r^{3}\right)^{2}=a^{2}\left(1-r^{3}\right)^{2}$

$=a^{2}\left(1+r^{6}-2 r^{2}\right)$

$=a^{2}\left[r^{6}-2 r^{2}+1\right]$

∴L.H.S=R.H.S

Question 30

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 2nd hour, 4th hour and nth hour?

Sol :

Bacteria in the beginning = 30 = a

After every hours, it doubles itself

After 1 hour it becomes = 30 x 2 = 60 = $a^r$

After 2 hours it will becomes = 60 x 2 = 120 = $a^2$

After 3 hours, it will becomes = 120 x 2 = 240 = $a^3$

After 4 hours it will becomes = 240 x 2 = 480 = $a^4$

∴ After n hour, it will become = $ar^n$

Question 31

The length of the sides of a triangle form a G.P. If the perimeter of the triangle is 37 cm and the shortest side is of length 9 cm, find the lengths of the other two sides.

Sol :

Lengths of a triangle are in GP. and its sum is 37 cm

Let sides be $a, a r, a r^{2}$ 

$a+a r+a r^{2}=37$

a=9

$\therefore 9+9 r+9 r^{2}=37$

$\Rightarrow 9 r+9 r^{2}=37-9=28$

$9 r^{2}+9 r-28=0$

$\Rightarrow 9 r^{2}+21 r-12 r=28=0$

$\left\{\begin{array}{l}\because 28 \times 9=252 \\ \therefore 252=21 \times-12 \\ 9=21-12\end{array}\right\}$

$\Rightarrow 3 r(3 r+7)-4(3 r+7)=0$

(3r+7)(3r-4)=0

3r+7=0 then  3r=-7, 

$r=\frac{-7}{3}$

or (3r-4)=0 then 3r=4$\Rightarrow r=\frac{4}{3}$

Sides are $9,9 \times \frac{4}{3}, 9 \times \frac{4}{3} \times \frac{4}{3}$

=9,12,16 cm