MCQ - MATHS STD 11 Science Questions - Vidyadip

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25 questions · auto-graded multiple-choice test.

MCQ 11 Mark

If in an infinite G.P., first term is equal to 10 times the sum of all successive terms, the its common ratio is:

A
$\frac{1}{10}$
✓
$\frac{1}{11}$
C
$\frac{1}{9}$
D
$\frac{1}{20}$

Answer

Correct option: B.

$\frac{1}{11}$

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]
$\text{a}=10\Big(\frac{\text{ar}}{1-\text{r}}\Big)$
$\Rightarrow\text{a}-\text{ar}=10\text{ar}$
$\Rightarrow11\text{ar}=\text{a}$
$\Rightarrow\text{r}=\frac{\text{a}}{11\text{a}}=\frac{1}{11}$

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MCQ 21 Mark

If $a, b, c$ are in A.P. and $x, y, z$ are in G.P., then the value of $x^{b-c} y^{c-a} z^{a-b}$ is:

A
0
✓
1
C
x y z
D
$x^a y^b z^c$

Answer

Correct option: B.

1

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)]
$=\text{x}^{\text{a}-\text{b}}\text{ y}^{2(\text{b}-\text{a})}\text{ z}^{\text{a}-\text{b}}$
$=(\text{xz})^{\text{a}-\text{b}}(\text{xz})^{\text{b}-\text{a}}$ $[\text{From}(\text{ii}),\text{y}^2=\text{xz}]$
$=(\text{xz})^0$
$=1$

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MCQ 31 Mark

If a, b, c are in G.P. and $\text{a}^{\frac{1}{\text{x}}}=\text{b}^{\frac{1}{\text{y}}}=\text{c}^{\frac{1}{\text{z}}},$ then xyz are in:

✓
AP
B
GP
C
HP
D
None of these.

Answer

Correct option: A.

AP

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.

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MCQ 41 Mark

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of original G.P. is:

✓
$\frac12$
B
$\frac{2}{3}$
C
$\frac13$
D
$\frac{-1}{2}.$

Answer

Correct option: A.

$\frac12$

$\frac{\text{a}}{1-\text{r}}=3$
$\text{a}=3-3\text{r}$
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),
$\text{a}=\frac32\text{ and r}=\frac12$

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MCQ 51 Mark

The nth term of a G.P. is $128$ and the sum of its $n$ terms is $225$ . If its common ratio is $2$, then its first term is:

✓
1
B
3
C
8
D
None of these.

Answer

Correct option: A.

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 = $\left(2^2\right) x \ldots$
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

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MCQ 61 Mark

If x is psitive, the sum to in $\frac{1}{1+\text{x}}-\frac{1-\text{x}}{(1+\text{x})^2}+\frac{(1-\text{x})^2}{(1+\text{x})^3}-\frac{(1-\text{x})^3}{(1+\text{x})^4}+\cdots\text{ is:}$

✓
$\frac{1}{2}$
B
$\frac{3}{4}$
C
1
D
None of these.

Answer

Correct option: A.

$\frac{1}{2}$

$\text{S}=\frac{1}{1+\text{x}}-\frac{1-\text{x}}{(1+\text{x})^2}+\frac{(1-\text{x})^2}{(1+\text{x})^3}-\frac{(1-\text{x})^3}{(1+\text{x})^4}+\cdots\infty$
It is clear that it is a G.P. with $\text{a}=\frac{1}{1+\text{x}}$ and $\text{r}=-\frac{(1-\text{x})}{(1+\text{x})}.$
$\therefore\text{S}=\frac{\text{a}}{(1-\text{r})}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1-\big(-\frac{(1-\text{x})}{(1+\text{x})}\big)\Big]}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1+\frac{(1-\text{x})}{(1+\text{x})}\Big]}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[\frac{(1+\text{x})+(1-\text{x})}{(1+\text{x})}\Big]}$
$\Rightarrow\text{S}=\frac12$

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MCQ 71 Mark

If a, b, c are in G.P. is 2 and x, y are AM's between a, b and b, c respectively, then:

A
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=2$
B
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{2}$
C
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{a}}$
✓
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$

Answer

Correct option: D.

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

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

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MCQ 81 Mark

The fractional value of 2.357 is:

A
$\frac{2355}{1001}$
B
$\frac{2379}{997}$
✓
$\frac{2355}{999}$
D
None of these.

Answer

Correct option: C.

$\frac{2355}{999}$

$2.\overline{357}=2.0+0.357+0.000357+0.000000357+\dots\infty$
$\Rightarrow2.\overline{357}=2+\Big[\frac{357}{10^3}+\frac{357}{10^6}+\frac{357}{10^9}+\dots\infty\Big]$
$\Rightarrow2.\overline{357}=2+\frac{\frac{357}{10^3}}{1-\frac{1}{10^3}}$
$\Rightarrow2.\overline{357}=2+\frac{357}{999}$
$\Rightarrow2.\overline{357}=\frac{2355}{999}$

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MCQ 91 Mark

Let $S$ be the sum, $P$ be the product and $R$ be the sum of the reciprocals of 3 terms of a G.P. then $P^2 R^3: S^3$ is equal to:

✓
1 : 1
B
$(\text { common ratio })^{\mathrm{n}}: 1$
C
$\left(\text { First term) }{ }^2 \text { (common ratio) }\right)^2$
D
None of these.

Answer

Correct option: A.

1 : 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).

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MCQ 101 Mark

If the first term of a G.P. $a_1, a_2, a_3, \ldots$ is unity such that $4 a_2+5 a_3$ is least, then common ratio of G.P. is:

✓
$\frac{-2}{5}$
B
$\frac{-3}{5}$
C
$\frac25$
D
None of these

Answer

Correct option: A.

$\frac{-2}{5}$

$-\frac{2}{5}$

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

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MCQ 111 Mark

Let x be the A.M. and y, z be two G.M.s between two positive numbers. Then, $\frac{\text{y}^3+\text{z}^3}{\text{xyz}}$ is equal to:

A
1
✓
2
C
$\frac12$
D
None of these.

Answer

Correct option: B.

2

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)]
$=2$

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MCQ 121 Mark

If a be A.M. and p, q be two G.M.'s between two numbers, then 2A is equal to:

✓
$\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
B
$\frac{\text{p}^3-\text{q}^3}{\text{pq}}$
C
$\frac{\text{p}^2+\text{q}^2}{2}$
D
$\frac{\text{pq}}{2}.$

Answer

Correct option: A.

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

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

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 $q = ar^2$ ____(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)]
$=\frac{\text{p}^3+\text{q}^3}{\text{pq}}$

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MCQ 131 Mark

Given that x > 0, the sum $\sum\limits^\infty_{\text{n}=1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{\text{n}-1}$ equals:

A
$\text{x}$
✓
$\text{x}+1$
C
$\frac{\text{x}}{2\text{x}+1}$
D
$\frac{\text{x}+1}{2\text{x}+1}$

Answer

Correct option: B.

$\text{x}+1$

$\sum\limits^\infty_{\text{x}+1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{(\text{n}-1)}$
$=1+\Big(\frac{\text{x}}{\text{x}+1}\Big)+\Big(\frac{\text{x}}{\text{x}+1}\Big)^2+\Big(\frac{\text{x}}{\text{x}+1}\Big)^3+\Big(\frac{\text{x}}{\text{x}+1}\Big)^4+\dots\infty$
$=\frac{1}{1-\big(\frac{\text{x}}{\text{x}+1}\big)}$ $\Big[\because$ it is a G.P. with a = 1 and $\text{r}=\Big(\frac{\text{x}}{\text{x}+1}\Big)\Big]$
$=\frac{(\text{x}+1)}{(\text{x}+1-\text{x})}$
$=\frac{(\text{x}+1)}{1}=(\text{x}+1)$

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MCQ 141 Mark

The first three of four given numbers are in G.P. and their last three are A.P. with common difference 6. If first and fourth numbers are equal, then the first number is:

A
2
B
4
C
6
✓
8

Answer

Correct option: D.

8

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, $2 \mathrm{r}=\mathrm{q}+\mathrm{p}$
Now, putting the values of $p$ and $r$ in (i):
$q^2=(q+12)(q+6)$
$\Rightarrow q^2=q^2+18 q+72$
$\Rightarrow 18 q+72=0$
$\Rightarrow q+4=0$
$\Rightarrow q=-4$
Now, putting the value of $q$ in $p=q+12$ :
$p=-4+12=8$

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MCQ 151 Mark

The value of $9^{\frac13}.9^{\frac19}.9^{\frac{1}{27}}\dots\text{to }\infty,$ is:

A
1
✓
3
C
9
D
None of these.

Answer

Correct option: B.

3

$9^{\frac13}\times9^{\frac19}\times9^{\frac{1}{27}}\times\dots\infty$
$=9^{\big(\frac13+\frac19+\frac{1}{27}+\dots\infty\big)}$
Here, it is a G.P. with $\text{a}=\frac13$ and $\text{r}=\frac13.$
$\therefore9^{\Bigg(\frac{\frac{1}{3}}{1-\frac13}\Bigg)}$
$=9^{\big(\frac12\big)}=3$

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MCQ 161 Mark

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is:

A
$\frac{\text{p}-\text{q}}{\text{q}-\text{r}}$
✓
$\frac{\text{q}-\text{r}}{\text{p}-\text{q}}$
C
pqr
D
None of these.

Answer

Correct option: B.

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

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

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.
$\therefore (\text{a} + (\text{q}−1)\text{d})^2=( \text{a} + (\text{p}−1)\text{d})\times(\text{a} + (\text{r}−1)\text{d})$
$\Rightarrow\text{a}^2+2\text{a} (\text{q}−1)\text{d}+( (\text{q}−1)\text{d})^2\\\ \ =\text{a}^2+\text{ad}(\text{r}−1+\text{p}−1)+(\text{p}−1) (\text{r}−1)\text{d}^2$
$\Rightarrow(2\text{q}−2−\text{r}−\text{p}+2)+\text{d}^2(\text{q}^2−2\text{q}+1−\text{pr}+\text{p}+\text{r}−1)=0$
$\Rightarrow(2\text{q}−\text{r}-\text{p})+\text{d}(\text{q}^2−2\text{q}−\text{pr}+\text{p}+\text{r})=0$
$\Rightarrow\text{a}=-\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(2\text{q}-\text{r}-\text{p})}$
$\therefore\text{Common ratio},\text{ r}=\frac{\text{a}_\text{q}}{\text{a}_\text{p}}$
$=\frac{\text{a}+(\text{q}-1)\text{d}}{\text{a}+(\text{p}-1)\text{d}}$
$=\frac{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{q}-1)\text{d}}{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{p}-1)\text{d}}$
$=\frac{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{pq}+\text{rq}+\text{rq}-2\text{q}^2-\text{p}-\text{r}+2\text{q}}{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{p}^2+\text{pr}-2\text{pq}-\text{p}-\text{r}+2\text{q}}$
$=\frac{\text{pq}-\text{pr}-\text{q}^2+\text{qr}}{\text{p}^2+\text{q}^2-2\text{pq}}$
$=\frac{\text{p}(\text{p}-\text{r})-\text{q}(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{p}-\text{q})(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{q}-\text{r})}{(\text{p}-\text{q})}$

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MCQ 171 Mark

If second term of a G.P. is 2 and the sum of its infinite terms is 8, then its first terms is:

A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$2$
✓
$4.$

Answer

Correct option: D.

$4.$

$\text{a}_2=2$
$\therefore\text{ar}=2\ \cdots(\text{i})$
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)]
$\Rightarrow\text{a}^2=8(\text{a}-2)$
$\Rightarrow\text{a}^2-8\text{a}+16=0$
$\Rightarrow(\text{a}-4)^2=0$
$\Rightarrow\text{a}=4$

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MCQ 181 Mark

In a G.P. of ever number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is:

A
$-\frac{4}{5}$
B
$\frac{1}{5}$
✓
4
D
None of these.

Answer

Correct option: C.

4

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.
$\therefore\text{r}=4$

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MCQ 191 Mark

The product $(32),(32)^{\frac{1}{6}}(32)^{\frac{1}{36}}\ \dots\text{ to }\infty$ is equal to:

✓
64
B
16
C
32
D
0

Answer

Correct option: A.

64

$32\times32^{\frac{1}{6}}\times32^{\frac{1}{36}}\times\ \cdots\infty$
$=32^{\big(1+\frac{1}{6}+\frac{1}{36}+\ \cdots\infty\big)}$
$=32^{\Bigg(\frac{1}{1-\frac{1}{6}}\Bigg)}$ $[\because$ it is a G.P. $]$
$=32^{\big(\frac65\big)}$
$=\big(2^5\big)^{\big(\frac65\big)}$
$=2^6$
$=64$

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MCQ 201 Mark

In a G.P. if the $(m + n)^{th}$ terms is p and $(m - n)^{th}$ term is q, then its $m^{th}$ term is:

A
$0$
B
$\text{pq}$
✓
$\sqrt{\text{pq}}$
D
$\frac12(\text{p}+\text{q})$

Answer

Correct option: C.

$\sqrt{\text{pq}}$

$\sqrt{\text{pq}}$

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

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MCQ 211 Mark

If $\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54},$ the value of x is:

A
7
✓
8
C
9
D
10.

Answer

Correct option: B.

8

$\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54}$
$\Rightarrow4^{(3+6+9+12+\ \dots+3\text{x})}=\Big(\frac{625}{10000}\Big)^{-54}$
$\Rightarrow4^{3(1+2+3+4+\dots+\text{x})}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(4^{-2}\Big)^{-54}$
Comparing both the sides:
$\Rightarrow3\Big(\frac{\text{x}(\text{x}+1)}{2}\Big)=108$
$\Rightarrow\text{x}(\text{x}+1)=72$
$\Rightarrow\text{x}^2+\text{x}-72=0$
$\Rightarrow\text{x}^2+9\text{x}-8\text{x}-72=0$
$\Rightarrow\text{x}(\text{x}+9)-8(\text{x}+9)=0$
$\Rightarrow(\text{x}+9)(\text{x}-8)=0$
$\Rightarrow\text{x}=8,-9$
$\Rightarrow\text{x}=8$ $[\because\text{ x}\text{ is psitive}]$

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MCQ 221 Mark

If p, q be two A.M.'s and G be one G.M. between two numbers, then $G^2 =$

✓
$(2\text{p}-\text{q})(\text{p}-2\text{q})$
B
$(2\text{p}-\text{q})(2\text{q}-\text{p})$
C
$(2\text{p}-\text{q})(\text{p}+2\text{q})$
D
None of these.

Answer

Correct option: A.

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

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

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

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MCQ 231 Mark

If the sum of first two terms of an infinite G.P. is 1 and every term is twice the sum of all the successive terms, then its first term is:

A
$\frac13$
B
$\frac23$
C
$\frac14$
✓
$\frac34.$

Answer

Correct option: D.

$\frac34.$

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):
$\text{a}+\frac{a}{3}=1$
$\Rightarrow\frac{4\text{a}}{3}=1$
$\Rightarrow4\text{a}=3$
$\Rightarrow\text{a}=\frac34$

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MCQ 241 Mark

If S be the sum, P the product and R be the sum of the reciprocals of n terms of a G.P. then $P^2$ is equal to:

A
$\frac{\text{S}}{\text{R}}$
B
$\frac{\text{R}}{\text{S}}$
C
$\Big(\frac{\text{R}}{\text{S}}\Big)^\text{n}$
✓
$\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$

Answer

Correct option: D.

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

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

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

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MCQ 251 Mark

The two geometric means between the numbers 1 and 64 are:

A
1 and 64
✓
4 and 16
C
2 and 16
D
8 and 16.

Answer

Correct option: B.

4 and 16

4 and 16

Solution:
Let the two G.M.s between 1 and 64 be $G_1$ and $G_2$.
Thus, 1, $G_1$, $G_2$ 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.

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