MCQ 511 Mark
In any case, the difference of the least and greatest term is:
MCQ 521 Mark
The product of $n$ positive numbers is unity. Their sum is:
AnswerCorrect option: D. never less than $n$
View full question & answer→MCQ 531 Mark
Choose the correct answer: The minimum value of $4^{\text{x}}+4^{1-\text{x}},\text{x}\in\text{R}$ is:
AnswerWe know that $\text{AM}\geq\text{GM}$
$\therefore\ \frac{4^\text{x}+4^{1-\text{x}}}{2}\geq\sqrt{4^\text{x}.4^{1-\text{x}}}$
$\Rightarrow4^\text{x}+4^{1-\text{x}}\geq2\sqrt{4^{\text{x}+1-\text{x}}}$
$4^\text{x}+4^{1-\text{x}}\geq2.2$
$\Rightarrow4^\text{x}+4^{1-\text{x}}\geq4$
Hence, the correct option is $(b).$
View full question & answer→MCQ 541 Mark
If $\log_{ax},\log_{bx}, \log_{cx}$ be in $HP,$ then $a, b, c$ are in:
View full question & answer→MCQ 551 Mark
If $a + 2b + 3c = 12 , (a, b, c \in R+),$ then $ab^2c^3$ is:
AnswerCorrect option: C. $\leq2^6$
View full question & answer→MCQ 561 Mark
A sequence is called $..........$ if $a_{n+1} = a_n + d$.
AnswerA sequence is called arithmetic progression if $a_{n+1} = a_n + d$ where $a_1$ is the first term and $d$ is common difference.
View full question & answer→MCQ 571 Mark
Sumof nterms of series $12 + 16 + 24 + 40 + …$ will be:
- A
$2(2n – 1) + 8n$
- B
$2(2n – 1) + 6 n$
- C
$3(2n – 1) + 8n$
- ✓
$4(2n – 1) + 8n$
AnswerCorrect option: D. $4(2n – 1) + 8n$
View full question & answer→MCQ 581 Mark
If $Tn$ denotes the $n^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + . . . ,$ then $T_{50}$ is:
- A
$49 2 - 1$
- B
$492$
- C
$502 + 1$
- ✓
$492 + 2$
AnswerCorrect option: D. $492 + 2$
View full question & answer→MCQ 591 Mark
How many terms of $\text{G.P.}\ 2,4,8,16, ………$ are required to give sum $254?$
Answer$=2$ and
$r=\frac{4}{2}=2.$
We know $\text{s}_\text{n}=\text{a}\frac{(\text{r}^\text{n}-1)}{\text{(r}-1)}$
$2\frac{(2^\text{n-1)}}{(2-1)}=254$
$\Rightarrow 2^n- 1 = 127$
$\Rightarrow 2^n = 128 = 2^7$
$\Rightarrow n = 7.$
View full question & answer→MCQ 601 Mark
Find the sum of series $6^3+ 7^3+…..…..+ 20^3$.
- ✓
$43875$
- B
$83775$
- C
$43775$
- D
$43975$
AnswerCorrect option: A. $43875$
$6^3+ 7^3+………..+ 20^3$
$= (1^3+ 2^3+ 3^3+……..+ 20^3) – (1^3+ 2^3+ 3^3+ 4^3+ 5^3)$
$=\Big(\frac{20\times21}{2}\Big)^2-\Big(\frac{5\times6}{2}\Big)^2$
$=(210)^2-(15)^2$
$=225\times195$
$=43875$
View full question & answer→MCQ 611 Mark
$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day, $4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is:
View full question & answer→MCQ 621 Mark
For an increasing $A.P.\ a_1, a_2, a_3..... a_n,$ if $a_1, a_3, a_5 = – 12$ and $a_1. a_3. a_5 = 80,$ then which of the following is/are true?
- ✓
$a_1 = -10$
- B
$a_2 = -1$
- C
$a_3= – 4$
- D
$a_5 = – 2 (a,c,d)$
AnswerCorrect option: A. $a_1 = -10$
View full question & answer→MCQ 631 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and $n^{th}$ term is $42,$ then find $n.$
AnswerWe know, $a = 20, d = 2, a_n = 42.$
$a + (n - 1) d = 42 20$
$\Rightarrow + 2(n - 1) = 42$
$\Rightarrow 2 (n - 1) = 42 - 20 = 22$
$\Rightarrow n - 1 = 11$
$\Rightarrow n = 12.$
View full question & answer→MCQ 641 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 $\ce{P^2R^3 : S^3}$ 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 651 Mark
If the first term of a $\ce{G.P. a_1, a_2, a_3, ...}$ is unity such that $4 a_2+ 5a_3$ is least, then common ratio of $\text{G.P.}$ 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 $\text{G.P.}$ 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 $\text{r}+\frac25=0,$
i.e. $\text{r}=-\frac25.$
View full question & answer→MCQ 661 Mark
Find the sum of series $1^3+ 3^3+ 5^3+………….+ 11^3$.
- ✓
$2556$
- B
$5248$
- C
$6589$
- D
$9874$
AnswerCorrect option: A. $2556$
$1^3+ 3^3+ 5^3+……………..+ 11^3$
$= (1^3+ 2^3+ 3^3+……+ 11^3) – (2^3+ 4^3+ 6^3+ 8^3+ 10^3)$
$= (1^3+ 2^3+ 3^3+……11^3) – 2^3(1^3+ 23 + 3^3+ 4^3+ 5^3)$
$\Big(\frac{11\times12}{2}\Big)^2-8\Big(\frac{5\times6}{2}\Big)^2$
$=66^2-8\times15^2$
$4356-1800=2556.$
View full question & answer→MCQ 671 Mark
The ratio of the $A.M.$ and $G.M.$ of two positive numbers $a$ and $b$ is $5 : 3.$ Find the ratio of $a$ to $b.$
- ✓
$9 : 1$
- B
$3 : 5$
- C
$1 : 9$
- D
$3 : 1$
AnswerCorrect option: A. $9 : 1$
$\frac{(\text{A.}\text{M.})}{\text{(G.}\text{M.)}}=\frac{5}{3}$
$\Rightarrow\frac{\text{a}+\text{b}}{2\sqrt{a}\text{b}}=\frac{5}{3}$
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
$\frac{\text{a}}{\text{b}}=3\Big(\frac{3}{1}\Big)^2=9.$
$\text{ so},\text{a}:\text{b}=9:1$
View full question & answer→MCQ 681 Mark
Choose the correct answer. If x, 2y and 3z are in A.P. where the distinct numbers x, y and z are in G.P., then the common ratio of the G.P. is:
- A
$3$
- B
$\frac{1}{3}$
- C
$2$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
Solution: (D) $\frac{1}{2}.$
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 $z = xr^2$.
Where 'r' is the common ratio.
$\therefore$ $4xr = x + 3xr^2$ [Using (1)]
$⇒ 4r = 1 + 3r^2$
$⇒ 3r^2 - 4r + 1 = 0$
$⇒ (3r - 1)(r - 1) = .0$
$\Rightarrow\text{r}=\frac{1}{3}$
(For r = 1; x, y, z are not distinct)
View full question & answer→MCQ 691 Mark
Find the sum to $6$ terms of each of the series $2*3+4*6+6*11+8*18+………..$
AnswerGeneral term of above series is $a_k = 2k*(k^2+2) = 2k^3+4k$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=2\sum\limits^\text{n}_\text{i}=0\text{ k}^3+4\sum\limits^\text{n}_\text{i}=0\text{ k}$
$=2\Big(\frac{(\text{n}(\text{n+1)}}{2}\Big)^2+4\frac{\text{n}(\text{n+1)}}{2}$
$=\text{n}^2\frac{(\text{n}+1)}{2}+2\text{n}(\text{n+1)}$
$=\frac{36\times49}{2}+2\times6\times7$
$=966.$
View full question & answer→MCQ 701 Mark
Find the sum $1^3+2^3+3^3+………+8^3$.
- A
$1225$
- B
$1184$
- C
$1475$
- ✓
$1296$
AnswerCorrect option: D. $1296$
We know, sum of cubes of first $n$ terms is given by $\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2.$
$\text{Here},\text{n}=8$
$\text{ so},\text{sum}=\Big(\frac{8\times9}{2}\Big)^2=1296.$
View full question & answer→MCQ 711 Mark
Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\text{ is}$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
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
$\text{S}_\text{n}=\sqrt{2}\sum\limits^{\text{n}}_{\text{k}=1}(\text{k})$
$\Rightarrow\text{S}_\text{n}=\sqrt{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
View full question & answer→MCQ 721 Mark
The sum of the infinity of the series $1+2/3 + 6/32 + 10/33 +14/34$ is:
View full question & answer→MCQ 731 Mark
Choose the correct answer. If $9$ times the $9^{th}$ term of an $A.P.$ is equal to $13$ times the $13^{th}$ term, then the $22^{nd}$ term of the $A.P.$ is:
AnswerLet the first term and common difference of given $A.P.$ be $a$ and $d$, respectively.
It is given that $9 \times t_9= 13 \times t_{13}$
$\Rightarrow 9(a + 8d) = 13(a + 12d)$
$\Rightarrow 9a + 72d = 13a + 156d$
$\Rightarrow 4a + 84d = 0$
$\Rightarrow 4(a + 21d) = 0$
$\Rightarrow t_{22}= 0$
View full question & answer→MCQ 741 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 751 Mark
$a_1 = a_2= 2, a_n = a_n- 1 - 1, n > 2$. Find $a_5$.
Answer$ \Rightarrow a_n=a_n-1-1, n>2 $
$ \Rightarrow a_3=a_2-1=2-1=1 $
$ \Rightarrow a_4=a_3 1=11=0 $
$\Rightarrow a_5=a_4-1=0-1=-1 $
View full question & answer→MCQ 761 Mark
If a be $\text{A.M.}$ and $p, q$ be two $\text{G.M.'s}$ 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 771 Mark
Jairam purchased a house in $Rs. 15000$ and paid $Rs. 5000$ at once. Rest money he promised to pay in annual instalment of $Rs. 1000$ with $10\%$ per annum interest. How much money is to be paid by Jairam?
- A
$Rs. 21555$
- B
$Rs. 20475$
- ✓
$Rs. 20500$
- D
$Rs. 20700$
AnswerCorrect option: C. $Rs. 20500$
View full question & answer→MCQ 781 Mark
Choose the correct answer. If in an $A.P., S_n= qn^2$ and $S_m = qm^2$, where $S_r$ denotes the sum of $r$ terms of the $AP,$ then $S_q$ equals:
- A
$\frac{\text{q}^3}{2}$
- B
$\text{mnq}$
- ✓
$q^3$
- D
$(m + n)q^2$
AnswerGiven,
$S n=q n^2$ and $S m=q m^2 $
$\therefore S_1=q, S_2=4 q, S_3=9 q$ and $S_4=16 q$
Now, $\mathrm{t}_1=\mathrm{q}$
$\therefore \mathrm{t}_2=S_2-S_1=4 q-q=3 q $
$t_3=S_3-S_2=9 q-4 q=5 q $
$ t_4=S_4-S_3=16 q-9 q=7 q$
So, the $A.P.$ is: $q, 3q, 5q, 7q, ....$
Thus, first term is $q$ and common difference is $3q - q = 2q.$
$\therefore\ \text{S}_\text{q}=\frac{\text{q}}{2}[2\times\text{q}+(\text{q}-1)2\text{q}]=\frac{\text{q}}{2}\times[2\text{q}+2\text{q}^2-2\text{q}]$
$=\frac{\text{q}}{2}\times2\text{q}^2$
$=\text{q}^3$
View full question & answer→MCQ 791 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 801 Mark
Find out next term of the series $2, 7, 28, 63, 126, ...:$
AnswerGiven the series' terms can be written as
$13 + 1, 23 - 1, 33 + 1, 43 - 1, 53 + 1, 63 - 1$ etc.
Hence the next number is $63 - 1 = 216 - 1 = 215$
View full question & answer→MCQ 811 Mark
If $a, b, c$ are in $AP,$ then the straight line $ax + by + c = 0$ will always pass through the point:
- A
$(-1, – 2)$
- ✓
$(1, – 2)$
- C
$(-1, 2)$
- D
$(1, 2)$
AnswerCorrect option: B. $(1, – 2)$
View full question & answer→MCQ 821 Mark
If $a_1, a_2, a-n$, are in $AP$ with common difference $d,$ then the sum of the series $\sin \ce{d(cosec{~a_1}\ cosec {~a_2} + \ cosec {~a_2}\ cosec {~a_3} +… + \ cosec {~a_{n-1}}\ cosec {~a_n})}$ is:
- A
$\sec {~a_1} \sec {~a_n} –$
- ✓
$\cot {~a_1}– \cot {~a_n}$
- C
$\tan {~a_1}– \tan {~a_n}$
- D
$\ce{cosec {~a_1} – cosec {~a_n}}$
AnswerCorrect option: B. $\cot {~a_1}– \cot {~a_n}$
View full question & answer→MCQ 831 Mark
In $G.P.\ 4, 8, 16, 32, …$ find the sum up to $5^{th}$ term.
AnswerIn the given $\text{G.P.}$
$=4$ and $\text{r}=\frac{8}{4}=2.$
We know, $\text{s}_\text{n}=a\frac{\text({r}_\text{n}-1)}{\text({r}-1)} $
$\Rightarrow{\text{s}}_5$
$=4\frac{(2^5-1)}{(2-1)}$
$=4\times31$
$=124.$
View full question & answer→MCQ 841 Mark
If the sum of the roots of the equation $\ce{ax^2bx + c} = 0$ is be equal to the sum of the reciprocals of their squares, then $\ce{bc^2,ca^2,ab^2}$ will be in:
View full question & answer→MCQ 851 Mark
The value of $\sum\limits^{\text{n}}_{\text{r}=1}\Big\{\big(2\text{r}-1\big)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$ is equal to:
- A
$\text{an}^2+\frac{\text{b}^{\text{n}-1}-1}{\text{b}^{\text{n}-1}(\text{b}-1)}$
- ✓
$\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
- C
$\text{an}^3+\frac{\text{b}^{\text{n}-1}-1}{\text{b}^{\text{n}}(\text{b}-1)}$
- D
AnswerCorrect option: B. $\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
We have,
$\sum\limits^{\text{n}}_{\text{r}=1}\Big\{(2\text{r}-1)\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$
$=\sum\limits^{\text{n}}_{\text{r}=1}\Big\{2\text{ra}-\text{a}+\frac{1}{\text{b}^\text{r}}\Big\}$
$=\sum\limits^{\text{n}}_{\text{r}=1}2\text{ar}-\sum\limits^{\text{n}}_{\text{r}=1}\text{a}+\sum\limits^{\text{n}}_{\text{r}=1}\frac{1}{\text{b}^{\text{r}}}$
$=\text{an}(\text{n}+1)-\text{an}+\frac{(1-\text{b}^\text{n})}{(1-\text{b})\text{b}^\text{n}}$
$=\text{an}^2+\frac{\text{b}^\text{n}-1}{\text{b}^\text{n}(\text{b}-1)}$
View full question & answer→MCQ 861 Mark
A sequence is called $............$ if $a_{n+1} = a_n + d.$
AnswerA sequence is called arithmetic progression if $a_{n+1} = a_n+ d$ where $a_1$ is the first term and $d$ is common difference.
View full question & answer→MCQ 871 Mark
If the altitudes of a triangle are in $AP,$ then the sides of the triangle are in:
View full question & answer→MCQ 881 Mark
If the sum of two numbers is $4$ times the geometric mean then find the ratio of numbers.
- ✓
$\frac{8\pm3\sqrt{5}}{1}$
- B
$\frac{8\pm3\sqrt{7}}{1}$
- C
$\frac{6\pm3\sqrt{5}}{1}$
- D
$\frac{6\pm3\sqrt{7}}{1}$
AnswerCorrect option: A. $\frac{8\pm3\sqrt{5}}{1}$
We know, $G.M.$ of two numbers $a$ and $b$ is $\sqrt{\text{ab.}}$
so, $\text{a}+\text{b}=\sqrt{\text{ab}}$
Squaring we get, $a^2+b^2= 16ab$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)+\Big(\frac{\text{a}}{\text{b}}\Big)$
Let $x=\frac{\text{a}}{\text{b}}.$
So $,x+\frac{1}{x}=16$
$\Rightarrow x^2 – 16x + 1 = 0$
$\Rightarrow x=\frac{16\pm\sqrt{256-4}}{4}$
$=\frac{16\pm\sqrt{252}}{2}$
$=\frac{16\pm6\sqrt{7}}{2}$
$=\frac{8\pm3\sqrt{7}}{1}.$
View full question & answer→MCQ 891 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 901 Mark
Which of the following relation gives Fibonacci sequence?
- ✓
${a}_{n}={a}_{{n}-1}+{a}_{t{n}-2}$
- B
${a}_{{n}-1}={a}_{n}+{a}_{{n}-2}$
- C
${a}_{{n}-2}={a}_{n}+{a}_{{n}-1}$
- D
${a}_{n}={a}_{{n}+1}+{a}_{{n}-2}$
AnswerCorrect option: A. ${a}_{n}={a}_{{n}-1}+{a}_{t{n}-2}$
This is a recurrence relation which gives Fibonacci sequence.
View full question & answer→MCQ 911 Mark
length of a side of $S_n$ equals the length of a diagonal of $S_n+1$. If the length of a side of $S_1$ is $10 \ cm$, then for which of the following values of $n,$ the area of $S_n$ less than $1\ \text{sq cm} ?$
View full question & answer→MCQ 921 Mark
In a $G.P., 5^{th}$ term is $27$ and $8^{th}$ term is $729.$ Find its $11^{th}$ term.
- A
$729$
- B
$2187$
- C
$6561$
- ✓
$19683$
AnswerCorrect option: D. $19683$
Given, $a_5 = 27$ and $a_8 = 729.$
$\Rightarrow ar^4 = 27$ and $ar^7 = 729$
On dividing we get, $r^3 = 27$
$\Rightarrow r=3$
$\Rightarrow\text{a}=\frac{23}{(3^4)}=\frac{1}{3}$
$\Rightarrow\text{a}_{11}=\text{a}^{10}$
$=(\frac{1}{3})(3^{10})$
$=39$
$=19683$
View full question & answer→MCQ 931 Mark
The sum of $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ ...\text{ is}$
- ✓
$121\big(\sqrt{6}+\sqrt{2}\big)$
- B
$243\big(\sqrt{3}+1\big)$
- C
$\frac{121}{\sqrt{3}-1}$
- D
$242\big(\sqrt{3}-1\big)$
AnswerCorrect option: A. $121\big(\sqrt{6}+\sqrt{2}\big)$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{S}_{10}=\sqrt{2}\sum\limits^{10}_{\text{k}=1}\Big(\sqrt{3^{(\text{k}-1)}}\Big)$
$\Rightarrow\text{S}_{10}=\sqrt{2}\Big(1+\sqrt{3}+\sqrt{3^2}+\ ...\ +\sqrt{3^9}\Big)$
$\Rightarrow\text{S}_{10}=\sqrt{2}\bigg(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\bigg)$
$\Rightarrow\text{S}_{10\text{n}}=\sqrt{2}\Big(\frac{3^5-1}{\sqrt{3}-1}\Big)\Big(\frac{\sqrt{3}+1}{\sqrt{3}+1}\Big)$
$\Rightarrow\text{S}_{10}=\frac{\sqrt{2}}{2}\big(3^5-1\big)\big(\sqrt{3}+1\big)$
$\Rightarrow\text{S}_{10}=\frac{1}{2}(242)\Big(\sqrt{6}+\sqrt{2}\Big)$
$\Rightarrow\text{S}_{10}=121\big(\sqrt{6}+\sqrt{2}\big)$
View full question & answer→MCQ 941 Mark
Find the sum to $n$ terms of the series whose $n^{th}$ term is $n (n-2).$
- A
$\frac{\text{n}(\text{n-1)}(2\text{n}+4)}{6}$
- ✓
$\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{6}$
- C
$\frac{(\text{n-1)}(2\text{n}-5)}{3}$
- D
$\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{3}$
AnswerCorrect option: B. $\frac{\text{n}(\text{n-1)}(2\text{n}-5)}{6}$
Given, $n^{th}$ term is $n(n-2)$
So, $a_k = k(k-2)$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\sum\limits^\text{n}_\text{i}$
$=0\text{ k}^2-2\sum\limits^\text{n}_\text{i}=0\text{ k}$
$=\frac{\text{n}(\text{n+1)}(2\text{n+1}}{6}-2\frac{\text{n}(\text{n+1)}}{2}$
$=\frac{\text{n}(\text{n+1)(2}\text{n}-5)}{6}.$
View full question & answer→MCQ 951 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 961 Mark
If $p^{th,} q^{th}$ and $r^{th}$ 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 $d$ 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 $G.P.$
$\therefore (\text{a} + (\text{q}−1)\text{d})^2=( \text{a} + (\text{p}−1)\text{d})\times(\text{a} + (\text{r}−1)\text{d})$
$\Rightarrow\text{a}^2+2\text{a} (\text{q}−1)\text{d}+( (\text{q}−1)\text{d})^2=\text{a}^2+\text{ad}(\text{r}−1+\text{p}−1)+(\text{p}−1) (\text{r}−1)\text{d}^2$
$\Rightarrow(2\text{q}−2−\text{r}−\text{p}+2)+\text{d}^2(\text{q}^2−2\text{q}+1−\text{pr}+\text{p}+\text{r}−1)=0$
$\Rightarrow(2\text{q}−\text{r}-\text{p})+\text{d}(\text{q}^2−2\text{q}−\text{pr}+\text{p}+\text{r})=0$
$\Rightarrow\text{a}=-\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(2\text{q}-\text{r}-\text{p})}$
$\therefore\text{Common ratio},\text{ r}=\frac{\text{a}_\text{q}}{\text{a}_\text{p}}$
$=\frac{\text{a}+(\text{q}-1)\text{d}}{\text{a}+(\text{p}-1)\text{d}}$
$=\frac{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{q}-1)\text{d}}{\frac{(\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r})\text{d}}{(\text{p}+\text{r}-2\text{q})}+(\text{p}-1)\text{d}}$
$=\frac{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{pq}+\text{rq}+\text{rq}-2\text{q}^2-\text{p}-\text{r}+2\text{q}}{\text{q}^2-2\text{q}-\text{pr}+\text{p}+\text{r}+\text{p}^2+\text{pr}-2\text{pq}-\text{p}-\text{r}+2\text{q}}$
$=\frac{\text{pq}-\text{pr}-\text{q}^2+\text{qr}}{\text{p}^2+\text{q}^2-2\text{pq}}$
$=\frac{\text{p}(\text{p}-\text{r})-\text{q}(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{p}-\text{q})(\text{q}-\text{r})}{(\text{p}-\text{q})^2}$
$=\frac{(\text{q}-\text{r})}{(\text{p}-\text{q})}$
View full question & answer→MCQ 971 Mark
Choose the correct answer. The lengths of three unequal edges of a rectangular solid block are in $G.P.$ If the volume of the block is $216\ cm^3$ and the total surface area is $252\ cm^2$, then the length of the longest edge is:
- ✓
$12\ cm$
- B
$6\ cm$
- C
$18\ cm$
- D
$3\ cm$
AnswerCorrect option: A. $12\ cm$
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 $= ar = 6 \times 2 = 12$
View full question & answer→MCQ 981 Mark
If $2p + 3q + 4r = 15,$ then the maximum value of $p_3\ q_5\ r_7$ is:
AnswerCorrect option: C. $5^5\times\frac{7^7}{2^{17}}\times9$
View full question & answer→MCQ 991 Mark
A series can also be denoted by symbol $.........$
- A
$\pi\text{a}_\text{n}$
- ✓
$\sum\text{a}_{\text{n}}$
- C
$\phi\text{a}_{\text{n}}$
- D
$\theta\text{a}_{\text{n}}$
AnswerCorrect option: B. $\sum\text{a}_{\text{n}}$
When we use addition between the terms of sequence, it is said to be series.
We know that addition can also be written in the form of sigma
so, series can also be denoted by $\sum\text{a}_{\text{n}}.$
View full question & answer→MCQ 1001 Mark
If $r = 1$ in a $G.P.$ then what is the sum to $n$ terms?
AnswerCorrect option: A. $n\times a$
If a is the first term of $G.P.,$ then $G.P.$
look like $a, a, a, a, …………$
Then sum to n terms becomes $n \times a.$
View full question & answer→