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.  $\frac12$

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),

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

4. 128

Solution:

Let G.P. be 4, G₁,  G₂,  G₃,  512.

⇒ a = 4 and a₅ = a × r⁴ = 512 × 4 × r⁴ = 512 ⇒  r⁴

^{$=\frac{512}{4}=128\Rightarrow\text{r=4.}$}

G₁ = a₂ = a × r = 4 × 4 = 16.

G₂ = G₁ × r = 16  × 4 = 64.

G₃ = G₂ × r = 64 × 4 = 256.

2. 1185

Solution:

6² + 7² +………………..….. + 15²

= (1² + 2² + 3² + …….. +15²) – (1² + 2² + 3² + 4² + 5²)

$=\frac{15\times16\times31}{6}-\frac{5\times6\times11}{6}$

$=1240-55=1185.$

3. $\text{n}-1-\frac{1}{2^{\text{n}}}$

Solution:

We have,

$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}}$

$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1(2^{\text{r}}-1)}{2\text{r}}$

$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\Big(1-\frac{1}{2^{\text{r}}}\Big)$

$\Rightarrow\text{S}_\text{n}=\text{n}-\sum\limits^{\text{n}}_{\text{r}=1}\Big(\frac{1}{2^{\text{r}}}\Big)$

$\Rightarrow\text{S}_\text{n}=\text{n}-\Bigg[\frac{\big(\frac{1}{2}\big)\big\{1-\big(\frac{1}{2}\big)^\text{n}\big\}}{1-\frac{1}{2}}\Bigg]$

$\Rightarrow\text{S}_\text{n}=\text{n}-\Big[1-\Big(\frac{1}{2}\Big)^\text{n}\Big]$

$\Rightarrow\text{S}_\text{n}=\text{n}-1+\frac{1}{2^\text{n}}$

4. 49² + 2.

Solution:

S_n = 2 + 3 + 6 + 11 + 18 + .... + t₅₀

Using method of difference, we get

S_n = 2 + 3 + 6 + 11 + 18 + .... + t₅₀ ....(1)

And S_n = 0 + 2 + 3 + 6 + 11 + .... + t₄₉ + t₅₀ ....(2)

Subtracting eq. (2) from eq. (2), we get

0 = 2 + 1 + 3 + 5 + 7 + .... -t₅₀ terms

⇒ t₅₀ = 2 + (1 + 3 + 5 + 7 + .... upto 49 terms)

$\Rightarrow\text{t}_{50}=2+\frac{49}{2}[2\times1+(49-1)2]=2+\frac{49}{2}[2+96]$

$\Rightarrow2+\frac{49}{2}\times98=2+49\times49=49^2+2$

Hence, the correct option is (d).

2. 12 and 3

Solution:

 We know, A.M. of two numbers a and b is 

$\frac{(\text{a}+\text{b)}}{2}$

$\Rightarrow\frac{(\text{a}+\text{b)}}{2}=\frac{15}{2}\Rightarrow\text{}a+\text{b}=15.$

Also, G.M. of two numbers a and b is $\sqrt{ab}$

$\Rightarrow\sqrt{ab}=6\Rightarrow\text{ab}=36.$

=> a(15-a) = 36 => a=3 or 12.

For a=3, b=12.

For a=12, b=3.

So, the two numbers are 3 and 12.

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

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. $\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}.$

Solution:

$\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\big[\frac{\text{r}(\text{r}+1)}{2}\big]^2}{\frac{\text{r}(\text{r}+1)}{2}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\text{r}(\text{r}+1)}{2}=\frac{1}{2}\bigg[\sum\limits^\text{n}_{\text{r}=1}\text{r}^2+\sum\limits^\text{n}_{\text{r}=1}\text{r}\bigg]$

$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}{+1})(2\text{n}{+1})}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$

$=\frac{1}{2}.\frac{\text{n}({\text{n}+1})}{2}\Big[\frac{2\text{n}{+1}}{3}+1\Big]=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$

$=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$

3. $\frac{\text{n}(\text{n}+1)}{4}$

Solution:

Let $\text{S}_\text{n}=\frac{1}{\log_24}+\frac{1}{\log_44}+\frac{1}{\log_84}+\ ...\ +\frac{1}{\log_2{^\text{n}4}}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log4}{\log4}+\frac{\log8}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{\log2^2}{\log4}+\frac{\log2^3}{\log4}+\ ...\ +\frac{\log2^\text{n}}{\log4}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}+\frac{2\log2}{\log4}+\frac{3\log2}{\log4}+\ ...\ +\frac{\text{n}\log2}{\log4}$

$\Rightarrow\text{S}_\text{n}=\frac{\log2}{\log4}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{\log4^{\frac{1}{2}}}{\log4}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{\frac{1}{2}\log4}{\log4}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{1}{2}(1+2+3+\ ...\ +\text{n})$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}$

2. 6

Solution:

Let the first term be a and common difference be d.

Then,

$\text{S}_{2\text{n}}=3\text{S}_{\text{n}}$

$\Rightarrow\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]=\frac{3\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$

$\Rightarrow2[2\text{a}+(2\text{n}-1)\text{d}]=3[2\text{a}+(\text{n}-1)\text{d}]$

$\Rightarrow4\text{a}+(4\text{n}-2)\text{d}=6\text{a}+(3\text{n}-3)\text{d}$

$\Rightarrow2\text{a}=(\text{n}+1)\text{d}$

Now,

$\frac{\text{S}_{3\text{n}}}{\text{S}_{\text{n}}}=\frac{\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}=\frac{3[2\text{a}+(3\text{n}-1)\text{d}]}{[2\text{a}+(\text{n}-1)\text{d}]}$

$=\frac{3[(\text{n}+1)\text{d}+(3\text{n}-1)\text{d}]}{[(\text{n}+1)\text{d}+(\text{n}-1)\text{d}]}=\frac{3[4\text{nd}]}{2\text{nd}}=6$

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:

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

2. 286

solution:

1² + 3² + 5² +…………………………..+ 11²

= (1² + 2² + 3² +……+ 11²) – (2² + 4² + 6² + 8² + 10²)

= (1² + 2² + 3² +……11²) – 2²(1² + 2² + 3² + 4² +5²)

$=\frac{16\times12\times23}{6}-\frac{4\times5\times6\times11}{6}$

$=506-220=286.$

2. 6

Solution:

Explanation: We know, geometric mean of two numbers a and b is given by

$\text{G}.\text{M}.=\sqrt{\text{a}\times\text{b}}$

$\text{SO},\text{G}.\text{M}.\text{ of }3\text{ and }12\text{ is }\sqrt{3}\times12=\sqrt{36}=6.$

1. $\frac{63}{32}$

Solution:

Given series is G.P. with first term 1 and common ratio

$\frac{1}{2}.$

We know,$\text{s}_{\text{n}}=\text{a}\frac{(1-\text{r}_\text{n})}{(1-\text{r)}}$ for r<1.

${\text{s}}_6=1\frac{(1-(\frac{1}{2})^6}{(1-\frac{1}{2})}=\frac{(1-\frac{1}{64})}{(1+2)}=63\times\frac{2}{64}=\frac{63}{32}.$

3. $\leq2^6$

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 $\text{r}+\frac25=0,\text{ i.e. }\text{r}=-\frac25.$

1. 2556

Solution:

1³ + 3³ + 5³ +…………………………..+ 11³

= (1³ + 2³ + 3³ +……+ 11³) – (2³ + 4³ + 6³ + 8³ + 10³)

= (1³ + 2³ + 3³+……11³) – 2³(1³ + 23 + 3³ + 4³ + 5³)

$\Big(\frac{11\times12}{2}\Big)^2-8\Big(\frac{5\times6}{2}\Big)^2$

$=66^2-8\times15^2$

$4356-1800=2556.$

1. 9 : 1

Soultion:

$\frac{(\text{A.}\text{M.})}{\text{(G.}\text{M.)}}=\frac{5}{3}$

$\Rightarrow\frac{\text{a}+\text{b}}{2\sqrt{a}\text{b}}=\frac{5}{3}$

Applying componendo and dividendo rule, we get

$\Rightarrow\frac{\text{a}+\text{b+2}\sqrt{\text{a}\text{b}}}{\text{a}+\text{b}-2\sqrt{a}\text{b}}=\frac{8}{2}$

$\Rightarrow\Big(\frac{\sqrt{\text{a}+\text{b}}}{\sqrt{\text{a}-\text{b}}}\Big)^2=4$

$\Rightarrow\Big(\frac{\sqrt{\text{a}+\text{b}}}{\sqrt{\text{a}-\text{b}}}\Big)^1=2$

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

Again applying componendo and dividendo rule, we get

$\frac{\text{a}}{\text{b}}=3\Big(\frac{3}{1}\Big)^2=9.\text{ so},\text{a}:\text{b}=9:1$

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

Solution:

Since, x, 2y and 3z are in A. P., we get

$\text{2y}=\frac{\text{x}+3\text{z}}{2}$

⇒ 4y = x + 3z

Also, x, y and z are ibn G.P.

Therefore, y = xr and z = xr².

Where 'r' is the common ratio.

$\therefore$ 4xr = x + 3xr² [Using (1)]

⇒ 4r = 1 + 3r²

⇒ 3r² - 4r + 1 = 0

⇒ (3r - 1)(r - 1) = .0

$\Rightarrow\text{r}=\frac{1}{3}$

(For r = 1; x, y, z are not distinct)

4. 966

Solution:

General term of above series is a_k = 2k*(k²+2) = 2k³+4k

Taking summation from k=1 to k=n on both sides, we get

$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=2\sum\limits^\text{n}_\text{i}=0\text{ k}^3+4\sum\limits^\text{n}_\text{i}=0\text{ k}=2\Big(\frac{(\text{n}(\text{n+1)}}{2}\Big)^2+4\frac{\text{n}(\text{n+1)}}{2}$

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

$=\frac{36\times49}{2}+2\times6\times7=966.$

4. 1296

Solution:

We know, sum of cubes of first n terms is given by

$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2.$

$\text{Here},\text{n}=8\text{ so},\text{sum}=\Big(\frac{8\times9}{2}\Big)^2+1296.$

3. $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$

Let T_n be the n^th term of the given series.

Thus, we have

$\text{T}_\text{n}=\sqrt{2\times\text{n}^2}=\text{n}\sqrt{2}$

Now, let S_n be the sum of n terms of the given series.

Thus, we have

$\text{S}_\text{n}=\sqrt{2}\sum\limits^{\text{n}}_{\text{k}=1}(\text{k})$

$\Rightarrow\text{S}_\text{n}=\sqrt{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$

$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$

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

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)]

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

1. $\frac{8\pm3\sqrt{7}}{1}$

Solution:

 We know, G.M. of two numbers a and b is $\sqrt{\text{ab.}}$

$\text{so},\text{a}+\text{b}=\sqrt{\text{ab}}$

Squaring we get, a²+b² = 16ab

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

$\text{Let}\times=\frac{\text{a}}{\text{b}}.$

$\text{So},\times+\frac{1}{\times}=16$

=> x² – 16x + 1 = 0

$\Rightarrow\times=\frac{16\pm\sqrt{256-4}}{4}=\frac{16\pm\sqrt{252}}{2}=\frac{16\pm6\sqrt{7}}{2}=\frac{8\pm3\sqrt{7}}{1}.$

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

1. $121\big(\sqrt{6}+\sqrt{2}\big)$

Solution:

Let T_n be the n^th term of the given series.

Thus, we have

$\text{S}_{10}=\sqrt{2}\sum\limits^{10}_{\text{k}=1}\Big(\sqrt{3^{(\text{k}-1)}}\Big)$

$\Rightarrow\text{S}_{10}=\sqrt{2}\Big(1+\sqrt{3}+\sqrt{3^2}+\ ...\ +\sqrt{3^9}\Big)$

$\Rightarrow\text{S}_{10}=\sqrt{2}\bigg(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\bigg)$

$\Rightarrow\text{S}_{10\text{n}}=\sqrt{2}\Big(\frac{3^5-1}{\sqrt{3}-1}\Big)\Big(\frac{\sqrt{3}+1}{\sqrt{3}+1}\Big)$

$\Rightarrow\text{S}_{10}=\frac{\sqrt{2}}{2}\big(3^5-1\big)\big(\sqrt{3}+1\big)$

$\Rightarrow\text{S}_{10}=\frac{1}{2}(242)\Big(\sqrt{6}+\sqrt{2}\Big)$

$\Rightarrow\text{S}_{10}=121\big(\sqrt{6}+\sqrt{2}\big)$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

1. 12cm.

Solution:

Let the length, breadth and height of rectangular solid block be$\frac{\text{a}}{\text{r}},$ a and ar, respectively.

$\therefore\ \text{Volume}=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=216\text{cm}^3$

$\Rightarrow\text{a}^3=216=6^3\Rightarrow\text{a}=6$

Also, Surface area $=2\Big(\frac{\text{a}}{\text{r}}.\text{a}+\text{a}.\text{ar}+\frac{\text{a}}{\text{r}}.\text{ar}\Big)=252$

$\Rightarrow2\text{a}^2\Big(\frac{1}{\text{r}}+\text{r}+1\Big)=252$

$\Rightarrow2\times36\Big(\frac{1+\text{r}^2+\text{r}}{\text{r}}\Big)=252$

$\Rightarrow2(1+\text{r}^2+\text{r})=7\text{r}$

$\Rightarrow2\text{r}^2-5\text{r}+2=0$

$\Rightarrow(2\text{r}-1)(\text{r}-2)=0$

$\therefore\ \text{r}=\frac{1}{2},2$

For $\text{r}=\frac{1}{2}:$ Length $=\frac{\text{a}}{\text{r}}=\frac{6\times2}{1}=12,$ Breadth = a = 6

Height $=\text{ar}=6\times\frac{1}{2}=3$

For r = 2: Length $=\frac{\text{a}}{\text{r}}=\frac{6}{2}=3,$ Breadth = a = 6

Height = ar = 6 × 2 = 12