MATHS M.C.Q (1 Marks)

2. $\frac{1}{11}$

Solution:

Let the first term of the G.P. be a.

Let its common ratio be r.

​According to the question, we have:

First term = 10 [Sum of all successive terms]

$\text{a}=10\Big(\frac{\text{ar}}{1-\text{r}}\Big)$

$\Rightarrow\text{a}-\text{ar}=10\text{ar}$

$\Rightarrow11\text{ar}=\text{a}$

$\Rightarrow\text{r}=\frac{\text{a}}{11\text{a}}=\frac{1}{11}$

2. 1

Solution:

a, b and c are in A.P.

$\therefore2\text{b}=\text{a}+\text{c}\ \cdots(\text{i})$

And, x, y and z are in G.P.

$\therefore\text{y}^2=\text{zy}$

Now, $\text{x}^{\text{b}-\text{a}}\text{ y}^{\text{c}-\text{a}}\text{ z}^{\text{a}-\text{b}}$

$=\text{x}^{\text{b}+\text{a}-2\text{b}}\text{ y}^{2\text{b}-\text{a}-\text{a}}\text{ z}^{\text{a}-\text{b}}$ [From (i)]

$=\text{x}^{\text{a}-\text{b}}\text{ y}^{2(\text{b}-\text{a})}\text{ z}^{\text{a}-\text{b}}$

$=(\text{xz})^{\text{a}-\text{b}}(\text{xz})^{\text{b}-\text{a}}$ $[\text{From}(\text{ii}),\text{y}^2=\text{xz}]$

$=(\text{xz})^0$

$=1$

1. AP

Solution:

a, b and c are in G.P.

$\therefore\text{b}^2=\text{ac}$

Taking $\log$ on both the sides:

$2\log\text{b}=\log\text{a}+\log\text{c}\ \cdots(\text{i})$

Now, $\text{a}^{\frac{1}{\text{x}}}=\text{b}^{\frac{1}{\text{y}}}=\text{c}^{\frac{1}{\text{z}}}$

Taking $\log$ on both the sides:

$\frac{\log\text{a}}{\text{x}}=\frac{\log\text{b}}{\text{y}}=\frac{\log\text{c}}{\text{z}}\ \cdots(\text{ii} )$

Now, comparing (i) and (ii):

$\frac{\log\text{a}}{\text{x}}=\frac{\log\text{a}+\log\text{c}}{2\text{y}}=\frac{\log\text{c}}{\text{z}}$

$\Rightarrow\frac{\log\text{a}}{\text{x}}=\frac{\log\text{a}+\log\text{c}}{2\text{y}}\text{ and }\frac{\log\text{a}}{\text{x}}=\frac{\log\text{c}}{\text{z}}$

$\Rightarrow\log\text{a}(2\text{y}-\text{x})=\text{x}\log\text{c}\text{ and }\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{\text{z}}$

$\Rightarrow\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{(2\text{y}-\text{x})}\text{ and }\frac{\log\text{a}}{\log\text{c}}=\frac{\text{x}}{\text{z}}$

$\Rightarrow\frac{\text{x}}{2\text{y}-\text{x}}=\frac{\text{x}}{\text{z}}$

$\Rightarrow2\text{y}=\text{x}+\text{z}$

Thus, x, y and z are in A.P.

1.  $\frac12$

Solution:

$\frac{\text{a}}{1-\text{r}}=3$

$\text{a}=3-3\text{r}$

Sum of square terms of G.P. is $\frac{\text{a}^2}{1-\text{r}^2}=3$

$\Rightarrow\frac{\text{a}}{1-\text{r}}=\frac{\text{a}^2}{1-\text{r}^2}$

or $\text{a}=1+\text{r}\ \dots(2)$

Solving (1) and (2),

$\text{a}=\frac32\text{ and r}=\frac12$

1.  1

Solution:

Let the firt term of the geometric progression = x

Common ration = 2

$\therefore$ 2nd term of the G.P. = 2x

$\therefore$ 3rd term = (2²)x ...

N th term can be written as $= (2^\text{n})\text{x}$

Sum of the n terms S = 255

as we can see, except x, all other terms in the G.P. are multiples of 2

and sum of all the terms is an odd number.

$\therefore$ x must be an odd number.

now n^th term

$(2^{\text{n}})\text{x}=128=\big(2^7\big)\times1$

There are no factors of odd numbers in 128, except 1

$\therefore$ x = 1

Series of G.P. is:

1, 2, 4, 8, 16, 32, 64, 128

Checking the sum of the n terms,

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

$\therefore$ First term of the G.P. = 1

4.  $\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$

Solution:

a, b and c are in G.P.

$\therefore\text{b}^2=\text{ac}\cdots(\text{i})$

a, x and b are in A.P.

$\therefore2\text{x}=\text{a}+\text{b}\ \cdots(\text{ii})$

Also, b, y and c are in A.P.

$\therefore2\text{y}=\text{b}+\text{c}$

$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}^2}{\text{a}}$ [Using (i)]

$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}^2}{(2\text{x}-\text{b})}$ [Using (ii)]

$\Rightarrow2\text{y}=\text{b}+\frac{\text{b}(2\text{x}-\text{b})+\text{b}^2}{(2\text{x}-\text{b})}$

$\Rightarrow2\text{y}=\frac{2\text{bx}-\text{b}^2+\text{b}^2}{(2\text{x}-\text{b})}$

$\Rightarrow2\text{y}=\frac{2\text{b}\text{x}}{(2\text{x}-\text{b})}$

$\Rightarrow\text{y}=\frac{\text{bx}}{(2\text{x}-\text{b})}$

$\Rightarrow\text{y}(2\text{x}-\text{b})=\text{bx}$

$\Rightarrow2\text{xy}-\text{by}=\text{bx}$

$\Rightarrow\text{bx}+\text{by}=2\text{xy}$

Dividing both the sides by xy:

$\Rightarrow\frac{1}{\text{y}}+\frac{1}{\text{x}}=\frac{2}{\text{b}}$

3.  $\frac{2355}{999}$

Solution:

$2.\overline{357}=2.0+0.357+0.000357+0.000000357+\dots\infty$

$\Rightarrow2.\overline{357}=2+\Big[\frac{357}{10^3}+\frac{357}{10^6}+\frac{357}{10^9}+\dots\infty\Big]$

$\Rightarrow2.\overline{357}=2+\frac{\frac{357}{10^3}}{1-\frac{1}{10^3}}$

$\Rightarrow2.\overline{357}=2+\frac{357}{999}$

$\Rightarrow2.\overline{357}=\frac{2355}{999}$ 

1. 1 : 1

Solution:

Let the three terms of the G.P. be $\frac{\text{a}}{\text{r}},\text{a},\text{ar}.$ Then

$\text{S}=\frac{\text{a}}{\text{r}}+\text{a}+\text{ar}$

$=\text{a}\Big(\frac{1}{\text{r}}+1+\text{r}\Big)$

$=\text{a}\Big(\frac{1+\text{r}+\text{r}^2}{\text{r}}\Big)$

$=\frac{\text{a}(\text{r}^2+\text{r}+1)}{\text{r}}$

Also,

$\text{P}=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=\text{a}^3$

And,

$\text{R}=\frac{\text{r}}{\text{a}}+\frac{1}{\text{a}}+\frac{1}{\text{ar}}$

$=\frac{1}{\text{a}}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)$

Now,

$\frac{\text{P}^2\text{R}^2}{\text{S}^3}=\frac{\big(\text{a}^3\big)^2\times\Big[\frac{1}{\text{a}}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)\Big]^3}{\Big[\text{a}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)\Big]^3}$

$=\frac{\text{a}^6\times\frac{1}{\text{a}^3}\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)^3}{\text{a}^3\Big(\frac{\text{r}^2+\text{r}+1}{\text{r}}\Big)^3}$

$=\frac11$

So, the ratio is 1 : 1.

Hence, the correct alternative is option (a).

1. $-\frac{2}{5}$

Solution:

If the first term is 1, then, the G.P. will be $1, \text{r},\text{r}^2,\text{r}^3,\ \dots$

Now, $5\text{r}^2+4\text{r}=5\Big(\text{r}^2+\frac45\text{r}\Big)$

$=5\Big(\text{r}^2+\frac45\text{r}+\frac{4}{25}-\frac{4}{25}\Big)$

$=5\Big(\text{r}+\frac25\Big)^2-\frac45$

This will be the least when $\text{r}+\frac25=0,\text{ i.e. }\text{r}=-\frac25.$

2. 2

Solution:

Let the two numbers be a and b.

a, x and b are in A.P.

$\therefore2\text{x}=\text{a}+\text{b}\ \cdots(\text{i})$

Also, a, y, z and b are in G.P.

$\therefore\frac{\text{y}}{\text{a}}=\frac{\text{z}}{\text{y}}=\frac{\text{b}}{\text{z}}$

$\Rightarrow\text{y}^2=\text{az},\text{yz}=\text{ab},\text{z}^2=\text{by}\ \cdots(\text{ii})$

Now, $\frac{\text{y}^3+\text{z}^3}{\text{xyz}}$

$=\frac{\text{y}^2}{\text{xz}}+\frac{\text{z}^2}{\text{xy}}$

$=\frac{1}{\text{x}}\Big(\frac{\text{y}^2}{\text{z}}+\frac{\text{z}^2}{\text{y}}\Big)$

$=\frac{1}{\text{x}}\Big(\frac{\text{az}}{\text{z}}+\frac{\text{by}}{\text{y}}\Big)$ [Using (ii)]

$=\frac{1}{\text{x}}(\text{a}+\text{b})$

$=\frac{2}{(\text{a}+\text{b})}(\text{a}+\text{b})$ [Using (i)]

$=2$

1.  $\frac12$

Solution:

$\text{S}=\frac{1}{1+\text{x}}-\frac{1-\text{x}}{(1+\text{x})^2}+\frac{(1-\text{x})^2}{(1+\text{x})^3}-\frac{(1-\text{x})^3}{(1+\text{x})^4}+\cdots\infty$

It is clear that it is a G.P. with $\text{a}=\frac{1}{1+\text{x}}$ and $\text{r}=-\frac{(1-\text{x})}{(1+\text{x})}.$

$\therefore\text{S}=\frac{\text{a}}{(1-\text{r})}$

$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1-\big(-\frac{(1-\text{x})}{(1+\text{x})}\big)\Big]}$

$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1+\frac{(1-\text{x})}{(1+\text{x})}\Big]}$

$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[\frac{(1+\text{x})+(1-\text{x})}{(1+\text{x})}\Big]}$

$\Rightarrow\text{S}=\frac12$

1.  $\frac{\text{p}^3+\text{q}^3}{\text{pq}}$

Solution:

Let the two positive numbers be a and b.

a, A and b are in A.P.

$\therefore2\text{A}=\text{a}+\text{b}\cdots(\text{i})$

Also, a, p, q and b are in G.P.

$\therefore\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}$

Again, p = ar and q = ar² ____(ii)

Now, 2A = a + b [From (i)]

$=\text{a}+\text{a}\Big(\frac{\text{b}}{\text{a}}\Big)$

$=\text{a}+\text{a}\Bigg(\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{3}}\Bigg)^3$

$=\text{a}+\text{ar}^3$

$=\frac{(\text{ar})^2}{\text{ar}^2}+\frac{\big(\text{ar}^2\big)^2}{\text{ar}}$

$=\frac{\text{p}^2}{\text{q}}+\frac{\text{q}^2}{\text{p}}$ [Using (ii)]

$=\frac{\text{p}^3+\text{q}^3}{\text{pq}}$

2.  $\text{x}+1$

Solution:

$\sum\limits^\infty_{​​\text{x}+1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{(\text{n}-1)}$

$=1+\Big(\frac{\text{x}}{\text{x}+1}\Big)+\Big(\frac{\text{x}}{\text{x}+1}\Big)^2+\Big(\frac{\text{x}}{\text{x}+1}\Big)^3+\Big(\frac{\text{x}}{\text{x}+1}\Big)^4+\dots\infty$

$=\frac{1}{1-\big(\frac{\text{x}}{\text{x}+1}\big)}$ $\Big[\because$ it is a G.P. with a = 1 and $\text{r}=\Big(\frac{\text{x}}{\text{x}+1}\Big)\Big]$

$=\frac{(\text{x}+1)}{(\text{x}+1-\text{x})}$

$=\frac{(\text{x}+1)}{1}=(\text{x}+1)$

4. 8

Solution:

The first and the last numbers are equal.

Let the four given numbers be p, q, r and p.

The first three of four given numbers are in G.P.

$\therefore\text{q}^2=\text{p}\cdot\text{r}\cdots(\text{i})$

And, the last three numbers are in A.P. with common difference 6.

We have:

First term = q

Second term = r = q + 6

Third term = p = q + 12

Also, 2r = q + p

Now, putting the values of p and r in (i):

q² = (q + 12)(q + 6)

⇒ q² = q² + 18q + 72

⇒ 18q + 72 = 0

⇒ q + 4 = 0

⇒ q = -4

Now, putting the value of q in p = q + 12:

p = -4 + 12 = 8

2.  3

Solution:

$9^{\frac13}\times9^{\frac19}\times9^{\frac{1}{27}}\times\dots\infty$

$=9^{\big(\frac13+\frac19+\frac{1}{27}+\dots\infty\big)}$

Here, it is a G.P. with $\text{a}=\frac13$ and $\text{r}=\frac13.$

$\therefore9^{\Bigg(\frac{\frac{1}{3}}{1-\frac13}\Bigg)}$

$=9^{\big(\frac12\big)}=3$

2.  $\frac{\text{q}-\text{r}}{\text{p}-\text{q}}$

Solution:

Let a be the first term and d be the common difference of the given A.P.

Then, we have:

p^th term, ap = a + (p−1)d

q^th term, aq = a + (q−1)d

r^th term, ar = a + (r−1)d

Now, according to the question the p^th, the q^th and the r^th terms are in G.P.

$\therefore (\text{a} + (\text{q}−1)\text{d})^2=( \text{a} + (\text{p}−1)\text{d})\times(\text{a} + (\text{r}−1)\text{d})$

$\Rightarrow\text{a}^2+2\text{a} (\text{q}−1)\text{d}+( (\text{q}−1)\text{d})^2\\\ \ =\text{a}^2+\text{ad}(\text{r}−1+\text{p}−1)+(\text{p}−1) (\text{r}−1)\text{d}^2$

$\Rightarrow(2\text{q}−2−\text{r}−\text{p}+2)+\text{d}^2(\text{q}^2−2\text{q}+1−\text{pr}+\text{p}+\text{r}−1)=0$

$\Rightarrow(2\text{q}−\text{r}-\text{p})+\text{d}(\text{q}^2−2\text{q}−\text{pr}+\text{p}+\text{r})=0$

$\Rightarrow\text{a}=-\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(2\text{q}-\text{r}-\text{p})}$

$\therefore\text{Common ratio},\text{ r}=\frac{\text{a}_\text{q}}{\text{a}_\text{p}}$

$=\frac{\text{a}+(\text{q}-1)\text{d}}{\text{a}+(\text{p}-1)\text{d}}$

$=\frac{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{q}-1)\text{d}}{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{p}-1)\text{d}}$

$=\frac{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{pq}+\text{rq}+\text{rq}-2\text{q}^2-\text{p}-\text{r}+2\text{q}}{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{p}^2+\text{pr}-2\text{pq}-\text{p}-\text{r}+2\text{q}}$

$=\frac{\text{pq}-\text{pr}-\text{q}^2+\text{qr}}{\text{p}^2+\text{q}^2-2\text{pq}}$

$=\frac{\text{p}(\text{p}-\text{r})-\text{q}(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$

$=\frac{(\text{p}-\text{q})(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$

$=\frac{(\text{q}-\text{r})}{(\text{p}-\text{q})}$ 

4.  $4.$

Solution:

$\text{a}_2=2$

$\therefore\text{ar}=2\ \cdots(\text{i})$

Also, $\text{S}_\infty=8$

$\Rightarrow\frac{\text{a}}{(1-\text{r})}=8$

$\Rightarrow\frac{\text{a}}{\Big(1-\frac{2}{\text{a}}\Big)}=8$ [Using (i)]

$\Rightarrow\text{a}^2=8(\text{a}-2)$

$\Rightarrow\text{a}^2-8\text{a}+16=0$

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

$\Rightarrow\text{a}=4$

3. 4

Solution:

Let there be 2n terms in a G.P.

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

$\because\text{ S}_{2\text{n}}=5(\text{S}_{\text{odd terms}})$

$\Rightarrow\frac{\text{a}\big(\text{r}^{2\text{n}-1}\big)}{(\text{r}-1)}=5\big(\text{a}+\text{ar}^2+\text{ar}^4+\text{ar}^6+\dots\text{ar}^{(2\text{n}-1)}\big)$

$\Rightarrow\frac{\text{a}\big(\text{r}^{2\text{n}}-1\big)}{(\text{r}-1)}=5\Bigg(\frac{\text{a}\big(\big(\text{r}^2\big)^\text{n}\big)}{\big(\text{r}^2-1\big)}\Bigg)$

$\Rightarrow\frac{\big(\text{r}^{2\text{n}}-1\big)}{(\text{r}-1)}=5\frac{\big(\big(\text{r}^2\big)^\text{n}-1\big)}{\big(\text{r}^2-1\big)}$

$\Rightarrow\frac{\big(\big(\text{r}^\text{n}\big)^2-1^2\big)}{(\text{r}-1)}=5\frac{\big(\big(\text{r}^\text{n}\big)^2-1^2\big)}{\big(\text{r}^2-1\big)}$

$\Rightarrow\frac{\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)}{(\text{r}-1)}=5\frac{\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)}{\big(\text{r}-1\big)\big(\text{r}+1\big)}$

$\Rightarrow\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)\big(\text{r}-1\big)\big(\text{r}+1\big)\\\ -5\big(\text{r}-1\big)\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)=0$

$\Rightarrow\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)\big(\text{r}-1\big)\big(\text{r}+1-5\big)=0$

But, r = 1 or −1 is not possible.

$\therefore\text{r}=4$

1. 64

Solution:

$32\times32^{\frac{1}{6}}\times32^{\frac{1}{36}}\times\ \cdots\infty$

$=32^{\big(1+\frac{1}{6}+\frac{1}{36}+\ \cdots\infty\big)}$

$=32^{\Bigg(\frac{1}{1-\frac{1}{6}}\Bigg)}$ $[\because$ it is a G.P. $]$

$=32^{\big(\frac65\big)}$

$=\big(2^5\big)^{\big(\frac65\big)}$

$=2^6$

$=64$

3. $\sqrt{\text{pq}}$

Solution:

Here, $\text{a}_{(\text{m}+\text{n})}=\text{p}$

$\Rightarrow\text{ar}^{(\text{m}+\text{n}-1)}=\text{p}\ \cdots(\text{i})$

Also, $\text{a}_{(\text{m}-\text{n})}=\text{q}$

$\Rightarrow\text{ar}^{(\text{m}-\text{n}-1)}=\text{q}\ \cdots(\text{ii})$

Mutliplying (i) and (ii):

$\Rightarrow\text{ar}^{(\text{m}+\text{n}-1)}\text{ar}^{(\text{m}-\text{n}-1)}=\text{pq}$

$\Rightarrow\text{a}^2\text{r}^{(2\text{m}-2)}=\text{pq}$

$\Rightarrow\Big(\text{ar}^{(\text{m}-1)}\Big)^2=\text{pq}$

$\Rightarrow\text{ar}^{(\text{m}-1)}=\sqrt{\text{pq}}$

$\Rightarrow\text{a}_\text{m}=\sqrt{\text{pq}}$

Thus, the m^th term is $\sqrt{\text{pq}}.$

1.  8

Solution:

$\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54}$

$\Rightarrow4^{(3+6+9+12+\ \dots+3\text{x})}=\Big(\frac{625}{10000}\Big)^{-54}$

$\Rightarrow4^{3(1+2+3+4+\dots+\text{x})}=\Big(\frac{1}{16}\Big)^{-54}$

$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(\frac{1}{16}\Big)^{-54}$

$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(4^{-2}\Big)^{-54}$

Comparing both the sides:

$\Rightarrow3\Big(\frac{\text{x}(\text{x}+1)}{2}\Big)=108$

$\Rightarrow\text{x}(\text{x}+1)=72$

$\Rightarrow\text{x}^2+\text{x}-72=0$

$\Rightarrow\text{x}^2+9\text{x}-8\text{x}-72=0$

$\Rightarrow\text{x}(\text{x}+9)-8(\text{x}+9)=0$

$\Rightarrow(\text{x}+9)(\text{x}-8)=0$

$\Rightarrow\text{x}=8,-9$

$\Rightarrow\text{x}=8$ $[\because\text{ x}\text{ is psitive}]$

1.  $(2\text{p}-\text{q})(\text{p}-2\text{q})$

Solution:

Let the two numbers be a and b.

a, p, q and b are in A.P.

$\therefore\text{ p}-\text{a}=\text{q}-\text{q}=\text{b}-\text{q}$

$\Rightarrow\text{ p}-\text{a}=\text{q}-\text{p}\text{ and}\text{ q}-\text{p}=\text{b}-\text{q}$

$\Rightarrow\text{ a}=2\text{p}-\text{q}\text{ and}\text{ b}=2\text{q}-\text{p}\cdots(\text{i})$

Also, a, G and b are in G.P.

$\therefore\text{G}^2=\text{ab}$

$\Rightarrow\text{G}^2=(2\text{p}-\text{q})(2\text{q}-\text{p})$ 

4.  $\frac34.$

Solution:

Let the terms of the G.P. be $\text{a},\text{a}_2,\text{a}_3,\text{a}_4.\text{a}_5,\ \dots,\infty.$

And, let the common ratio be r.

Now, $\text{a}+\text{a}_2=1$

$\therefore\text{a}+\text{ar}=1\dots(\text{i})$

Also, $\text{a}=2(\text{a}_2+\text{a}_3+\text{a}_4+\text{a}_5+\dots\infty)$

$\Rightarrow\text{a}=2\big(\text{ar}+\text{ar}^2+\text{ar}^3+\text{ar}^4+\ \dots\infty\big)$

$\Rightarrow\text{a}=2\Big(\frac{\text{ar}}{1-\text{r}}\Big)$

$\Rightarrow1-\text{r}=2\text{r}$

$\Rightarrow3\text{r}=1$

$\Rightarrow\text{r}=\frac13$

Putting the value of r in (i):

$\text{a}+\frac{a}{3}=1$

$\Rightarrow\frac{4\text{a}}{3}=1$

$\Rightarrow4\text{a}=3$

$\Rightarrow\text{a}=\frac34$

4. $\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}$

Solution:

Sum of n terms of the G.P., $\text{S}=\frac{\text{a}(\text{r}^{\text{n}}-1)}{(\text{r}-1)}$

Product of n terms of the G.P., $\text{P}=\text{a}^{\text{n}}\text{r}^{\big[\frac{\text{n}(\text{n}-1)}{2}\big]}$

Sum of the reciprocals of n terms of the G.P., $\text{R}=\frac{\Big[\frac{1}{\text{r}^\text{n}}-1\Big]}{\text{a}\big(\frac{1}{\text{r}}-1\big)}=\frac{(\text{r}^{\text{n}}-1)}{\text{ar}^{(\text{n}-1)}(\text{r}-1)}$

$\therefore\text{P}^2=\bigg\{\text{a}^2\text{r}^\frac{2(\text{n}-1)}{2}\bigg\}^\text{n}$

$\Rightarrow\text{P}^2=\Bigg\{\frac{\frac{\text{a}(\text{r}^\text{n}-1)}{(\text{r}-1)}}{\frac{(\text{r}^\text{n}-1)}{\text{ar}^{(\text{n}-1)(\text{r}-1)}}}\Bigg\}^\text{n}$

$\Rightarrow\text{P}^2=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$

Let the first term of the G.P. be a and the common ratio be r.

Sum of n terms, $\text{S}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}$

Product of the G.P., $\text{P}=\text{a}^{\text{n}}\text{r}^{\frac{\text{n}(\text{n}+1)}{2}}$

Sum of the reciprocals of n terms, $\Rightarrow\text{R}=\frac{\big(\frac{1}{\text{r}^{\text{n}-1}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}=\frac{\big(\frac{1-\text{r}^{\text{n}}}{\text{r}^\text{n}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}$

$\text{P}^2=\bigg\{\text{a}^2\text{r}^{\frac{(\text{n}+1)}{2}}\bigg\}^{\text{n}}$

$\text{P}^2=\begin{Bmatrix} \frac{\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}}{\frac{\Big(\frac{1-\text{r}^\text{n}}{\text{r}^\text{n}}\Big)}{\text{a}\Big(\frac{1-\text{r}}{\text{r}}\Big)}}\end{Bmatrix}=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$

2. 4 and 16

Solution:

Let the two G.M.s between 1 and 64 be G₁ and G₂.

Thus, 1, G₁, G₂ and 64 are in G.P.

$64=1\times\text{r}^3$

$\Rightarrow\text{r}=\sqrt[3]{64}$

$\Rightarrow\text{r}=4$

$\Rightarrow\text{G}_1=\text{ar}=1\times4=4$

And, $\text{G}_2=\text{ar}^2=1\times4^2=16$

Thus, 4 and 16 are the required G.M.s.