MCQ 11 Mark
The first three terms of an $AP$ respectively are $3 y-1,3 y+5$ and $5 y+1$. Then $y$ equals:
AnswerIf $a, b$ and $c$ are in $AP,$
$b-a=c-b$
$2 b=a+c$
$2(3 y+5)=3 y-1+5 y+1$
$6 y+10=8 y$
$10=8 y-6 y$
$2 y=10$
$y=5$
Hence the correct option is $(c).$
View full question & answer→MCQ 21 Mark
If k, $2 k-1$ and $2 k+1$ are three consecutive terms of an $A.P.,$ the value of $k$ is.
AnswerLet $a, b$ and $c$ are in $A.P.$ then $b-a=c-b$
Here the three consecutive terms of an $A.P.$ are $k, 2 k -1$ and $2 k+1$
$\Rightarrow 2 k-1-k=2 k+1-(2 k-1)$
$\Rightarrow 2 k-1-k=2 k+1-2 k+1$
$\Rightarrow k-1=2$
$\Rightarrow k=2+1=3$
The correct answer is $(b).$
View full question & answer→MCQ 31 Mark
The common difference of the $A.P. \frac{1}{p}, \frac{(1-p)}{p}, \frac{(1-2 p)}{p} \ldots \ldots \ldots$ is:
AnswerThe given $A.P.$ is
$\frac{1}{p}, \frac{(1-p)}{p}, \frac{(1-2 p)}{p} \ldots \ldots \ldots$
And,
First term $\left(a_1\right)=\frac{1}{p}$
Second term $\left(a_2\right)=\frac{(1-p)}{p}$
Common difference $(d)=a_2-a_1$
$=\frac{(1-p)}{p}-\frac{1}{p}$
$=\frac{1-p-1}{p}$
$\Rightarrow d=-1$
Thus the common difference of the given $A.P.$ is $"-1".$
Hence the correct option is $(c).$
View full question & answer→MCQ 41 Mark
If the $n^{\text {th }}$ term of an $A.P.$ is $(2 n+1)$, then the sum of its first three terms is
AnswerWe have,
$a_n=(2 n+1)$
$\Rightarrow a_1=2 \times 1+1=3$
So, the given sequence is an $A.P.$ with first term $a=a_1=3$.
And the second term, $a_2=2 \times 2+1=5$.
So, the common difference,
$d=a_2-a_1=5-3=2$
Therefore, the sum of first $3$ terms of the $A.P.$ is given by
$S_n=\frac{n}{2}\{2 a+(n-1) d\}$
$=\frac{3}{2}\{6+(3-1) 2\}$
$=\frac{3}{2}(6+4)$
$=\frac{3}{2}(10)$
$=15$
Hence the correct option is $(b).$
View full question & answer→MCQ 51 Mark
In an $AP,$ if $d=-2, n=5$ and $a_n=0$ then the value of $a$ is
AnswerGiven that,
$d=-2, n=5 \text { and } a_n=0$
We know that
$a_n=a+(n-1) d$
Substituting the given values in above equation.
$\Rightarrow 0=a+(5-1)(-2)$
$\Rightarrow 0=a+(4)(-2)$
$\Rightarrow 0=a-8$
$\Rightarrow a=8$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 61 Mark
Three alarm clocks ring their alarms at regular intervals of $20\ min, 25\ min$ and $30\ min$ respectively. If they first beep together at $12$ noon, at what time will they beep again for the first time?
- A
$4: 00\ pm$
- B
$4: 30\ pm$
- ✓
$5: 00\ pm$
- D
$5: 30\ pm$
AnswerCorrect option: C. $5: 00\ pm$
Time when they ring together
$=\operatorname{LCM}(20,25,30)$
According to prime factorisation,
$20=2 \times 2 \times 5$
$25=5 \times 5$
$30=2 \times 3 \times 5$
$\operatorname{LCM}(20,25,30)=2 \times 2 \times 3 \times 5 \times 5=300$
Thus, $3$ bells ring together after $300$ minutes or $5$ hours.
Since, they rang together first at $12$ noon, then they ring together agian at $5\ pm.$
View full question & answer→MCQ 71 Mark
If the sum of the first n terms of an A.P be $3 n ^2+ n$ and its common difference is 6 , then its first term is
Answer(d)
In the given problem, the sum of $n$ terms of an A.P. is given by the expression,
$
S_n=3 n^2+n
$
Here, we can find the first term by substituting $n$ $=1$ as sum of first term of the A.P. will be the same as the first term. So we get,
$
\begin{aligned}
S_{n} & =3 n^2+n \\
S_1 & =3(1)^2+(1) \\
& =3+1 \\
& =4
\end{aligned}
$
Therefore, the first term of this A.P is $a=4$.
View full question & answer→MCQ 81 Mark
The $n^{\text {th }}$ term of the A.P. a, $3 a , 5 a , \ldots \ldots$ is
- A
- ✓
$(2 n-1) a$
- C
$(2 n +1) a$
- D
AnswerCorrect option: B. $(2 n-1) a$
b
$\begin{aligned} A P & : a, 3 a, 5 a \ldots \ldots \ldots \\ a & =a \\ d & =3 a-a=2 a \\ T_n & =a+(n-1) d \\ & =a+(n-1) 2 a \\ & =a+2 a n-2 a \\ & =2 a n-a=(2 n-1) a\end{aligned}$
View full question & answer→MCQ 91 Mark
The common difference of the $A.P. \frac{1}{ p }, \frac{1- p }{ p }, \frac{1-2 p }{ p }$, is
- A
- B
$\frac{1}{ p }$
- ✓
- D
$-\frac{1}{ p }$
Answer$\text { } A P=\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p} \ldots \ldots . .$
$a_2-a_1=a_3-a_2$
$\Rightarrow \frac{1-p}{p}-\frac{1}{p}=\frac{1-2 p}{p}+\frac{1+p}{p}$
$\Rightarrow \frac{1-p-1}{p}=\frac{1-2 p-1+p}{p}$
$\Rightarrow \frac{-b c}{b c}=\frac{-b c}{b c}$
$\Rightarrow-1$
View full question & answer→MCQ 101 Mark
How many terms of the $AP 3,7,11,15, \ldots-$ will make the sum $406$ ?
Answer Let $S _{ n }=406$
Then, $\frac{ n }{2}[2 \times 3+( n -1) \times 4]=406$
$\Rightarrow \frac{n}{2}(6+4 n-4)=406$
$\Rightarrow \frac{n}{2}(4 n+2)=406$
$\Rightarrow n(2 n+1)=406 \Rightarrow 2 n^2+n-406=0$
$\therefore n=\frac{-1 \pm \sqrt{1+3248}}{4}=\frac{-1+\sqrt{3249}}{4}$
$\text { (neglecting negative value) }$
$\therefore n=\frac{-1+57}{4}=\frac{56}{4}=14$
View full question & answer→MCQ 111 Mark
The sum of first $16$ terms of the $AP \ 10,6,2, \cdots$ is
- A
$320$
- ✓
$-320$
- C
$-352$
- D
$-440$
AnswerCorrect option: B. $-320$
$ l = T _{16}=10+15 \times(-4)=10-60=-50$
$\therefore \text { Sum }=\frac{ n }{2}( a + l )$
$=\frac{16}{2}(10-50)$
$=8 \times(-40)$
$=-320$
View full question & answer→MCQ 121 Mark
$(5+13+21+\ldots .+181)=$ ?
- A
$2476$
- B
$2337$
- C
$2219$
- ✓
$2139$
AnswerCorrect option: D. $2139$
Let the number of terms be $n$.
$\text {Then, } T_{n}=181$
$\therefore 5+(n-1) \times 8=181$
$\Rightarrow(n-1) \times 8=176 $
$\Rightarrow n-1=22 $
$\Rightarrow n=23$
$\therefore \text { Sum }=\frac{n}{2}(a+l)$
$=\frac{23}{2}(5+181)$
$=23 \times 93$
$=2139$
View full question & answer→MCQ 131 Mark
The sum of first $40$ positive integers divisible by $6$ is.
- A
$2460$
- B
$3640$
- ✓
$4920$
- D
$4860$
AnswerCorrect option: C. $4920$
$\text { Required sum }=6+12+18+\cdots+240$
$=\frac{40}{2}(6+240)$
$=20 \times 246$
$=4920 .$
View full question & answer→MCQ 141 Mark
The sum of first $20$ odd natural number is
Answer$ S _{20}=1+3+5+7+\cdots$
$\text { Here } a =1, d=2, S_0, T_{20}=( a +19 d)$
$=1+19 \times 2=38=\text { last term (l) }$
$\therefore S _{20}=\frac{ n }{2}( a + l )=\frac{20}{2}(1+39)$
$=20 \times 20$
$=400$
View full question & answer→MCQ 151 Mark
The $13^{\text {th }}$ term of an $AP$ is $4$ times its $3^{\text {rd }}$ term. If its $5^{\text {th }}$ term is $16$ then the sum of its first $10$ terms is.
Answer$T_{13}=4 T_3 $
$\Rightarrow a+12 d=4(a+2 d)$
$\Rightarrow a+12 d=4 a+8 d$
$\Rightarrow 3 a-4 d=0 \ldots \ldots(1)$
and $ T_5=16$
$a+4 d=16 \ldots \ldots(2)$
On solving equation $(1)$ and $(2),$
$3 a-4 d=0$
$\frac{a+4 d=16}{4 a=16 \Rightarrow} a=\frac{16}{4}=4$
Putting the value of a in equation $(ii),$
We get, $4+4 d=16$
$\Rightarrow 4 d=16-4 $
$\Rightarrow d=\frac{12}{4}=3$
$\therefore S_{10}=\frac{10}{2}(2 a+9 d)$
$=5(2 \times 4+9 \times 3)$
$=5(8+27)=35 \times 5=175 .$
View full question & answer→MCQ 161 Mark
The $5^{\text {th }}$ term of an $A P$ is $20$ and the sum of its $7^{\text {th }}$ and $11^{\text {th }}$ terms is $64 .$ The common difference of the $A.P.$ is.
AnswerGiven, $T_5=20$
$\therefore a+4 d=20$
and $ T_7+T_{11}=64$
$a+6 d+a+10 d=64$
$2 a+16 d=64$
$\Rightarrow a+8 d=32$
On solving equation $(i)$ and $(ii)$
View full question & answer→MCQ 171 Mark
The $5^{\text {th}}$ term of an $AP$ is $-3$ and its common difference is $-4$. The sum of its first $10$ terms is.
Answer$\because a +4 d=-3$
$\because d=-4$ given
$\therefore a +4 \times-4=-3$
$\Rightarrow a =-3+16=13 .$
$\therefore S _{10}=\frac{10}{2}[2 a +9 d]$
$=5[2 \times 13+9 \times(-4)]$
$=5[26-36]$
$=5 \times(-10)$
$=-50$
View full question & answer→MCQ 181 Mark
The sum of first $n$ terms of an $AP$ is $\left(4 n^2+2 n\right)$. The $n ^{\text {th }}$ term of this $AP$ is
- A
$(6 n-2)$
- B
$(7 n-3)$
- ✓
$(8 n -2)$
- D
$(8 n +2)$
AnswerCorrect option: C. $(8 n -2)$
$T_n=\left(S_n-S_n-1\right)$
$=\left(4 n^2+2 n\right)-\left\{4(n-1)^2\right.+2(n-1)\}$
$=\left(4 n^2+2 n\right)-\left(4 n^2-6 n+2\right)$
$=(8 n-2)$
View full question & answer→MCQ 191 Mark
The sum of first $n$ terms of an $AP$ is $\left(5 n-n^2\right)$. The $n ^{\text {th }}$ term of the $AP$ is.
- A
$(5-2 n )$
- ✓
$(6-2 n )$
- C
$(2 n-5)$
- D
$(2 n -6)$
AnswerCorrect option: B. $(6-2 n )$
$T_n=\left(S_n-S_{n-1}\right)$
$=\left(5 n-n^2\right)-\left\{5(n-1)-(n-1)^2\right\}$
$=\left(5 n-n^2\right)-\left(7 n-n^2-6\right)$
$=6-2 n$
View full question & answer→MCQ 201 Mark
The sum of first $n$ terms of an $AP$ is $\left(3 n^2+6 n\right)$. The common difference of the $A.P$ is
Answer$\text { Given, } S_{n}=3 n^2+6 n$
$\therefore T_1=S_1=\left(3 \times 1^2+6 \times 1\right)=9,$
$S_2=\left(3 \times 2{ }^2+6 \times 2\right)=24 .$
$\therefore T_2=\left(S_2-S_1\right)=24-9=15$
Hence, common difference $(d)=T_2-T_1$
$=15-9=6 .$
View full question & answer→MCQ 211 Mark
If the $n ^{\text {th }}$ term of an $A P$ is $(2 n+1)$ then the sum of its first three term is
Answer$T_1+T_2+T_3$
$=(2 \times 1+1)+(2 \times 2+1)+(2 \times 3+1) $
$=3+5+7$
$=15$
View full question & answer→MCQ 221 Mark
What is the common difference of an $AP$ in which $a _{18}- a _{14}=32$ ?
Answer$a_{18}-a_{14}=32$
$a+17 d-a-13 d=32$
$\Rightarrow 4 d=32 $
$\Rightarrow d=\frac{32}{4}=8$
Hence, common difference is $8 .$
View full question & answer→MCQ 231 Mark
The $2^{nd}$ term of an $AP$ is $13$ and its $5^{th}$ term is $25 $. What is its $17^{th}$ term?
AnswerAccording to the question,
$a+d=13$
$a+4 d=25$
On solving equation $(i), (ii)$
$a+d=13$
$-a \pm 4 d=-25$
$\Rightarrow d=4.$
Putting the value of $d$ in equation $(i)$
$a+4=13$
$\Rightarrow a=13-4=9$
$\therefore T_{17}=(a+16 d)$
$=9+16 \times 4$
$=9+64$
$=73$
View full question & answer→MCQ 241 Mark
What is the $20^{th}$ term from the end of the $\text{AP}\ 3,8$, $13, \cdots-253$ ?
Answer$20^{\text {th }}$ term from the end $=[1-( n -1) d ]$
$=(253-19 \times 5)$
$=253-95=158 .$
View full question & answer→MCQ 251 Mark
What term of the $AP\ 21,42,63,84, \ldots$ is $210$.
- A
$9^{th}$
- ✓
$10^{th}$
- C
$11^{th}$
- D
$12^{th}$
AnswerCorrect option: B. $10^{th}$
Let $ T_{n}=210$
Then, $ 21+(n-1) \times 21=210$
$\therefore \quad(n-1) \times 21=210-21$
$\Rightarrow(n-1)=\frac{189}{21}=9$
$\Rightarrow n=9+1=10$
So$, 10^{\text {th }}$ term of the given $AP$ is $210$ .
View full question & answer→MCQ 261 Mark
Which term of the $ AP\ 25,20,15,-$ is the first negative term?
- A
$10^{th}$
- B
$9^{th}$
- C
$8^{th}$
- ✓
$7^{th}$
AnswerCorrect option: D. $7^{th}$
$\text { Let } T_{n}<0 .$
Then $25+(n-1) \times(-5)<0$
$\therefore 30<5 n$
$\Rightarrow 5 n>30$
$\Rightarrow n>6$
So$,$ the required term is $7^{\text {th }}$.
View full question & answer→MCQ 271 Mark
The $8^{th}$ term of an $A P$ is $17$ and its $14^{th}$ term is $29.$ The common difference of the $A P$ is
AnswerAccording to the question$,$
$T_{14}-T_8=29-17$
$\Rightarrow(a+13 d)-(a+7 d)=12$
$\Rightarrow a+13 d-a-7 d=12$
$\Rightarrow 6 d=12 \Rightarrow d=\frac{12}{6}=2$
Hence$,$ the common difference $=2$
View full question & answer→MCQ 281 Mark
The $7^{th}$ term of an $A P$ is $4$ and its common difference is $-4$. What is its first term?
Answer$\because 7^{\text {th }} \text { term }=4$
$\therefore a +6 d=4$
$\Rightarrow a +6 \times(-4)=4$
$\Rightarrow a -24=4$
$\Rightarrow a =4+24=28$
View full question & answer→MCQ 291 Mark
How many two$-$digit numbers are divisible by $3$ ?
AnswerTwo$-$digit numbers divisible by $3$ are $12,15,18, ...., 99 .$
Let $T _{ n }=99$, Then $12+( n -1) \times 3=99$
$\Rightarrow(n-1) 3=99-12$
$\Rightarrow(n-1)=\frac{87}{3}=29$
$\Rightarrow n=29+1$
$=30$
View full question & answer→MCQ 301 Mark
An $AP 5, 12, 19, ...$ has $50$ terms. Its last term is
Answer$\text { Given, first term }(a)=5 \text {, }$
$\text { Common difference } d =12-5=7$
$\therefore 50^{\text {th }} \text { term }= a +( n -1) d$
$=5+(50-1) \times 7$
$=5+49 \times 7$
$=348$
View full question & answer→MCQ 311 Mark
The $7^{\text {th }}$ term of an $A P$ is $-1$ and its $16^4$ term is $17$. The $n^{\text {th }}$ term of the $A P$ is.
- A
$(3 n+8)$
- B
$(4 n-7)$
- C
$(15-2 n)$
- ✓
$(2 n-15)$
AnswerCorrect option: D. $(2 n-15)$
Given, $T _7=-1$
$ \Rightarrow a +6 d=-1$ and $T _{16}=17$
$ \Rightarrow a +15 d=17$
On solving equation $(i)$ and $(ii)$.
$a+6 d=-1$
$-a \pm 15 d=-17$
$-9 d=-18d=\frac{18}{9}=2$
Putting the value of $d$ in equation $(i),$
We get $a +6 \times 2=-1$
$\Rightarrow a=-1-12=-13$
$\therefore T_{n}=a+(n-1) d$
$=-13+(n-1) \times 2$
$=(2 n-15) .$
View full question & answer→MCQ 321 Mark
If $4, x _1, x _2, x _3, 28$ are in $AP$ , then $x _2=$ ?
AnswerGiven $AP$ is $4, x _1, x _2, x _3, 28$.
Clearly, first term $a =4$
and fifth term $T_5=28$
Now, $T _5= a +4 d$
$\Rightarrow 28=4+4 d$
$\Rightarrow 4 d=24$
$\Rightarrow d=\frac{24}{4}=6$
$\therefore x_3=T_4$
$=a+3 d$
$=4+3 \times 6$
$=22$
View full question & answer→MCQ 331 Mark
The next term of the A.P $\sqrt{7}, \sqrt{28}, \sqrt{63}, \cdots$ is.
- A
$\sqrt{70}$
- B
$\sqrt{84}$
- C
$\sqrt{98}$
- ✓
$\sqrt{112}$
AnswerCorrect option: D. $\sqrt{112}$
(d)
Given terms are $\sqrt{7}, \sqrt{4 \times 7}, \sqrt{9 \times 7}, \cdots$
View full question & answer→MCQ 341 Mark
The common difference of the A.P $\frac{1}{ P }, \frac{1- P }{ P }, \frac{1-2 P }{ P },- is$
Answer(c)
$d=\left\{\frac{1-P}{P}-\frac{1}{P}\right\}=\frac{1-P-1}{P}=\frac{-P}{P}=-1$
View full question & answer→