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
$\frac{1}{20}$
AnswerCorrect 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]
$\text{a}=10\Big(\frac{\text{ar}}{1-\text{r}}\Big)$
$\Rightarrow\text{a}-\text{ar}=10\text{ar}$
$\Rightarrow11\text{ar}=\text{a}$
$\Rightarrow\text{r}=\frac{\text{a}}{11\text{a}}=\frac{1}{11}$
View full question & answer→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$a, b$ 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 $(i)]$
$=\text{x}^{\text{a}-\text{b}}\text{ y}^{2(\text{b}-\text{a})}\text{ z}^{\text{a}-\text{b}}$
$=(\text{xz})^{\text{a}-\text{b}}(\text{xz})^{\text{b}-\text{a}}$ $[\text{From}(\text{ii}),\text{y}^2=\text{xz}]$
$=(\text{xz})^0$
$=1$
View full question & answer→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:
Answera, 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.
View full question & answer→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
$\frac{-1}{2}.$
AnswerCorrect 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),
$\text{a}=\frac32\text{ and r}=\frac12$
View full question & answer→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:
AnswerLet the firt term of the geometric progression $= x$
Common ration $= 2$
$\therefore 2^{nd}$ term of the $G.P. = 2x$
$\therefore 3^{rd}$ term $= (2^2)x ...$
$N^{th}$ 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 $G.P. = 1$
View full question & answer→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
- D
AnswerCorrect 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 $\text{r}=-\frac{(1-\text{x})}{(1+\text{x})}.$
$\therefore\text{S}=\frac{\text{a}}{(1-\text{r})}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1-\big(-\frac{(1-\text{x})}{(1+\text{x})}\big)\Big]}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[1+\frac{(1-\text{x})}{(1+\text{x})}\Big]}$
$\Rightarrow\text{S}=\frac{\frac{1}{(1+\text{r})}}{\Big[\frac{(1+\text{x})+(1-\text{x})}{(1+\text{x})}\Big]}$
$\Rightarrow\text{S}=\frac12$
View full question & answer→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
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{a}}$
- ✓
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=\frac{2}{\text{b}}.$
AnswerCorrect 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:
$\Rightarrow\frac{1}{\text{y}}+\frac{1}{\text{x}}=\frac{2}{\text{b}}$
View full question & answer→MCQ 81 Mark
The fractional value of 2.357 is:
- A
$\frac{2355}{1001}$
- B
$\frac{2379}{997}$
- ✓
$\frac{2355}{999}$
- D
AnswerCorrect option: C. $\frac{2355}{999}$
$2.\overline{357}=2.0+0.357+0.000357+0.000000357+\dots\infty$
$\Rightarrow2.\overline{357}=2+\Big[\frac{357}{10^3}+\frac{357}{10^6}+\frac{357}{10^9}+\dots\infty\Big]$
$\Rightarrow2.\overline{357}=2+\frac{\frac{357}{10^3}}{1-\frac{1}{10^3}}$
$\Rightarrow2.\overline{357}=2+\frac{357}{999}$
$\Rightarrow2.\overline{357}=\frac{2355}{999}$
View full question & answer→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 $P^2R^3 : S^3$ is equal to:
AnswerCorrect option: A. $1 : 1$
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 $(a).$
View full question & answer→MCQ 101 Mark
If the first term of a $G.P. \ a_1, a_2, a_3, ...$is unity such that $4 a_2 + 5a_3$ is least, then common ratio of $G.P.$ is:
- ✓
$\frac{-2}{5}$
- B
$\frac{-3}{5}$
- C
$\frac25$
- D
AnswerCorrect option: A. $\frac{-2}{5}$
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 $\text{r}+\frac25=0,\text{ i.e. }\text{r}=-\frac25.$
View full question & answer→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:
AnswerLet 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)]
$=2$
View full question & answer→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
$\frac{\text{pq}}{2}.$
AnswerCorrect option: A. $\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
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 $(ii)]$
$=\frac{\text{p}^3+\text{q}^3}{\text{pq}}$
View full question & answer→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:
AnswerCorrect 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 $\text{r}=\Big(\frac{\text{x}}{\text{x}+1}\Big)\Big]$
$=\frac{(\text{x}+1)}{(\text{x}+1-\text{x})}$
$=\frac{(\text{x}+1)}{1}=(\text{x}+1)$
View full question & answer→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:
AnswerThe 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 $(i):$
$q^2 = (q + 12)(q + 6)$
$\Rightarrow q^2 = q^2 + 18q + 72$
$\Rightarrow 18q + 72 = 0$
$\Rightarrow q + 4 = 0$
$\Rightarrow q = -4$
Now, putting the value of $q$ in $p = q + 12:$
$p = -4 + 12 = 8$
View full question & answer→MCQ 151 Mark
The value of $9^{\frac13}.9^{\frac19}.9^{\frac{1}{27}}\dots\text{to }\infty,$ is:
Answer$9^{\frac13}\times9^{\frac19}\times9^{\frac{1}{27}}\times\dots\infty$
$=9^{\big(\frac13+\frac19+\frac{1}{27}+\dots\infty\big)}$
Here, it is a G.P. with $\text{a}=\frac13$ and $\text{r}=\frac13.$
$\therefore9^{\Bigg(\frac{\frac{1}{3}}{1-\frac13}\Bigg)}$
$=9^{\big(\frac12\big)}=3$
View full question & answer→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:
AnswerCorrect option: B. $\frac{\text{q}-\text{r}}{\text{p}-\text{q}}$
Let a be the first term and dbe 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 $G.P.$
$\therefore (\text{a} + (\text{q}−1)\text{d})^2=( \text{a} + (\text{p}−1)\text{d})\times(\text{a} + (\text{r}−1)\text{d})$
$\Rightarrow\text{a}^2+2\text{a} (\text{q}−1)\text{d}+( (\text{q}−1)\text{d})^2\\\ \ =\text{a}^2+\text{ad}(\text{r}−1+\text{p}−1)+(\text{p}−1) (\text{r}−1)\text{d}^2$
$\Rightarrow(2\text{q}−2−\text{r}−\text{p}+2)+\text{d}^2(\text{q}^2−2\text{q}+1−\text{pr}+\text{p}+\text{r}−1)=0$
$\Rightarrow(2\text{q}−\text{r}-\text{p})+\text{d}(\text{q}^2−2\text{q}−\text{pr}+\text{p}+\text{r})=0$
$\Rightarrow\text{a}=-\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(2\text{q}-\text{r}-\text{p})}$
$\therefore\text{Common ratio},\text{ r}=\frac{\text{a}_\text{q}}{\text{a}_\text{p}}$
$=\frac{\text{a}+(\text{q}-1)\text{d}}{\text{a}+(\text{p}-1)\text{d}}$
$=\frac{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{q}-1)\text{d}}{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{p}-1)\text{d}}$
$=\frac{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{pq}+\text{rq}+\text{rq}-2\text{q}^2-\text{p}-\text{r}+2\text{q}}{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{p}^2+\text{pr}-2\text{pq}-\text{p}-\text{r}+2\text{q}}$
$=\frac{\text{pq}-\text{pr}-\text{q}^2+\text{qr}}{\text{p}^2+\text{q}^2-2\text{pq}}$
$=\frac{\text{p}(\text{p}-\text{r})-\text{q}(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{p}-\text{q})(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{q}-\text{r})}{(\text{p}-\text{q})}$
View full question & answer→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$\text{a}_2=2$
$\therefore\text{ar}=2\ \cdots(\text{i})$
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)]
$\Rightarrow\text{a}^2=8(\text{a}-2)$
$\Rightarrow\text{a}^2-8\text{a}+16=0$
$\Rightarrow(\text{a}-4)^2=0$
$\Rightarrow\text{a}=4$
View full question & answer→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
$\frac{1}{5}$
- ✓
- D
AnswerLet 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.
$\therefore\text{r}=4$
View full question & answer→MCQ 191 Mark
The product $(32),(32)^{\frac{1}{6}}(32)^{\frac{1}{36}}\ \dots\text{ to }\infty$ is equal to:
Answer$32\times32^{\frac{1}{6}}\times32^{\frac{1}{36}}\times\ \cdots\infty$
$=32^{\big(1+\frac{1}{6}+\frac{1}{36}+\ \cdots\infty\big)}$
$=32^{\Bigg(\frac{1}{1-\frac{1}{6}}\Bigg)}$ $[\because$ it is a G.P. $]$
$=32^{\big(\frac65\big)}$
$=\big(2^5\big)^{\big(\frac65\big)}$
$=2^6$
$=64$
View full question & answer→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:
AnswerCorrect option: C. $\sqrt{\text{pq}}$
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 $\sqrt{\text{pq}}.$
View full question & answer→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:
Answer$\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54}$
$\Rightarrow4^{(3+6+9+12+\ \dots+3\text{x})}=\Big(\frac{625}{10000}\Big)^{-54}$
$\Rightarrow4^{3(1+2+3+4+\dots+\text{x})}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(4^{-2}\Big)^{-54}$
Comparing both the sides:
$\Rightarrow3\Big(\frac{\text{x}(\text{x}+1)}{2}\Big)=108$
$\Rightarrow\text{x}(\text{x}+1)=72$
$\Rightarrow\text{x}^2+\text{x}-72=0$
$\Rightarrow\text{x}^2+9\text{x}-8\text{x}-72=0$
$\Rightarrow\text{x}(\text{x}+9)-8(\text{x}+9)=0$
$\Rightarrow(\text{x}+9)(\text{x}-8)=0$
$\Rightarrow\text{x}=8,-9$
$\Rightarrow\text{x}=8$ $[\because\text{ x}\text{ is psitive}]$
View full question & answer→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
AnswerCorrect option: A. $(2\text{p}-\text{q})(\text{p}-2\text{q})$
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 $G.P.$
$\therefore\text{G}^2=\text{ab}$
$\Rightarrow\text{G}^2=(2\text{p}-\text{q})(2\text{q}-\text{p})$
View full question & answer→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
$\frac14$
- ✓
$\frac34.$
AnswerCorrect 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):
$\text{a}+\frac{a}{3}=1$
$\Rightarrow\frac{4\text{a}}{3}=1$
$\Rightarrow4\text{a}=3$
$\Rightarrow\text{a}=\frac34$
View full question & answer→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
$\Big(\frac{\text{R}}{\text{S}}\Big)^\text{n}$
- ✓
$\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
AnswerCorrect option: D. $\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
Sum of n terms of the $\text{G.P.}$, $\text{S}=\frac{\text{a}(\text{r}^{\text{n}}-1)}{(\text{r}-1)}$
Product of n terms of the $\text{G.P.}$, $\text{P}=\text{a}^{\text{n}}\text{r}^{\big[\frac{\text{n}(\text{n}-1)}{2}\big]}$
Sum of the reciprocals of n terms of the $\text{G.P.}$, $\text{R}=\frac{\Big[\frac{1}{\text{r}^\text{n}}-1\Big]}{\text{a}\big(\frac{1}{\text{r}}-1\big)}=\frac{(\text{r}^{\text{n}}-1)}{\text{ar}^{(\text{n}-1)}(\text{r}-1)}$
$\therefore\text{P}^2=\bigg\{\text{a}^2\text{r}^\frac{2(\text{n}-1)}{2}\bigg\}^\text{n}$
$\Rightarrow\text{P}^2=\Bigg\{\frac{\frac{\text{a}(\text{r}^\text{n}-1)}{(\text{r}-1)}}{\frac{(\text{r}^\text{n}-1)}{\text{ar}^{(\text{n}-1)(\text{r}-1)}}}\Bigg\}^\text{n}$
$\Rightarrow\text{P}^2=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$
Let the first term of the $\text{G.P.}$ 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 $\text{G.P.}$, $\text{P}=\text{a}^{\text{n}}\text{r}^{\frac{\text{n}(\text{n}+1)}{2}}$
Sum of the reciprocals of n terms, $\Rightarrow\text{R}=\frac{\big(\frac{1}{\text{r}^{\text{n}-1}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}=\frac{\big(\frac{1-\text{r}^{\text{n}}}{\text{r}^\text{n}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}$
$\text{P}^2=\bigg\{\text{a}^2\text{r}^{\frac{(\text{n}+1)}{2}}\bigg\}^{\text{n}}$
$\text{P}^2=\begin{Bmatrix} \frac{\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}}{\frac{\Big(\frac{1-\text{r}^\text{n}}{\text{r}^\text{n}}\Big)}{\text{a}\Big(\frac{1-\text{r}}{\text{r}}\Big)}}\end{Bmatrix}=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$
View full question & answer→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 $16.$
AnswerCorrect option: B. $4$ and $16$
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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