MCQ 11 Mark
The $7^{\text {th }}$ term of the given Arithmetic Progression (A.P.):
$\frac{1}{a},\left(\frac{1}{a}+1\right) \cdot\left(\frac{1}{a}+2\right), \ldots$ is
- ✓
$\left(\frac{1}{a}+6\right)$
- B
$\left(\frac{1}{a}+7\right)$
- C
$\left(\frac{1}{a}+8\right)$
- D
$\left(\frac{1}{a}+7^7\right)$
AnswerCorrect option: A. $\left(\frac{1}{a}+6\right)$
(a) $\left(\frac{1}{a}+6\right)$
Explanation:
Given A.P. is $\frac{1}{a} \cdot\left(\frac{1}{a}+1\right) \cdot\left(\frac{1}{a}+2\right)$
Here, first term, $A =\frac{1}{a}$
Common difference $D =\frac{1}{a}+1-\frac{1}{a}=1$
Then, $7^{\text {th }}$ term of A.P. $= A +(n-1) D$
$\begin{array}{l}=\frac{1}{a}+(7-1) \times 1 \\ =\frac{1}{a}+6\end{array}$
View full question & answer→MCQ 21 Mark
The $n^{\text {th }}$ term of an Arithmetic Progression (A.P.) is $2 n+5$. The $10^{\text {th }}$ term is:
Answer(c) 25
Explanation:
$\because n^{\text {th }}$ term of A.P. $=2 n+5$
Put $n=10$
$\therefore 10^{\text {th }}$ term of A.P. $=2 \times 10+5$
$=20+5=25$ Ans.
View full question & answer→MCQ 31 Mark
Statement (A): The 10th term of the AP $8,10,12, \ldots 126$ is 36
Statement (B): The sum of the 10th terms of the AP 8, 10, 12, ..., 126 is 170
Which of the statement is valid?
Answer(b) Only B
Explanation:
We have, $8,10,12, \ldots .126$
$\begin{array}{l}a=8, d=10-8=2, n=10 \\ a_n=a+(n-1) d \\ a_{10}=8+(10-1)^2=8+18=26 \\ a_{10}=26\end{array}$
$\therefore$ Statement A is incorrect
$\begin{array}{l}S n=\frac{n}{2}[2 a+(n-11) d] \\ S_{10}=\frac{10}{2}[2 \times 8+(10-1) \times 2]=5(16+18) \\ =5 \times 34=170\end{array}$
$\therefore$ Statement B is correct.
View full question & answer→MCQ 41 Mark
Statement (A): The first term and the common difference of the A.P. $2,7,12, \ldots \ldots$ is 2 and 5 respectively.
Statement (B): The 10th term of the A.P. 2, 7, $12 \ldots$ is 47
Which of the statement is valid?
Answer(c) Both A and B
Explanation:
We have $2,7,12, \ldots$
first term $(a)=2$
common difference $( d )=7-2=5$
$n=10$
nth term of an A.P. is
$a_n=a+(n-1) d$
10th term of an A.P. is
$\begin{array}{l}a_{10}=2+(10-1) \times 5 \\ =2+45 \Rightarrow a_{10}=47\end{array}$
View full question & answer→MCQ 51 Mark
Which of the following is/are an Arithmetic Progression (A.P.)?
$1.1,4,9,16$,.........
2. $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, 4 \sqrt{3}$..........
3.8,6,4,2, ..........
Answer(c) only 2 and 3
Explanation:
1. Here,
$\begin{array}{l}a_2-a_1=4-1=3, a_3-a_2=9-4=5, a_4-a_3=16-9=7 \text { etc. } \\ \Rightarrow a_2-a_1 \neq a_3-a_2\end{array}$
Hence, the given list of numbers does not form an A.P.
2. Here,
$a_2-a_1=2 \sqrt{3}-\sqrt{3}=\sqrt{3}, a_3-a_2=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}, a_4-a_3=4 \sqrt{3}-3 \sqrt{3}=\sqrt{3}$, etc.
Any term - Preceding term $=\sqrt{3}$, a fixed number.
Hence, the given list of numbers form an A.P.
3. Here,
$a_2-a_1=6-8=-2, a_3-a_2=4-6=-2, a_4-a_3=4-2=-2$
Any term - Preceding term $=-2$, a fixed number
Hence, the given list of numbers forms an A.P.
View full question & answer→MCQ 61 Mark
The $n^{\text {th }}$ term of an Arithmetic progression (A.P.) is $3 n-4$. The $14^{\text {th }}$ term is:
Answer(b) 38
Explanation:
$\ldots n^{\text {th }}$ term of A.P. $=3 n-4$
Put $n=14$
$\begin{array}{l}\therefore 14^{\text {th }} \text { term of A.P. }=3(14)-4 \\ =42-4 \\ =38\end{array}$
View full question & answer→MCQ 71 Mark
Eighth term from the end of the A.P. $41,22,03, ...... -266$ is:
Answer(d) -133
Explanation:
The reverse A.P. of the given A.P. is
-266,-247, ...... , 03,22,41
Now, for this A.P.,
a=-266, d=-247-(-266)=-247+266=19
$\therefore$ Required term $=8$ th term from the end of the $41,22, \ldots$.
$\begin{array}{l}=8 \text { th term from the beginning }-266,-247, \ldots, \\ =t_8 \\ =a+7 d \\ =-266+7(19) \\ =-266+133 \\ =-133\end{array}$
View full question & answer→MCQ 81 Mark
The sum of first twenty five terms of the A.P. $6,2,-2,-6, \ldots$ is:
Answer(c) -420
Explanation:
Here, $a=6, d=2-6=-4$ and $n=25$
We know, $S _n=\frac{n}{2}[2 a+(n-1) a]$
$S_{10}=\frac{10}{2}[2 \times 6+(25-1)(-4)]$
$\begin{array}{l}=5 \times(12-96) \\ =5 \times(-84) \\ =-420\end{array}$
View full question & answer→MCQ 91 Mark
The number of terms in the A.P. $-25,-29 \frac{1}{2},-34, \ldots . .,-70$ is:
Answer(a) 11
Explanation:
Here, $a=-25, d=-29 \frac{1}{2}-(-25)$
$\begin{array}{l}=-\frac{59}{2}+25 \\ =\frac{-59+50}{2}=\frac{-9}{2}\end{array}$
and $l=-70$
We know, $l=a+(n-1) d$
$\begin{array}{l}\Rightarrow-70=-25+(n-1)\left(\frac{-9}{2}\right) \\ \Rightarrow-140=-50+(n-1)(-9) \\ \Rightarrow-140+50=(n-1)(-9) \\ \Rightarrow-90=(n-1)(-9) \\ \Rightarrow n-1=\frac{-90}{-9} \\ \Rightarrow n-1=10 \\ \Rightarrow n=11 .\end{array}$
View full question & answer→MCQ 101 Mark
If the common difference of an A.P. is 8 , then the difference in its $27^{\text {th }}$ and $14^{\text {th }}$ terms is:
Answer(c) 104
Explanation:
$\begin{array}{l}t_{27}-t_{14}=(a+26 d)-(a+13 d) \\ =13 d \\ =13 \times 8 \\ =104\end{array}$
View full question & answer→MCQ 111 Mark
The $12^{\text {th }}$ term of the A.P. $25,17,9, \ldots$...is:
Answer(b) -63
Explanation:
In the given A.P.,
$\begin{array}{l}a=25 \text { and } d =17-25=-8 \\ \therefore 12^{\text {th }} \text { term }=t 12=a+(12-1) \\ =25+11 \times(-8) \\ =25-88 \\ =-63\end{array}$
View full question & answer→MCQ 121 Mark
If the $n^{\text {th }}$ term of an Arithmetic Progression (A.P.) is $( n +3)$, then the first three terms of the A.P. are:
- A
$1,2,3$
- B
$2,4,6$
- ✓
$4,5,6$
- D
$7,8,9$
AnswerCorrect option: C. $4,5,6$
(c) 4, 5,6
Explanation:
The $n^{\text {th }}$ term of an $( A \cdot P )=(n+3)$
Then, First term of an A.P $=(1+3)=4$
Second term of an A.P $=(2+3)=5$
Third term of A.P $=(3+3)=6$
The first three terms of the A.P are $(4,5,6)$.
View full question & answer→MCQ 131 Mark
$57,54,51,48, \ldots$ are in Arithmetic Progression. The value of the $8^{\text {th }}$ term is:
Answer(a) 36
Explanation :
$\begin{array}{l}57,54,51,48, \ldots A \cdot P \\ a=57, d=54-57=-3, n=8 \\ T_n=a+(n-1) d \\ T_8=57+(8-1) \times(-3) \\ =57+7 \times(-3)=57-21=36 \\ T_8=36\end{array}$
View full question & answer→MCQ 141 Mark
The sum of first 16 terms of the A.P., $9,6,3, \ldots$ is :
Answer(d) -216
Explanation:
Here, $a=9, d=6-9=-3$ and $n=16$
$\begin{array}{l}\therefore S_{16}=\frac{n}{2}[2 a+(n-1) d] \\ =\frac{16}{2}[2 \times 9+(16-1) \times(-3)] \\ =8 \times(-27) \\ =-216\end{array}$
View full question & answer→MCQ 151 Mark
The $15^{\text {th }}$ term of the A.P., $-1 \cdot 2,0 \cdot 8,2 \cdot 8$, is :
Answer(a) 26.8
Explanation:
Here, $a=-1.2$ and $d=0.8-(-1.2)=2.0$
$\begin{array}{l}\therefore 15^{\text {th }} \text { term }=t 15=a+14 d \\ =-1.2+14 \times 2.0 \\ =26.8 .\end{array}$
View full question & answer→MCQ 161 Mark
The $n^{\text {th }}$ term of the A.P. $a, 3 a, 5 a, \ldots$ is :
- ✓
$a(2 n-1)$
- B
$a(2 n-2)$
- C
$a(n-1)$
- D
$a(2 n-3)$
AnswerCorrect option: A. $a(2 n-1)$
(a) $a(2 n-1)$
Explanation:
Here,
first term $=a$, common difference $=3 a-a=2 a$
$\begin{array}{l}\therefore n^{\text {th }} \text { term }=t_n=a+(n-1) d \\ =a+(n-1) \times 2 a \\ =2 a n-a=a(2 n-1) .\end{array}$
View full question & answer→MCQ 171 Mark
The value of $x$ in the equation $(x+1)+(x+4)+(x+7)+\ldots+(x+28)=165$ is:
Answer(b) 2
Explanation:
Given equation is
$(x+1)+(x+4)+(x+7)+\ldots(x+28)=165$
We observe that the terms $(x+1),(x+4),(x+7) \ldots$ are in A.P.
$\therefore a=x+1, d=(x+4)-(x+1)=3 \text { and } l=x+28$
We know, $l=a+(n-1) d$
$\begin{array}{l}\Rightarrow x+28=(x+1)+(n-1) \times 3 \\ \Rightarrow n=10\end{array}$
So, there are 10 terms in this series.
Now, we know
$\begin{array}{l}S_n=\frac{n}{2}(a+1) \\ \Rightarrow 165=\frac{10}{2}[(x+1)+(x+28)] \\ \Rightarrow 33=2 x+29 \\ \Rightarrow x=2 .\end{array}$
View full question & answer→MCQ 181 Mark
If in an A.P. first term is 1 , last term is 20 and the sum of all terms is 399 , then the number of terms in the A.P. is:
Answer(b) 38
Explanation:
Here, $a=1, l=20$ and $S _n=399$
We know, $S _n=\frac{n}{2}(a+i)$
$\begin{array}{l}\Rightarrow 399=\frac{n}{2}(1+20) \\ \Rightarrow n=\frac{399 \times 2}{21}=38 .\end{array}$
View full question & answer→MCQ 191 Mark
The $n$th term of an A.P. is given by $(5 n-3)$, then the sum of its first five terms is:
Answer(c) 60
Explanation:
We have, $t_n=5 n-3$
$\therefore$ For $n=1, t_1=5(1)-3=2$
For $n=5, t_5=5(5)-3=22$
Now, $S _5=\frac{n}{2}\left(t_1+t_5\right)$
$\begin{array}{l}=\frac{5}{2}(2+22) \\ =60\end{array}$
View full question & answer→MCQ 201 Mark
If $4^{\text {th }}$ and $6^{\text {th }}$ terms of an A.P. are 8 and 14 respectively, then the sum of the first 20 terms is:
Answer(c) 550
Explanation:
We have, $t_4=8$ and $t_6=14$
$\Rightarrow a+3 d=8 \text { and } a+5 d=14$
Solving the two equations for $a$ and $d$, we get
d=3, a=-1
Now, $S _{20}=\frac{\pi}{2}[2 a+(n-1) a]$
$\begin{array}{l}=\frac{20}{2}[2 \times(-1)+(20-1) \times 3] \\ =10 \times 55=550\end{array}$
View full question & answer→MCQ 211 Mark
The sum of first ten terms of the A.P. $5,8,11, \ldots$ is:
Answer(b)185
Explanation:
Here, $a=5, d=8-5=3$ and $n=10$
We know, $S _n=\frac{n}{2}[2 a+(n-1) d]$
$S_{10}=\frac{10}{2}[2 \times 5+(10-1) \times 3]$
$\begin{array}{l}=5 \times 37 \\ =185 .\end{array}$
View full question & answer→MCQ 221 Mark
The value of $n$, for which the $n^{\text {th }}$ term of A.P. $63,65,67, \ldots$ is equal to the $n^{\text {th }}$ term of A.P. $3,10,17, \ldots$ are equal to each other is:
Answer(a) 13
Explanation:
For the A.P. $63,65,67, \ldots$
a=63 and d=65-63=2
$\begin{array}{l}\therefore n^{\text {th }} \text { term }=t_n=a+(n-1) d \\ =63+(n-1) \times 2 \\ =2 n+61\end{array}$
For the A.P. $3,10,17, \ldots$
a=3 and d=10-3=7
$\begin{array}{l}\therefore n^{\text {th }} \text { term }=t_n \\ =a+(n-1) d \\ =3+(n-1) \times 7 \\ =7 n-4\end{array}$
Now, according to question,
$\begin{array}{l}2 n+61=7 n-4 \\ \Rightarrow n=13 .\end{array}$
View full question & answer→MCQ 231 Mark
The $72^{\text {nd }}$ term of the A.P. $7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \cdots$ is:
- A
- B
$124 \frac{1}{4}$
- ✓
- D
$133 \frac{3}{4}$
Answer(c) 132
Explanation:
Here, $a=7 \frac{3}{4}-\frac{31}{4}$
and $d=9 \frac{1}{2}-7 \frac{3}{4}=\frac{19}{2}-\frac{31}{4}=\frac{7}{4}$
Now, $t_{72}=a+(72-1) d$
$\begin{array}{l}=\frac{31}{4}+71 \times \frac{7}{4} \\ =\frac{528}{4} \\ =132 .\end{array}$
View full question & answer→MCQ 241 Mark
The number of terms in the A.P. $18,15 \frac{1}{2}, 13, \ldots,-47$ is:
Answer(a) 27
Explanation:
Here, $a=18, d=15 \frac{1}{2}-18=\frac{31}{2}-18=-\frac{5}{2}$ and $l=-47$
We know, $l=a+(n-1) d$
$\begin{array}{l}\Rightarrow-47=18+(n-1) \times\left(\frac{-5}{2}\right) \\ \Rightarrow n-1=-65 \times\left(\frac{-2}{5}\right)=26 \\ \Rightarrow n=27 .\end{array}$
View full question & answer→MCQ 251 Mark
The $n^{\text {th }}$ term of the A.P. $\frac{1}{m}, \frac{1+m}{m}, \frac{1+2 m}{m}, \ldots$ is:
- A
$\frac{1+n}{m}$
- B
$\frac{1+m(n+1)}{m}$
- C
$\frac{1+m n}{m}$
- ✓
$\frac{1+m(n-1)}{m}$
AnswerCorrect option: D. $\frac{1+m(n-1)}{m}$
(d) $\frac{1+m(n-1)}{m}$
Explanation:
Here, $a=\frac{1}{m}, d=\frac{1+m}{m}-\frac{1}{m}=\frac{1+m-1}{m}=1$
$\therefore n^{\text {th }}$ term $=t_n=a+(n-1) d$
$\begin{array}{l}=\frac{1}{m}+(n-1) \times 1 \\ =\frac{1+m(n-1)}{m}\end{array}$
View full question & answer→MCQ 261 Mark
Tenth term from the end of the A.P. $18,16,14, \ldots,-10$ is:
Answer(b) 8
Explanation:
The reverse A.P. of the given A.P. is
$-10,-8, \ldots, 14,16,18$
Now, for this A.P.,
$a=-10, d=-8-(-10)=2$
$\therefore$ Required term
$\begin{array}{l}=10^{\text {th }} \text { term from the end of the } 18,16,14, \ldots \\ =10^{\text {th }} \text { term from the beginning }-10,-8, \ldots \\ =t_{10}=a+9 d \\ =-10+9 \times 2 \\ =8\end{array}$
View full question & answer→MCQ 271 Mark
Which term of the A.P. $21,18,15, \ldots$ is -81 ?
- A
$32^{\text {nd }}$ term
- B
$33^{ rd }$ term
- C
$34^{\text {th }}$ term
- ✓
$35^{\text {th }}$ term
AnswerCorrect option: D. $35^{\text {th }}$ term
(d) $35^{\text {th }}$ term
Explanation:
a=21, d=18-21=-3
Let -81 be the $n^{\text {th }}$ term of the A.P.
Then, $t_n=a+(n-1) d$
$\begin{array}{l}\Rightarrow-81=21+(n-1) \times(-3) \\ \Rightarrow n-1=\frac{-102}{-3} \\ \Rightarrow n=35\end{array}$
Hence, $35^{\text {th }}$ term of the A.P. is -81 .
View full question & answer→MCQ 281 Mark
If 8 times the eighth term of an A.P. is 15 times the fifteenth term, then the $23^{\text {rd }}$ term of the A.P. is:
Answer(a) 0
Explanation:
$8 t_8=15 t_{15}$
$\begin{array}{l}\Rightarrow 8(a+7 d)=15(a+14 d) \\ \Rightarrow 8 a-15 a=210 d-56 d \\ \Rightarrow-7 a=154 d \\ \Rightarrow a=-22 d \ldots( i )\end{array}$
Now, $23^{\text {rd }}$ term $=t_{23}$
$\begin{array}{l}=a+22 d \\ =-22 d+22 d=0[\text { Using }(i)]\end{array}$
View full question & answer→MCQ 291 Mark
If in an A.P., last term is zero with common difference -2 and total number of terms 5 , then the first term of the A.P. is:
Answer(c) 8
Explanation:
Here, $l=0, d=-2$ and $n=5$
We know, $l=a+(n-1) d$
$\begin{array}{l}\Rightarrow 0=a+(5-1) \times(-2) \\ \Rightarrow a=8\end{array}$
View full question & answer→MCQ 301 Mark
If $3 x-1,3 x+5$ and $5 x+1$ are the three consecutive terms of an A.P., then the value of $x$ is:
Answer(d) 5
Explanation:
Since, the given terms are consecutive terms of A.P., so the difference between two consecutive terms must be same.
$\therefore(3 x+5)-(3 x-1)=(5 x+1)-(3 x+5)$
$\begin{array}{l}\Rightarrow 6=2 x-4 \\ \Rightarrow x=5 .\end{array}$
View full question & answer→MCQ 311 Mark
The number of terms of an A.P. $5,9,13, \ldots, 185$ is:
Answer(b) 46
Explanation:
The A.P. is $5,9,13, \ldots, 185$.
$\therefore a=5, d=9-5=4$ and $l=185$
We know, $l=a+(n-1) d$
$\begin{array}{l}\Rightarrow 185=5+(n-1) \times 4 \\ \Rightarrow n-1=\frac{180}{4}=45 \\ \Rightarrow n=46\end{array}$
View full question & answer→MCQ 321 Mark
If there are 15 terms in an A.P. whose first term is $\sqrt{2}$ and common difference is $2 \sqrt{2}$, then the last term is:
- A
$31 \sqrt{2}$
- B
$30 \sqrt{2}$
- ✓
$29 \sqrt{2}$
- D
$28 \sqrt{2}$
AnswerCorrect option: C. $29 \sqrt{2}$
(c) $29 \sqrt{2}$
Explanation:
Here, $n=15, a=\sqrt{2}, d=2 \sqrt{2}$
$\therefore$ Last term $=t_{15}=a+14 d$
$\begin{array}{l}=\sqrt{2}+14 \times 2 \sqrt{2} \\ =29 \sqrt{2} .\end{array}$
View full question & answer→MCQ 331 Mark
If the common difference of an A.P. is 5 , then the difference in its $30^{\text {th }}$ and $15^{\text {th }}$ terms is:
Answer(d) 75
Explanation:
$t_{30}-t_{15}=(a+29 d)-(a+14 d)$
$=15 d=15 \times 5=75$
$[\because-d=5($ given $)]$
View full question & answer→MCQ 341 Mark
If the fifth and sixth terms of an A.P. are 6 and 5 , respectively, then its first term is:
Answer(c) 10
Explanation:
According to question,
$\begin{array}{l}t_5=6 ; t_6=5 \\ \Rightarrow a+4 d=6 ; a+5 d=5\end{array}$
Solving the two equations for $a$ and $d$, we get
d=-1, a=10
View full question & answer→MCQ 351 Mark
If the second term of an A.P. is 5 and the seventh term is 20 , then the A.P. is:
- A
$3,5,7, \ldots$
- B
$2,4,6, \ldots$
- ✓
$2,5,8, \ldots$
- D
$3,6,9, \ldots$
AnswerCorrect option: C. $2,5,8, \ldots$
(c) 2, 5, 8,
Explanation:
According to question,
$\begin{array}{l}t_2=5 ; t_7=20 \\ \Rightarrow a+d=5 ; a+6 d=20\end{array}$
Solving these two equations for $a$ and $d$, we get
d=3, a=2
$\therefore$ The A.P. is $2,(2+3),(2+(2(3))$, $\ldots$ i.e., $2,5,8, \ldots$
View full question & answer→MCQ 361 Mark
The $8^{\text {th }}$ term of the A.P. $12,8,4,0, \ldots$ is:
Answer(a) -16
Explanation:
In the given A.P.,
$a=12$ and $d=8-12=-4$
$\therefore 8^{\text {th }}$ term $=t_8=a+(8-1) d$
$\begin{array}{l}=12+7 \times(-4) \\ =-16 .\end{array}$
View full question & answer→