MCQ 11 Mark
If $k, 2k - 1$ and $2k + 1$ are three consecutive terms of an $A.P.,$ the value of $k$ is :
AnswerIf $a, b, c$ are in $A.P.$
$b - a = c - b ....(1)$
According to the given problem
$a = k, b = 2k - 1, c = 2k + 1$
Using $(1)$
$(2k - 1) - k = (2k + 1) - (2k - 1)$
$2k - 1 - k = 2k + 1 - 2k - 1$
$k - 1 = 2$
$k = 2 + 1$
$k = 3$
View full question & answer→MCQ 21 Mark
If the $n^{th}$ term of an $A.P$. is $(2n + 1),$ then the sum of its first three terms is :
AnswerGiven the $n^{th}$ term of an $A.P$. is $2n + 1$.
i.e. $, T_n= 2n + 1$
So,
$ T_1=2 \times 1+1=2+1=3 $
$ T_2=2 \times 2+1=4+1=5 $
$ T_3=2 \times 3+1=6+1=7 $
Now, sum of its first three term $=T_1+T_2+T_3=3+5+7=15$
Hence, sum of first three term of an $A.P$. is $15.$
View full question & answer→MCQ 31 Mark
If the common difference of an $A.P$. is $3,$ then $a_{20}- a_{15}$ is :
AnswerLet the first term of the $A.P$. be $a$.
$ a_n=a(n-1) d $
$ a_{20}-a_{15}=[a+(20-1) d]-[a+(15-1) d] $
$= 19d - 14d$
$= 5d$
$= 5 \times 3$
$= 15$
View full question & answer→MCQ 41 Mark
The first three terms of an $AP$ respectively are $3y - 1, 3y + 5$ and $5y + 1$. Then $y$ equals :
AnswerThe first three terms of an $AP$ are $3y - 1, 3y + 5$ and $5y + 1,$ respectively.
We need to find the value of $y$.
We know that if $a, b$ and $c$ are in $AP,$ then:
$\text{b} - \text{a} = \text{c - b }$
$\Rightarrow2\text{b} = \text{a + c} $
$\therefore\ 2(3\text{y} + 5) = 3\text{y} - 1 + 5\text{y} + 1 $
$\Rightarrow6\text{y} + 10 = 8\text{y}$
$\Rightarrow10 = 8\text{y} - 6\text{y}$
$\Rightarrow2\text{y} = 10$
$\Rightarrow\text{y} = 5$
Hence, the correct option is $C$.
View full question & answer→MCQ 51 Mark
The first three terms of an $AP$ respectively are $3y - 1, 3y + 5$ and $5y + 1$. Then $y$ equals :
AnswerThe first three terms of an $AP$ are $3y - 1, 3y + 5$ and $5y + 1,$ respectively.
We need to find the value of $y$.
We know that if $a, b$ and $c$ are in $AP,$ then:
$\text{b} - \text{a} = \text{c - b }$
$\Rightarrow2\text{b} = \text{a + c} $
$\therefore\ 2(3\text{y} + 5) = 3\text{y} - 1 + 5\text{y} + 1 $
$\Rightarrow6\text{y} + 10 = 8\text{y}$
$\Rightarrow10 = 8\text{y} - 6\text{y}$
$\Rightarrow2\text{y} = 10$
$\Rightarrow\text{y} = 5$
Hence, the correct option is $C.$
View full question & answer→MCQ 61 Mark
The common difference of the $AP\ \frac{1}{\text{p}},\frac{1 - \text{p}}{\text{p}}.\frac{1-2\text{p}}{\text{p}}..........\text{is}$ :
AnswerIt is given that,
$\frac{1}{\text{p}},\frac{1-\text{p}}{\text{p}},\frac{1-2\text{p}}{\text{p}}\ ....\text{is an A.P.}$
Common difference $= a_2-a_1$
$\text{d}=\Big[\frac{(1-\text{p})}{\text{p}}\Big]-\frac{1}{\text{p}}$
$=\frac{(1-\text{p}-1)}{\text{p}}$
$=\frac{\text{p}}{\text{p}}$
$=-1$
Therefore,
common difference $= d = -1.Z$
View full question & answer→MCQ 71 Mark
The sum of first $20$ odd natural numbers is :
AnswerThe first $20$ odd no, is given by :
$\{1, 3, 5, 7, 9, 11, ....,37, 39\}$
These form an $A.P.,$ where
$a_1= 1, d = 2, n = 20$
So,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{20}=\frac{20}{2}[2(1)+(20-1)2]$
$=\frac{20}{2}[2+(19\times2)]$
$=10[2+38]$
$=10\times40$
$\therefore\ \text{S}_{20}=400$
The sum of first $20$ odd no. is $400$.
View full question & answer→MCQ 81 Mark
In an $AP,$ if $d = -2, n = 5$ and $a_n= 0$, then the value of a is :
AnswerWe know that $T_n=a+(n-1) d$
Where $d=-2, n=5, T_n=0$
So $,T_n=a+(n-1) d$
$a + (5 - 1)(-2) = 0$
$a + 4(-2) = 0$
$a + (-8) = 0$
$a - 8 = 0$
$a = 8$
View full question & answer→MCQ 91 Mark
Which of the following is not an $A.P.?$
- A
$-1.2,0.8,2.8,\ \dots$
- B
$3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\ \dots$
- ✓
$\frac43,\frac73,\frac93,\frac{12}3,\ \dots$
- D
$\frac{-1}{5},\frac{-2}{5},\frac{-3}{5},\ \dots$
AnswerCorrect option: C. $\frac43,\frac73,\frac93,\frac{12}3,\ \dots$
For Arithmetic progression the series have common difference in two consecutive terms i.e. $\text{a}_2 - \text{a}_1 = \text{a}_3 - \text{a}_2$
- $-1.2,0.8,2.8,\ \dots$
Here, $\text{a}_1=1.2,\text{a}_2=0.8,\text{a}_3=2.8,$
$0.8-(-1.2)=2.8-0.8$
$2=2$
So, the series given is an $A.P$.
- $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\ \dots$
$\text{a}_2 - \text{a}_1 = \text{a}_3 - \text{a}_2$
$3+\sqrt{2}-3=3+3\sqrt{2}-3+2\sqrt{2}$
$\sqrt{2}=\sqrt{2}$
So, the series given is an $A.P$.
- $\frac43,\frac73,\frac93,\frac{12}3,\ \dots$
$\text{a}_2 - \text{a}_1 = \text{a}_3 - \text{a}_2$
$\frac73,\frac{-4}3,\frac{-9}{3},\frac{-7}3$
$\frac33=\frac23$
$3\neq2$
So, the series given is not an $A.P$.
- $\frac{-1}{5},\frac{-2}{5},\frac{-3}{5},\ \dots$
$\text{a}_2 - \text{a}_1 = \text{a}_3 - \text{a}_2$
$\frac{-2}5-\Big(\frac{-1}{5}\Big)-\Big(\frac{-3}{5}\Big)-\Big(\frac{-2}{5}\Big)$
$\frac{-2}{5}+\frac15=\frac{-3}5+\frac25$
$\frac{-2+1}{5}=\frac{-3+2}{5}$
$-1=-1$
So, the series given is an $A.P.$
Therefore, option $ C$ is correct. View full question & answer→MCQ 101 Mark
The value of $k$ for which the points $A (0, 1), B (2, k)$ and $C(4, –5)$ are collinear is :
Answergiven points are $A(0, 1), B(2, k)$ and $C(4, -5)$
Since , The points are collinear therefore , the area of triangle is equal to $0$
then,
$ \mathrm{x}_1=0, \mathrm{y}_1=1 $
$ \mathrm{x}_2=2, \mathrm{y}_2=\mathrm{k} $
$ \mathrm{x}_3=4, \mathrm{y}_3=-5 $
area of triangle $=\frac{1}{2}(\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2))$
$\frac{1}{2}(0(\text{k+5})+2(-5-1)+4(1-\text{k}))=0$
$\frac{1}{2}(-12+4-4\text{k})=0$
$\frac{1}{2}(-8-4\text{k})=0$
$-8-4\text{k}=0$
$\text{k}=-2$
hence, The value of $k$ is $-2$
View full question & answer→MCQ 111 Mark
The value of $k,$ for which the pair of linear equations $k x+y=k^2$ and $x+k y=1$ have infinitely many solutions is :
View full question & answer→MCQ 121 Mark
The exponent of $2$ in the prime factorization of $144,$ is :
Answer$144=2 \times 2 \times 2 \times 2 \times 3 \times 3$
$=2^4 \times 3^2$
$=4$
View full question & answer→MCQ 131 Mark
The common difference of the $A.P. \frac{1}{\text{p}},\frac{1-\text{p}}{\text{p}},\frac{1-2\text{p}}{\text{p}}, .......... $ is :
AnswerLet $\text{a}_1=\frac{1}{\text{p}},\text{a}_2=\frac{1-\text{p}}{\text{p}},\text{a}_3\frac{1-2\text{p}}{\text{p}}$
$\therefore\text{common difference}=\text{a}_2-\text{a}_1;\ \text{a}_3- \text{a}_2;$
$=\frac{(1-\text{p})-(1)}{\text{p}};\frac{(1-2\text{p})-(1-\text{p})}{\text{p}}$
$=-1$
View full question & answer→MCQ 141 Mark
The $n^{th}$ term of the $A.P. a, 3a, 5a, .........$ is.
- A
$na$
- B
$(2n – 1) a$
- ✓
$(2n + 1) a$
- D
$2na$
AnswerCorrect option: C. $(2n + 1) a$
$a = a, d = 3a - a = 2a$
$A_n= a + (n - 1)d$
$= a + (n - 1)2a$
$= a{1 + 2(n - 1)}$
$= a{1 + 2n - 2}$
$= a(2n - 1) = (2n - 1)a$
View full question & answer→MCQ 151 Mark
The first term of an $AP$ is $p$ and the common difference is $q,$ then its $10^{th}$ term is :
- A
$q + 9p.$
- B
$p - 9q.$
- ✓
$p + 9q.$
- D
$2p + 9q.$
AnswerCorrect option: C. $p + 9q.$
$10^{th}$ term $= p + (10 - 1)q$
$a_{10}= p + 9q.$
View full question & answer→MCQ 161 Mark
The sum of first $n$ positive integers is :
- A
$\frac{{\text{n(n - 1)}}}{2}$
- B
$\frac{{\text{n(n - 1)}}}{3}$
- C
$\frac{{\text{n(n + 1)}}}{3}$
- ✓
$\frac{{\text{n(n + 1)}}}{2}$
AnswerCorrect option: D. $\frac{{\text{n(n + 1)}}}{2}$
$n$ positive integers are $\{1, 2, 3, 4, ... n\}$
Here $, a = 1, d = 2 - 1 = 1$ and $n = n$
$\text{s}_{\text{n}} = \frac{\text{n}}{2}(\text{2a}+(\text{n - 1})\text{d})$
$\text{s}_{\text{n}} = \frac{\text{n}}{2}(\text{2a}+(\text{nd} - \text{d})$
$=\frac{\text{n}}{2}(1 + \text{n})$
$ = \frac{\text{n(n + 1)}}{2}$
View full question & answer→MCQ 171 Mark
Mark the correct alternative in the following: If the sum of three consecutive terms of an increasing $A.P$. is $51$ and the product of the first and third of these terms is $273,$ then the third term is :
AnswerGiven,
$a_1+a_2+a_3=51 .....(i)$
and $a_1 \times a_3=273 ....(ii)$
We know, $a_n=a+(n-1) d$
$1^{st }$ term, $a_1=a+(1-1) d$
$\Rightarrow a_1=a+0$
$\Rightarrow a_1=a$
$2^{nd }$ term $, a_2=a+(2-1) d$
$\Rightarrow a_2=a+d$
$3^{rd}$ term $, a_3a + (3 - 1)d$
$\Rightarrow a_3 = a + 2d .....(iii)$
Put the values in eq $(i)$
$\Rightarrow a + a + d + a + 2d = 51$
$\Rightarrow 3a + 3d = 51$
$\Rightarrow 3(a + d) = 51$
$\Rightarrow\ \text{a}+\text{d}=\frac{51}{3}$
$\Rightarrow a + d = 17$
$\Rightarrow a = 17 - d$
Put the value of $a_1$ and $a_3$ in eq. $(ii)$
$\Rightarrow (a)(a + 2d) = 273$
$\Rightarrow (17 - d)(17 - d + 2d) = 273$
$\Rightarrow (17 - d)(17 + d) = 273$
$\Rightarrow 17^2- d^2= 273$
$[\because (a - b)(a + b) = a2 - b2]$
$\Rightarrow 289 - d^2= 273$
$\Rightarrow 289 - 273 = d^2$
$\Rightarrow d^2= 16$
$\Rightarrow\ \text{d}=\sqrt{16}=4$
$d = 4$ because $d$ never be negative in incueasing $A.P$.
To find $a_3$ put the value of $d $ and a eq. $(iii)$
$ a_3=a+2 d $
$ \Rightarrow a_3=17-d+2 d $
$ \Rightarrow a_3=17-4+2(4) $
$ \Rightarrow a_3=13+8 $
$ \Rightarrow a_3=21 $
Hence, carrect chice is $(C)$.
View full question & answer→MCQ 181 Mark
The common difference of the $\text{AP}\frac{1}{\text{3}},\frac{1-3\text{b}}{\text{3}},\frac{1-6\text{b}}{\text{3}},....$ is :
- A
$\frac{1}{3}$
- B
$\frac{-1}{3}$
- C
$\text{b}$
- ✓
$-\text{b}$
AnswerCorrect option: D. $-\text{b}$
Common difference
$=\frac{1-3\text{b}}{3{}}-\frac{1}{3}$
$=\frac{1-3\text{b}-1}{3}$
$=\frac{-3\text{b}}{3}$
$= -\text{b}$
View full question & answer→MCQ 191 Mark
Mark the correct alternative in the following : If four numbers in $A.P$. are such that their sum is $50$ and the greatest number is $4$ times, the least, then the numbers are :
- ✓
$\{5, 10, 15, 20\}$
- B
$\{4, 101, 16, 22\}$
- C
$\{3, 7, 11, 15\}$
- D
AnswerCorrect option: A. $\{5, 10, 15, 20\}$
Here, we are given that four numbers are in $A.P$., such that their sum is $50$ and the greatest number is $4$ times the smallest.
So, let us take the four terms as $a - d, a, a + d, a + 2d.$
Now, we are given that sum of these numbers is $50,$ so we get,
$(a - d) + (a) + (a + d) + (a + 2d) = 50$
$a - d + a + a + d + a + 2d = 50$
$4a + 2d = 50$
$2a + d = 50 .....(i)$
Also, the greatest number is $4$ times the smallest, so we get,
$a + 2d = 4(a - d)$
$a + 2d = 4a - 4d$
$4d + 2d = 4a - a$
$6d = 3a$
$\text{d}=\frac{3}{6}\text{d}\ .....(\text{ii})$
Now, using $(ii)$ in $(i),$ we get,
$2\text{a}+\frac{3}{6}\text{a}=25$
$\frac{12\text{a}+3\text{a}}{6}=25$
$15\text{a}=150$
$\text{a}=\frac{150}{15}$
$\text{a}=10$
Now, using the value of a in $(ii),$ We get,
$\text{d}=\frac{3}{6}(10)$
$\text{d}=\frac{10}{2}$
$\text{d}=5$
So, first term is given by,
$a - d = 10 - 5$
$= 5$
Second term is given by,
$a = 10$
Third term is given by,
$a + d = 10 + 5$
$= 15$
Fourth term is given by,
$a + 2d = 10 + (2)(5)$
$= 10 + 10$
$= 20$
Therefore, the four terms are $\{5, 10, 15, 20\}.$
Hence, the correct opting is $(A)$.
View full question & answer→MCQ 201 Mark
Mark the correct alternative in the following : The common difference of the $A.P$. is $\frac{1}{2\text{q}},\frac{1-2\text{q}}{2\text{q}},\frac{1-4\text{q}}{2\text{q}}, .....$ is
AnswerLet a be the first term and d be the common difference.
The given $A.P$. is $\frac{1}{2\text{q}},\frac{1-2\text{q}}{2\text{q}},\frac{1-4\text{q}}{2\text{q}}, .....$
Common difference $= d =$ Second term $-$ First term
$=\frac{1-2\text{q}}{2\text{q}}-\frac{1}{2\text{q}}$
$=\frac{-2\text{q}}{2\text{q}}=-1$
Hence, the correct option is $(A).$
View full question & answer→MCQ 211 Mark
Choose the correct answer from the given four options : The famous mathematician associated with finding the sum of the first $100$ natural numbers is :
AnswerGauss is the famous mathematician associated with finding the sum of the first natural numbers
i.e., $\{1, 2, 3 ............ 100.\}$
View full question & answer→MCQ 221 Mark
The $17^{th}$ term of an $A.P.$ exceeds its $10^{th}$ term by $7,$ then the common difference is :
AnswerAccording to question,
Given that the $17^{th}$ term of an $A.P$ exceeds its $10^{th}$ term by $7$.
$d = ?$
$\Rightarrow a + 16d = a + 9d + 7$
$\Rightarrow 16d - 9d = 7$
$\Rightarrow 7d = 7$
$\Rightarrow\text{d} = \frac{7}{7} = 1$
$\therefore$ common difference $= 1.$
View full question & answer→MCQ 231 Mark
Mark the correct alternative in the following : The next term of the $A.P. \sqrt{7},\sqrt{28},\sqrt{63},\ .....$
- A
$\sqrt{70}$
- B
$\sqrt{84}$
- C
$\sqrt{97}$
- ✓
$\sqrt{112}$
AnswerCorrect option: D. $\sqrt{112}$
Let a be the first term and $d$ be the common difference.
The given $A.P$. is $\sqrt{7},\sqrt{28},\sqrt{63},\ .....$
$\text{i.e.},\sqrt{7},\sqrt{4\times7},\sqrt{9\times7}$
$\text{i.e.},\sqrt{7},2\sqrt{7},3\sqrt{7},\ .....$
Common difference $= d =$ Second term $-$ First term
$=2\sqrt{7}-\sqrt{7}$
$=\sqrt{7}$
$\therefore$ Next term of the $A.P. =3\sqrt{7}+\sqrt{7}$
$=4\sqrt{7}$
$=\sqrt{16\times7}$
$=\sqrt{112}$
Hence, the correct option is $(D).$
View full question & answer→MCQ 241 Mark
The common difference of the $A.P.\ \frac{1}{\text{p}}, \frac{1 - \text{p}}{\text{p}}, \frac{1 - \text{2p}}{\text{p}},$ is :
Answer$\text{d} = \frac{(1-\text{p)}}{\text{p}}-\frac{1}{\text{p}} $
$= \big(\frac{1-\text{p - 1}}{\text{p}}\big) = \frac{\text{-p}}{\text{p}}=-1$
View full question & answer→MCQ 251 Mark
If $a, 7, b, 23, c$ are in $A.P$. then the value of $c$ is :
AnswerLet $d$ be a common difference. Then,
$a_5 = c = a + 4d ... (i)$
$a_2= a + d = 7 ... (ii)$
$a_4= a + 3d = 23 ... (iii)$
Solving eq. $(ii)$ and $(iii),$
we get $a = -1$ and $d = 8$
$\therefore c = a + 4d$
$= -1 + 4 \times 8 = 31$
View full question & answer→MCQ 261 Mark
Which term of the $AP\ \ \{ 72, 63 54, ....\}$ is $0$ ?
- A
$8^{\text {th }}$
- ✓
$9^{\text {th }}$
- C
$10^{\text {th }}$
- D
$11^{\text {th }}$
AnswerCorrect option: B. $9^{\text {th }}$
The given $AP$ is $\{72, 63, 54, .....\}$
$a = 72$ and $d = 63 - 72 = -9$
$a_n= a + (n - 1)d$
$\Rightarrow 0 = 72 + (n -1)(-9)$
$\Rightarrow -72 = (n - 1)(-9)$
$\Rightarrow 8 = n - 1$
$\Rightarrow n = 9$
So, the $9^{th}$ term is $0$.
View full question & answer→MCQ 271 Mark
Choose the correct answer from the given four options : The list of numbers $\{– 10, – 6, – 2, 2,...\}$ is :
- A
An $AP$ with $d = -16$
- ✓
An $AP$ with $d = 4$
- C
An $AP$ with $d = -4$
- D
Not an $AP$
AnswerCorrect option: B. An $AP$ with $d = 4$
The given numbers are $\{-10, -6, -2, 2..........\}$
Here, $a,=-10, a_2=-6, a_3=-2$ and $a_4=2 \ldots \ldots \ldots$
Since, $a_2-a_1=-6-(-10)$
$ a_2-a_1=-6+10=4 $
$ a_3-a_2=-2-(-6) $
$ a_4-a_3=2-(-2) $
$ a_4-a_3=2+2=4 ..............$
Each successive term of given list has same difference i.e. $4$.
So, the given list forms an $AP$ with common difference, $d = 4$.
View full question & answer→MCQ 281 Mark
The $5^{th}$ term of an $AP$ is $20$ and the sum of its $7^{th}$ and $11^{th}$ terms is $64$. The common difference of the $AP$ is :
AnswerLet a be frist term and $d$ be the common difference.
$a_{5=20}$
$\Rightarrow a + 4d= 20 .....(i)$
$S_7+ S_{11}= 64$
$\Rightarrow a + 6d + a + 10d = 64$
$\Rightarrow 2a + 16d = 64$
$\Rightarrow a + 8d = 32 .....(ii)$
Subtracting $(i)$ from $(ii),$ we get
$4d = 12$
$\Rightarrow d = 3$
View full question & answer→MCQ 291 Mark
A sum of $Rs. 700$ is to be used to award $7$ prizes. If each prize is $Rs. 20$ less than its preceding prize, then the value of the first prize is :
- A
$Rs. 100$
- ✓
$Rs. 160$
- C
$Rs. 200$
- D
$Rs. 180$
AnswerCorrect option: B. $Rs. 160$
Let the first prize be a.
The seven prizes form an $AP$ with first term a and common difference $, d = -20, n = 7$
Now the sum of all seven prizes $= Rs. 700$
$\therefore S_n= 700$
$\Rightarrow\frac{\text{n}}{2}[\text{2a}+(\text{n - 1})\text{d}] = 700$
$\Rightarrow\frac{7}{2}[\text{2a} + (7-1)(-20)] = 700$
$\Rightarrow 2a - 120 = 200$
$\Rightarrow 2a = 320$
$\Rightarrow a = 160$
View full question & answer→MCQ 301 Mark
Mark the correct alternative in the following : The common difference of the $A.P. \ \frac{1}{3},\frac{1-3\text{b}}{3},\frac{1-6\text{b}}{3}, ....$ is
- A
$\frac{1}{3}$
- B
$-\frac{1}{3}$
- ✓
$-\text{b}$
- D
$\text{b}$
AnswerCorrect option: C. $-\text{b}$
$A.P$. is $\frac{1}{3},\frac{1-3\text{b}}{3},\frac{1-6\text{b}}{3}, .....\ \text{is}$
Common difference $\text{d}=\frac{1-3\text{b}}{3}-\frac{1}{3}=\frac{1-6\text{b}}{3}-\frac{(1-3\text{b})}{3}$
$\frac{1-3\text{b}-1}{3}=\frac{1-6\text{b}-1+3\text{b}}{3}$
$-3\text{b}=-3\text{b}$
$-\text{b}=-\text{b}$
View full question & answer→MCQ 311 Mark
Mark the correct alternative in the following: If in an $A.P. \ S_n=n^2 p$ and $S_m=m^2 p,$ where $S_r$ denotes the sum of $r$ terms of the $A.P.,$ then $S_p$ is equal to :
- A
$\frac{1}{2}\text{p}^2$
- B
$\text{mnp}$
- ✓
$p^3$
- D
$(m + n)p^2$.
Answer$ S_n=n^2 p, S_m=m^2 p$
$ \therefore S_r=r^2 p $ and $ S_p=p^2 q=p^3$
Hence $S_p=p^3$.
View full question & answer→MCQ 321 Mark
Mark the correct alternative in the following : The number of terms of the $A.P. \ \{3, 7, 11, 15, .....\}$ to be taken so that the sum is $406$ is :
AnswerGiven, $A.P. \ \{3, 7, 11, 15, .....\}$
$S_n= 406$
First term $, a = 3$
and Difference $, d = 7 - 3 = 4$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 406=\frac{\text{n}}{2}[2(3)+(\text{n}-1)4]$
$\Rightarrow 406 \times 2 = n[6 + 4n - 4]$
$\Rightarrow 812 = n[4n + 2]$
$\Rightarrow 812 = n \times 2[2n + 1]$
$\Rightarrow 406 = n[2n + 1]$
$ \Rightarrow 406=2 n^2+n $
$ \Rightarrow 2 n^2+n-406=0 $
$ \Rightarrow 2 n^2+29 n-28 n-406=0 $
$\Rightarrow n(2n + 29) - 14(2n + 29) = 0$
$\Rightarrow (n - 14)(2n + 29) = 0$
Here,
$n - 14 = 0$
$\Rightarrow n = 14$
and $2n + 29 = 0$
$2n = -29$
$\text{n}=\frac{-29}{2}$
We know term never ne negative
So, number of terms is $14$
Hence, correct choice is $(D).$
View full question & answer→MCQ 331 Mark
The sum of first $20$ odd natural numbers is :
AnswerThe frist $20$ odd natural numbers will be $\{1, 3, 5, 7, ......\}$
Here,
$a = 1$
$d = 3 - 1 = 2$
$n = 20$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_{20}=\frac{20}{2}\big[2(1)+19(2)\big]$
$\Rightarrow\text{S}_{20}=10[2+38]$
$\Rightarrow\text{S}_{20}=400$
View full question & answer→MCQ 341 Mark
The value of $k$ for which the numbers $x, 2x + k, 3x + 6$ are in $A.P$. is :
AnswerIf the numbers $x, 2x + k, 3x + 6 $ are in $A.P.,$
then $, 2x + k - x = 3x + 6 - 2x - k$
$\Rightarrow x + k = x + 6 - k$
$\Rightarrow 2k = 6$
$\Rightarrow k = 3$
View full question & answer→MCQ 351 Mark
If $a, b, c, l, m$ are in $A.P$. then the value of $a - 4b + 6c - 4l + m$ is :
AnswerGiven: $a, b, c, l, m$ are in $A.P.$
$\therefore a + m = 2c ... (i)$
$b + l = 2c ... (ii)$
$a - 4b + 6c - 4l + m$
$= a + m + 6c - 4b - 4l$
$= a + m + 6c - 4(b + l)$
substituting from $(i)$ and $(ii)$
$= 2c + 6c - 8c$
$= 0$
View full question & answer→MCQ 361 Mark
Mark the correct alternative in the following: If the first, second and last term of an $A.P,$ are $a, b$ and $2a$ respectively, its sum is
- A
$\frac{\text{ab}}{2(\text{b}-\text{a})}$
- B
$\frac{\text{ab}}{(\text{b}-\text{a})}$
- ✓
$\frac{3\text{ab}}{2(\text{b}-\text{a})}$
- D
AnswerCorrect option: C. $\frac{3\text{ab}}{2(\text{b}-\text{a})}$
In the given problem, we are given first, second and last term of an $A.P$. We need to find its sum.
So, here
First term $= a$
Second term $(a_2) = b$
Last term $(l) = 2a$
Now, using the formula $a_n= a + (n - 1)d$
$a_2= a + (2 - 1)d$
$b = a + d$
$d = b - a .....(i)$
Also,
$a_n= a + (n - 1)d$
$2a = a + nd - d$
$2a - a = nd - d$
$\frac{\text{a}+\text{d}}{\text{d}}=\text{n}\ .....\text{(ii)} $
Furthere as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
Substituting $(ii) $ in the above equation, we get
Using $(i),$ we get
$\text{S}_\text{n}=\frac{\text{a}+(\text{b}-\text{a})}{2(\text{b}-\text{a})}(3\text{a})$
$\text{S}_\text{n}=\frac{\text{b}}{2(\text{b}-\text{a})}(3\text{a})$
Thus,
$\text{S}_\text{n}=\frac{3\text{ab}}{2(\text{b}-\text{a})}$
Therefore, the correct option is $(C)$.
View full question & answer→MCQ 371 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : If $S_n$ is the sum of the first $n$ terms of an $A.P,$ then its $n^{\text {th }}$ term $a_n$ is given by $a_n=S_n-S_{n-1}$.
Reason : The $10^{\text {th }}$ term of the $A.P. 5,8,11,14, \ldots \ldots$ is $35$ .
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ isthe correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
$a_{10} = a + 9d$
$= 5 + 9(3)$
$= 5 + 27 = 32$
View full question & answer→MCQ 381 Mark
In an $A.P$., the third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12,$ then its first term is :
AnswerAccording to question,
$a_3=16$
$\Rightarrow a + 2d = 16 ... (i)$
And $a_7=a_5+12$
$\Rightarrow a + 6d = a + 4d + 12$
$\Rightarrow a + 6d - a - 4d = 12$
$\Rightarrow 2d = 12$
$\Rightarrow d = 6$
Putting value of $d$ in eq. $(i),$
we get $a + 2 \times 6 = 16$
$\Rightarrow a = 4$
View full question & answer→MCQ 391 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : Sum of first hundred even natural numbers divisible by $5$ is $500.$
Reason : Sum of first $n-$ terms of an $A.P$. is given by $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a+}\ell] \text{ where }\ell =$ last term.
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ isthe correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A).$
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true
$\{10, 20, 30, 40, ..........\}$
They form an $A.P.$ with,
$\text{a}=10,\text{d}=10$
$\text{s}_{100}=\frac{100}{2}[2(10)+99(10)]50500$
Reason is correct.
View full question & answer→MCQ 401 Mark
Mark the correct alternative in the following : It the sums of n terms of two arithmetic progressions are in the ratio $\frac{3\text{n}+5}{5\text{n}-7},$ then their $n^{th}$ terms are in the ratio.
- A
$\frac{3\text{n}-1}{5\text{n}-1}$
- ✓
$\frac{3\text{n}+1}{5\text{n}+1}$
- C
$\frac{5\text{n}+1}{3\text{n}+1}$
- D
$\frac{5\text{n}-1}{3\text{n}-1}$
AnswerCorrect option: B. $\frac{3\text{n}+1}{5\text{n}+1}$
Given,
$\frac{\text{Sum of A.P}_1}{\text{Sum of A.P}_2}=\frac{\text{S}_\text{n}}{\text{S}^1_\text{n}}=\frac{3\text{n}+5}{5\text{n}+7}\ .....\text{(i)}$
We know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P._1$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P._2$ $\text{S}'_\text{n}=\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']$
Put the value in eq. $(i)$
$\Rightarrow\ \frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{3\text{n}+5}{5\text{n}+7}=\frac{[2\text{a}+(\text{n}-1)\text{d}]}{[2\text{a}'+(\text{n}-1)\text{d}']}$
Now, put the $n = 2n - 1$
$\Rightarrow\ \frac{3(2\text{n}-1)+5}{5(2\text{n}-1)+7}=\frac{[2\text{a}+(2\text{n}-1-1)\text{d}]}{[2\text{a}'+(2\text{n}-1-1)\text{d}']}$
$\Rightarrow\ \frac{6\text{n}-3+5}{10\text{n-5+7}}=\frac{[2\text{a}+(2\text{n}-2)\text{d}]}{[2\text{a}'+(2\text{n}-2)\text{d}']}$
$\Rightarrow\ \frac{6\text{n}+2}{10\text{n}+2}=\frac{2\text{a}+2(\text{n}-1)\text{d}}{2\text{a}'+2(\text{n}-1)\text{d}'}$
$\Rightarrow\ \frac{2(3\text{n}+1)}{2(5\text{n}+1)}=\frac{2[\text{a}+(\text{n}-1)\text{d}]}{2[\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{3\text{n}+1}{5\text{n}+1}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n-1})\text{d}']}$
$\Rightarrow\ \frac{3\text{n}+1}{5\text{n}+1}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n}-1)\text{d}']}$
We know, $a_n= a + (n - 1)d$
$\Rightarrow\ \frac{3\text{n}+1}{5\text{n}+1}=\frac{\text{a}_\text{n}}{\text{a}'_\text{n}}$
Hence, correct choice is $(B).$
View full question & answer→MCQ 411 Mark
The next term of the $\text{AP}\sqrt{7},\sqrt{28},\sqrt{63},\ ...$ is :
- A
$\sqrt{70}$
- B
$\sqrt{84}$
- C
$\sqrt{98}$
- ✓
$\sqrt{112}$
AnswerCorrect option: D. $\sqrt{112}$
The $AP$ is given to be $\sqrt{7},\sqrt{28},\sqrt{63}.$
Let $d$ be the common difference.
$\text{d}=\sqrt{28}-\sqrt{7}=2\sqrt{7}-\sqrt{7}=\sqrt{7}$
So, the next term $= \sqrt{63}+\sqrt{7}$
$=3\sqrt{7}+\sqrt{7}$
$=4\sqrt{7}$
$=\sqrt{112}$
View full question & answer→MCQ 421 Mark
Mark the correct alternative in the following : The common difference of an $A.P.,$ the sum of whose $n$ terms is $S_n$, is
AnswerCorrect option: A. $ S_n-2 S_{n-1}+S_{n-2} $
The sum of $u$ term of an $A.P. = =S_n$
There fore sum of $n - 1$ term of $A.P. =S_{n-1}$
Similarly sum of $n - 2$ term of $A.P. =S_{n-2}$
an $={S}_{n}-{S}_{n}-1$ and ${a}_{n}-1={S}_{n}-1-{S}_{n}-2$
Common difference,
$d=a_n-a_{n-1}=S_n-S_{n-1}-\left[S_{n-1}-s_{n-2}\right] $
$ =S_n-S_{n-1}-S_{n-1}+S_{n-2} $
$=S_n-2 S_{n-1}+S_{n-2}$
Corrent option is $(a)$.
View full question & answer→MCQ 431 Mark
Choose the correct answer from the given four options : In an $AP$ if $a = 1, a_n= 20$ and $S_n= 399,$ then $n$ is :
Answer$\because\text{S}_{\text{n}}= \frac{\text{n}}{2}\big[2\text{a}+\big(\text{n}-1\big)\text{d}\big] $
$399= \frac{\text{n}}{2}\big[2\times1+\big(\text{n}-1\big)\text{d}\big] $
$798 = 2n + n(n - 1)d ........(i)$
and $an = 20 [\because an = a + (n - 1)d]$
$\Rightarrow a + (n - 1)d = 20$
$\Rightarrow (n - 1)d = 19 .......(ii)$
Using Eq. $(ii)$ in Eq. $(i),$ we get
$\Rightarrow 798 = 2n + 19n$
$\Rightarrow 798 = 21n$
$\therefore\text{n}=\frac{498}{21}=38$
View full question & answer→MCQ 441 Mark
Mark the correct alternative in the following : The first and last term of an $A.P$. are $a$ and $l$ respectively. If $S$ is the sum if all the terms of the $A.P$. and the common difference is given by $\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})}$, then $k =$
AnswerGiven,
First term $, a$
Last term $, l$
Sum of $l$ terms, $S_l$
and Difference, $\text{d}=\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})}\ .....(\text{i})$
We know $, l = a + (n - 1)d$
put in eq. $(i) \ \text{d}=\frac{(\text{a}+(\text{n}-1)\text{d})^2-\text{a}^2}{\text{k}-[(\text{a}+(\text{n}-1)\text{d})+\text{a}]}$
$\Rightarrow\ \text{d}=\frac{\text{a}^2+((\text{n}-1)\text{d})^2+2\text{a}(\text{n}-1)\text{d}-\text{a}^2}{\text{k}-[\text{a}+\text{nd}-\text{d}+\text{a}]}$
$\text{d}=\frac{(\text{n}-1)^2\text{d}^2+2\text{a}(\text{n}-1)\text{d}}{\text{k}-[2\text{a}+(\text{n}-1)\text{d}]}$
$\Rightarrow\ \text{d}=\frac{\text{d}[(\text{n}-1)^2\text{d}+2\text{a}(\text{n}-1)]}{\text{k}-[2\text{a}+(\text{n}-1)\text{d}]}$
$\Rightarrow k - [2a+ (n - 1)d] = (n - 1)2 d + 2a (n - 1)$
$\Rightarrow k = (n - 1)2 d + 2a (n - 1) + [2a + (n - 1)d]$
$\Rightarrow k = (n - 1)[(n - 1)d + 2a] + [2a + (n - 1)d]$
$\Rightarrow k = 2a + (n - 1)d [n - 1 + 1]$
$\Rightarrow k = (2a + (n - 1)d)n$
We know, $\text{S}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Now divide by $2$ both side
$\Rightarrow\ \frac{\text{k}}{2}=(2\text{a}+(\text{n}-1)\text{d})\frac{\text{n}}{2}$
$\Rightarrow\ \frac{\text{k}}{2}=\text{S}$
$\Rightarrow\ \text{k}=2\text{S}$
Hence, correct chice is $(B)$.
View full question & answer→MCQ 451 Mark
Mark the correct alternative in the following : If the first term of an $A.P$. is a and $n^{th}$ term is $b,$ then its common difference is :
- A
$\frac{\text{b}-\text{a}}{\text{n}+1}$
- ✓
$\frac{\text{b}-\text{a}}{\text{n}-1}$
- C
$\frac{\text{b}-\text{a}}{\text{n}}$
- D
$\frac{\text{b}+\text{a}}{\text{n}-1}$
AnswerCorrect option: B. $\frac{\text{b}-\text{a}}{\text{n}-1}$
Here,
We are given the first term of the $A.P.$ as a and the $n^{th}$ term $(a_n)$ as $b$.
So, let us take the common difference of the $A.P$. as $d$.
Now, as we know,
$a_n= a + (n - 1)d$
On substituting the values given in the question, we get.
$b = a + (n - 1)d$
$(n - 1)d = b - a$
$\text{d}=\frac{\text{b}-\text{a}}{\text{n}-1}$
$\therefore\text{d}=\frac{\text{b}-\text{a}}{\text{n}-1}$
Hence the correct option is $(B)$.
View full question & answer→MCQ 461 Mark
Mark the correct alternative in the following : The sum of first $20$ odd natural numbers is :
AnswerLet a be the first term and $d$ be the common difference.
We know that, sum of first $n$ terms $=\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{a}]$
The given series is $1 + 3 + 5 + .....$
First term $= a = 1$
Common difference $= d = 3 - 1 = 2$
$\therefore\ \text{S}_{20}=\frac{20}{2}[2\times1+(20-1)2]$
$= 10(2 + 19 \times 2)$
$= 10(40)$
$= 400$
Hence, the correct option is $(C)$.
View full question & answer→MCQ 471 Mark
The common difference of an $A.P$. in which $a_{18}- a_{14}= 32$ is :
AnswerGiven : $a_{18}- a_{14}= 32$
$\Rightarrow a + (18 - 1)d - [a + (14 - 1)d] = 32$
$\Rightarrow a + 17d - a - 13d = 32$
$\Rightarrow 4d = 32$
$\Rightarrow d = 8$
View full question & answer→MCQ 481 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : $a_n- a_{n-1}$ is not independent of $n$ then the given sequence is an $AP$.
Reason : Common difference $d = a_n - a_{n-1}$ is constant or independent of $n$.
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ isthe correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true
Assertion is incorrect.
We have, common difference of an $AP$
$d =a_n- a_{n-1}$ is independent of $n$ or constant.
So, $A$ is correct but $R$ is incorrect.
View full question & answer→MCQ 491 Mark
How many three $-$ digit numbers are divisible by $9$ ?
AnswerThe two $-$ digit numbers divisible by $9$ start from
$\{108, 117, 126, 135, ......., 999\}$
Here,
$a = 108$
$d = 9$
$a_n= a + (n - 1)$
$\Rightarrow 999 = 108 + (n - 1)(9)$
$\Rightarrow 999 = 108 + 9n - 9$
$\Rightarrow 900 = 9n$
$\Rightarrow n = 100$
View full question & answer→MCQ 501 Mark
Which of the following is not an $A.P.?$
- A
$\{-1.2, -3.2, -5.2, -7.2, ...\}$
- B
$\{a, 2a, 3a, 4a, ...\}$
- ✓
$\{2, 4, 8, 16, ...\}$
- D
$\{2, 5252, 3, 7272, ...\}$
AnswerCorrect option: C. $\{2, 4, 8, 16, ...\}$
In $\{2, 4, 8, 16, ...\}$
$d=a_2-a_1=4-2=2$
And $d=a_3-a_2=8-4=4$
Also $d=a_4-a_3=16-8=8$
Here, the common difference is not the same for all terms,
$\therefore$ it is not an $AP.$
View full question & answer→