MCQ 1511 Mark
The $31^{st}$ term of an $A.P$. whose first two terms are $-3$ and $4$ is :
AnswerGiven : $a = -3, d = 4 - (-3) = 7$
$\therefore a_{31}= -3 + (31 - 1) \times 7$
$= -3 + 30 \times 7$
$= -3 + 210 = 207$
View full question & answer→MCQ 1521 Mark
The common difference of the $A.P$ whose an $= -3n + 7$ is :
AnswerGiven : $an = -3n + 7$
Putting $n = 1, 2, 3$ we get
$a = -3 \times 1 + 7 = -3 + 7 = 4$
$a_2= -3 \times 2 + 7 = -6 + 7 = 1$
$a_3= -3 \times 3 + 7 = -9 + 7 = -2$
$\therefore$ Common difference $(d) $
$= a_2- a = 1 - 4 = -3$
View full question & answer→MCQ 1531 Mark
If $4, x_1, x_2, x_3, 28$ are in $AP$ then $x_3= ?$
AnswerGiven that $4,\text{x}_1,\text{x}_2,\text{x}_3,28$ are in $AP$.
Let $d$ be the common difference.
Since $28$ is the $5^{th}$ term,
$28 = 4 + 4d$
$\Rightarrow 4d = 24$
$\Rightarrow d = 6$
$x_3= a + (3)d .....$
$(x_3$ is the fourth term$)$
$\Rightarrow x_3= 4 + 3(6)$
$\Rightarrow x_3= 22$
View full question & answer→MCQ 1541 Mark
$200$ logs are stacked in a such a way that $20$ logs in the bottom row, $19$ logs in the next row, $18$ logs in the row next to it and So on. The number of logs in the top row is :
AnswerThe number of logs in the row from bottom to the top is $20, 19, 18, …$
which form an $A.P.$ with first term $20$ and common difference $19 - 20 = -1$
Let the $200$ logs be arranged in rows. $nn$
Then, $S_n= 200, a = 20, d = 19 - 20 = -1$
$\Rightarrow200=\frac{\text{n}}{2}[\text{2a} + ({\text{n - 1}})\text{d}]$
$\Rightarrow 400 = n[2 \times 20 + (n - 1)(-1)]$
$ \Rightarrow n_2-41 n+400=0 $
$ \Rightarrow n_2-16 n-25 n+400=0 $
$\Rightarrow n(n - 16) - 25(n - 16) = 0$
$\Rightarrow (n - 25) (n - 16) = 0$
$\Rightarrow n - 25 = 0$ and $n - 16 =0$
$\Rightarrow n = 25$ and $n = 16$
$n = 25$ is not possible as on calculating the number of logs in $25^{th}$ row, there is a negative number of logs, which is not possible.
$\therefore$ the number of logs is $16$.
Now, the number of logs in $16^{th}$ row $, a16 = 20 + (16 - 1)(-1)$
$= 20 - 15 = 5$
$\therefore$ The number of logs in the top row is $5.$
View full question & answer→MCQ 1551 Mark
Choose the correct answer from the given four options : The $11^{th}$ term of the $AP: -5, \frac{-5}{2}, 0, \frac{5}{2},.........$ is :
AnswerGiven ${AP},-5,\frac{-5}{2},0,\frac{5}{2}$
Here $, a = -5$
$\text{d}=\frac{-5}{2}+5=\frac{5}{2}$
$a11 = a + (11 - 1)d$
$[\therefore a_n= a + (n - 1)d]$
$=-5+\big(10\big)\times\frac{5}{2}$
$= -5 + 25 = 20$
View full question & answer→MCQ 1561 Mark
The $9^{th}$ term of an $A.P$. is $499$ and the $499^{th}$ term is $9$. The term which is equal to zero is :
- A
$504^{th}$ term
- B
$510^{th}$ term
- C
$500^{th}$ term
- ✓
$508^{th}$ term
AnswerCorrect option: D. $508^{th}$ term
$a_9= a + 8d$
$a_9= 499\ ($given$)$
$\therefore a + 8d = 499 .... (i)$
$a_{499}= a + 498d$
$a_{499}= 9$
$\therefore a + 498d = 9 ...... (ii)$
Subtract $(i)$ from $(ii)$
$\Rightarrow 490d = -490$
$\Rightarrow d = -1$
Substitute the value of $d$
$\Rightarrow a + 8(-1) = 499$
$\Rightarrow a = 507$
$\therefore$ For $a_n= 0$
$\Rightarrow a + (n - 1)d = 0$
$\Rightarrow 507 + (n - 1) (-1) = 0$
$\Rightarrow n = 508$
So $,508^{th}$ term is equal to zero.
View full question & answer→MCQ 1571 Mark
The $7^{th}$ term of an $AP$ is $-1$ its $16^{th}$ term is $17$. The nth term of the $AP$ is :
- A
$(3n + 8)$
- B
$(4n - 7)$
- C
$(15 - 2n)$
- ✓
$(2n - 15)$
AnswerCorrect option: D. $(2n - 15)$
Let a be the frist term and $d$ be the common difference.
$a_7= -1$
$\Rightarrow a + 6d = -1 .....(i)$
$a_{16}= 17$
$\Rightarrow a + 15d = 17 .....(ii)$
Subtracting $(i)$ from $(ii),$ we get
$9d = 18$
$\Rightarrow d = 2$
Substituting in $(i),$ we get $a = - 13$
So,$ a_n=a+(n-1) d $
$ \Rightarrow a_n=-13+(n-1) 2 $
$ \Rightarrow a_n=-13+2 n-2 $
$ \Rightarrow a_n=2 n-15 $
View full question & answer→MCQ 1581 Mark
Choose the correct answer from the given four options : The $21^{st}$ term of the $AP$ whose first two terms are $–3$ and 4 is :
AnswerMain concept used: $a_n= a + (n - 1)d$
Here, $a=a_1=-3, a_2=4$
$\therefore d=a_2-a_1=4-(-3)=4+3=7$
Hence, $d=7$
Now, $a_n=a+(n-1) d$
$ \Rightarrow a_{21}=-3+(21-1) \times 7 $
$ \Rightarrow a_{21}=-3+20 \times 7 $
$ \Rightarrow a_{21}=-3+140 $
$ \Rightarrow a_{21}=137 $
View full question & answer→MCQ 1591 Mark
In an $A.P. S_p= q, S_q= p$ and $Sr$ denotes the sum of first $r$ terms. Then, $S_{p+q}$ is equal to :
- ✓
$p + q$
- B
$0$
- C
$pq$
- D
$-(p + q)$
AnswerCorrect option: A. $p + q$
Here, $S_p= q, S_q= p$
$S_{p+q}=$ Sum of $(p + q)$ terms
$=$ Sum of $p$ term $+$ Sum of $q$ terms
$= q + p$
View full question & answer→MCQ 1601 Mark
$200$ logs are stacked in such a way that $20$ logs in the bottom row, $19 $logs in the next row, $18$ logs in the row next to it, and so on. The total number of rows is :
AnswerThe number of logs in the row from bottom to the top are $20, 19, 18, …$
which form an $A.P$. with first term $20$ and common difference $19 - 20 = -1$
Let the $200$ logs be arranged in $n$ rows.
Then $, n= 200$
$\Rightarrow200=\frac{\text{n}}{2}[2\text{a} + (\text{n + 1})\text{d}]$
$\Rightarrow 400 = n[2 \times 20 + (n - 1)(-1)]$
$ \Rightarrow n_2-41 n+400=0 $
$ \Rightarrow n_2-16 n-25 n+400=0 $
$\Rightarrow n(n - 16) - 25(n - 16) = 0$
$\Rightarrow (n - 25) (n - 16) = 0$
$\Rightarrow n - 25 = 0$ or $ n - 16 = 0$
$\Rightarrow n = 25$ or $n = 16$
$n = 25$ is not possible as on calculating the number of logs in $25^{th}$ row,
There is a negative number of logs, which is not possible.
$\therefore$ The number of rows is $16.$
View full question & answer→MCQ 1611 Mark
Mark the correct alternative in the following : Two $A.P.'s$ have the same common difference. The first term of one of these is $8$ and that of the other is $3$. The difference between their $30^{th}$ term is :
AnswerIn two $A.P.'s$ common difference is same
Let $A$ and $a$ are two $A.P.'s$
First term of $A$ is $8$ and first term of $a$ is $3$
$A_{30}- a_{30}= 8 + (30 - 1)d - 3 - (30 – 1)d$
$= 5 + 29d - 29d = 5$
View full question & answer→MCQ 1621 Mark
The sum of first $n$ terms of an $AP$ is $(5n - n^2).$ The $n^{th}$ term of the $A.P$. is :
- A
$(2n - 6)$
- ✓
$(6 - 2n)$
- C
$(2n - 5)$
- D
$(5 - 2n)$
AnswerCorrect option: B. $(6 - 2n)$
$ 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) $
$= (6 - 2n)$
View full question & answer→MCQ 1631 Mark
Mark the correct alternative in the following : If $\mathrm{S}_{\mathrm{r}}$ denotes the sum of the first $r$ terms of an $A.P$. Then, $S_{3 n}:\left(S_{2 n}-S_n\right)$ is :
Answer$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}], \text{S}_{2\text{n}}=\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]$
and $\text{S}_{3\text{n}}=\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]$
Now $\text{S}_{2\text{n}}-\text{S}_\text{n}=\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]-\frac{\text{n}}{2}$
$[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[4\text{a}+(4\text{n}-2)\text{d}]-]2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[4\text{a}-2\text{a}+(4\text{n}-2-\text{n}+1)\text{d}]$
$=\frac{\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}] =\frac{1}{3}(\text{S}_{3\text{n}})$
$\therefore\ \text{S}_{3\text{n}}:(\text{S}_{2\text{n}}-\text{S}_\text{n})=3:1$ or $\frac{3}{1}=3$
View full question & answer→MCQ 1641 Mark
Mark the correct alternative in the following : If the sum of first $n$ even natural numbers is equal to $k$ times the sum of first $n$ odd natural numbers, then $k =$
- A
$\frac{1}{\text{n}}$
- B
$\frac{\text{n}-1}{\text{n}}$
- C
$\frac{\text{n}+1}{2\text{n}}$
- ✓
$\frac{\text{n}+1}{\text{n}}$
AnswerCorrect option: D. $\frac{\text{n}+1}{\text{n}}$
Sun of $n$ even natural number $=n(n+1)$
and sum of $n$ odd natural numbers $=n^2$
$\therefore \mathrm{n}(\mathrm{n}+1)=\mathrm{kn}^2$
$\Rightarrow\ \text{k}=\frac{\text{n}(\text{n}+1)}{\text{n}^2}=\frac{\text{n}+1}{\text{n}}$
View full question & answer→MCQ 1651 Mark
The sum of the first $20$ terms of the $A.P.\{ 10, 6, 2, ...\}$ is :
- A
$-480$
- B
$-500$
- C
$-400$
- ✓
$-560$
AnswerCorrect option: D. $-560$
Given : $a = 10, d = 6 - 10 = -4$ and $n = 20$
$\therefore\text{S}_{\text{n}} = \frac{\text{n}}{2} [2\text{a} +({\text{n}-1})\text{d}]$
$\Rightarrow\text{S}_{20} = \frac{20}{2} [2\times10+(20-1)\times(-4)]$
$\Rightarrow S_{20}=10[20-76]$
$=10 \times(-56)=-560$
View full question & answer→MCQ 1661 Mark
The first term of an $A.P.$ if its $S_n=n^2+2 n$ is :
AnswerGiven : $S_n=n^2+2 n$
Putting $n=1,$ we get
$S=a=(1)^2+2 \times 1$
$=1+2=3$
View full question & answer→MCQ 1671 Mark
Choose the correct answer from the given four options : If the $2^{nd}$ term of an $AP$ is $13$ and the $5^{th}$ term is $25,$ what is its $7^{th}$ term?
AnswerGiven, $a_2=13$ and $a_5=25$
${\left[\because a_n=-a+(n-1) d\right]}$
$\Rightarrow a+(2-1) d=25$
and $a + (5 + 1)d = 25$
$\Rightarrow a + d = 13 .........(i)$
and $a + 4d = 25 ........(ii)$
on subtracting Eq. $(i)$ from Eq. $(ii),$ we get
$\Rightarrow 3d = 25 - 13 = 12 $
$\Rightarrow d =4$
From Eq. $(i), a = 13 - 4 = 9$
$\therefore a_7= a + (7 - 1)d$
$\Rightarrow a_7=9+6 \times 4$
$\Rightarrow a_7=33$
View full question & answer→MCQ 1681 Mark
The $8^{th}$ term of an $AP$ is $17$ and its $14^{th}$ term is $29$. The common difference of the $AP$ is :
AnswerLet a be the frist term and $d$ be the common difference.
$a_8= 17 $
$\Rightarrow a + 7d = 17 .....(i)$
$a_{14}= 29 $
$\Rightarrow a + 13d = 29 .....(ii)$
Subtracting $(i)$ from $(ii),$ we get
$6d = 12$
$\Rightarrow d = 2$
View full question & answer→MCQ 1691 Mark
If the first term of an $A.P.$ is a and nth term is $b,$ then its common difference is :
- ✓
$\frac{\text{b}-\text{a}}{\text{n}-1}$
- B
$\frac{\text{b}-\text{a}}{\text{n}}$
- C
$\frac{\text{b}-\text{a}}{\text{n + 1}}$
- D
$\frac{\text{b + a}}{\text{n }-1}$
AnswerCorrect option: A. $\frac{\text{b}-\text{a}}{\text{n}-1}$
In the given $A.P.$
First term $= a$ and nth term $= b$
$\therefore a + (n - 1) d = b$
$\Rightarrow (n - 1) d = b - a$
$\Rightarrow\text{d} = \frac{\text{b}-\text{a}}{\text{n}-1}$
View full question & answer→MCQ 1701 Mark
How many natural numbers between 1 and 1000 are divisible by 5 ?
Answer(c) : Numbers lying between 1 and 1000 which are divisible by 5 are : 5, 10, 15, 20, ....., 995 .
Let the numbers be $n$. Then, $5+(n-1) \times 5=995$ $\Rightarrow 5(n-1)=990 \Rightarrow(n-1)=198 \Rightarrow n=199$
View full question & answer→MCQ 1711 Mark
The first and last torm of an A.P. are $a$ and $I$ and sum of the A.P. is $S$, then the common difference is $\frac{l^2-a^2}{k-(l+a)}$. Here $k$ is equal to
Answer(b): We have, $S=\frac{n}{2}(a+l) \Rightarrow \frac{2 S}{a+l}=n$
Also, $l=a+(n-1) d$
$
\Rightarrow d=\frac{l-a}{n-1}=\frac{l-a}{\frac{2 S}{a+l}-1} \Rightarrow d=\frac{l^2-a^2}{2 S-(l+a)}=\frac{l^2-a^2}{k-(l+a)}
$
$
\therefore \quad k=2 S
$
View full question & answer→MCQ 1721 Mark
Which of the following is not an A.P.?
- A
$-3,-\frac{5}{2},-2,-\frac{3}{2}, \ldots \ldots$
- ✓
$0.3,0.33,0.333, \ldots \ldots$
- C
$\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$
- D
$p, 2 p+1,3 p+2,4 p+3, \ldots \ldots$
AnswerCorrect option: B. $0.3,0.33,0.333, \ldots \ldots$
(b) : We have, $0.33-0.3=0.03,0.333-0.33=0.003$
$
\Rightarrow 0.33-0.3 \neq 0.333-0.33
$
So, $0.3,0.33,0.333, \ldots .$. is not an A.P.
View full question & answer→MCQ 1731 Mark
$\frac{3}{\sqrt{5}}+\sqrt{5}+\frac{7}{\sqrt{5}}+\ldots$ to 15 terms is equal to
- ✓
$51 \sqrt{5}$
- B
$17 \sqrt{5}$
- C
$81 \sqrt{5}$
- D
$9 \sqrt{5}$
AnswerCorrect option: A. $51 \sqrt{5}$
(a) : The given series is an A.P. with $a=\frac{3}{\sqrt{5}}$,
$
\begin{aligned}
d & =\sqrt{5}-\frac{3}{\sqrt{5}}=\frac{2}{\sqrt{5}} \\
\therefore \quad S_{15} & =\frac{15}{2}\left[2 \times \frac{3}{\sqrt{5}}+14 \times \frac{2}{\sqrt{5}}\right]=\frac{15}{2}\left[\frac{6}{\sqrt{5}}+\frac{28}{\sqrt{5}}\right] \\
& =\frac{255}{\sqrt{5}}=51 \sqrt{5}
\end{aligned}
$
View full question & answer→MCQ 1741 Mark
Find the sixteenth term of the A.P. $-10,-6,-2,2, \ldots$
Answer(d) : We have, $a=-10, d=-6+10=4$
$
\therefore \quad a_{16}=a+15 d=-10+15 \times 4=50
$
View full question & answer→MCQ 1751 Mark
Three numbers in an A.P. have sum 18. Its middle term is
Answer(a) : Let the numbers be $a-d, a$ and $a+d$.
Given, $a-d+a+a+d=18 \Rightarrow 3 a=18 \Rightarrow a=6$
$\therefore$ Middle term $=6$
View full question & answer→MCQ 1761 Mark
Find the sum of first 40 positive integers divisible by 6 .
Answer(d) : Numbers divisible by 6 are $6,12,18, \ldots$
The series form an A.P. with first term $(a)=6$ and common difference $(d)=6$ and $n=40$
Now, sum of $n$ terms $S_n=\frac{n}{2}\{2 a+(n-1) d\}$
$
\begin{aligned}
\therefore \quad S_{40} & =\frac{40}{2}\{2 \times 6+(40-1) 6\}=20\{12+234\} \\
& =20 \times 246=4920
\end{aligned}
$
View full question & answer→MCQ 1771 Mark
Find the sum of first $10$ terms of the $A.P. x-8, x-2, x+4, \ldots$
- A
$10 x+210$
- ✓
$10 x+190$
- C
$5 x+190$
- D
$5 x+210$
AnswerCorrect option: B. $10 x+190$
Here, $a=x-8, d=x-2-x+8=6$
$\therefore S_{10}=\frac{10}{2}[2(x-8)+(10-1) 6]=5[2 x-16+54]$
$=5(2 x+38)=10 x+190$
View full question & answer→MCQ 1781 Mark
Find the sum of first 15 multiples of 8.
Answer(c) : Multiples of 8 are $8,16,24,32, \ldots$
It forms an A.P.
Here, $a=d=8$
$
\begin{aligned}
\therefore & S_{15}=\frac{15}{2}[2 \times 8+(15-1) 8]\left[\because S_n=\frac{n}{2}(2 a+(n-1) d)\right] \\
& =\frac{15}{2}[16+112]=\frac{15}{2} \times 128=15 \times 64=960
\end{aligned}
$
View full question & answer→MCQ 1791 Mark
If the first, second and last terms of an $A.P.$ are $a, b$ and $2 a$ respectively, its sum is
- A
$\frac{a b}{2(b-a)}$
- B
$\frac{a b}{b-a}$
- ✓
$\frac{3 a b}{2(b-a)}$
- D
AnswerCorrect option: C. $\frac{3 a b}{2(b-a)}$
We have, $2 a=a+(n-1)(b-a)$
$\Rightarrow n-1=\frac{a}{b-a} $
$\Rightarrow n=\frac{b}{b-a}$
$\therefore S_n=\frac{b}{2(b-a)}(a+2 a)=\frac{3 a b}{2(b-a)}$
View full question & answer→MCQ 1801 Mark
Find how many terms are there in the $A.P. 16,24,32, \ldots \ldots, 96$.
AnswerHere $a=16, d=24-16=8$ and last term $=96$.
$\Rightarrow l=96$
$\Rightarrow a+(n-1) d=96 $
$\Rightarrow 16+(n-1) 8=96$
$\Rightarrow \quad(n-1) 8=80 $
$\Rightarrow n-1=10 $
$\Rightarrow n=11$
So there are $11$ terms in the $A.P.$
View full question & answer→MCQ 1811 Mark
If the sum of first $7$ terms of an $A.P$. is $49$ and that of $17$ terms is $289 ,$ then, its first term is
AnswerLet $a$ and $d$ be respectively the first term and common difference of given $A.P$.
$S_7=49 $
$\Rightarrow \frac{7}{2}(2 a+6 d)=49$
$\Rightarrow 2 a+6 d=14\ldots(i)$
and $ S_{17}=289 $
$\Rightarrow \frac{17}{2}(2 a+16 d)=289$
$\Rightarrow 2 a+16 d=34\ldots(ii)$
Subtracting $(i)$ from $(ii),$ we get
$10 d=20$
$ \Rightarrow d=2$ and by $(i) a=1$
View full question & answer→MCQ 1821 Mark
If $x \neq y$ and the sequences $x, a_1, a_2, y$ and $x, b_1, b_2, y$ each are in $A.P.,$ then $\left(\frac{a_2-a_1}{b_2-b_1}\right)$ is
- A
$\frac{2}{3}$
- B
$\frac{3}{2}$
- ✓
$1$
- D
$\frac{3}{4}$
AnswerWe have, $x, a_1, a_2, y$ are in $A.P.$
$\therefore y=x+3 d$
$\Rightarrow y-x=3\left(a_2-a_1\right)\left[\because a_2-a_1=a_1-x=d\right]$
$\Rightarrow a_2-a_1=\frac{(y-x)}{3}$
Similarly, $b_2-b_1=\left(\frac{y-x}{3}\right)$
$\therefore \frac{a_2-a_1}{b_2-b_1}=1$
View full question & answer→MCQ 1831 Mark
If $9$ times the $9^{\text {th }}$ term in an arithmetic progression is equal to $15$ times of its $15^{\text {th }}$ term, then what is the $24^{\text {th }}$ term?
AnswerLet $a$ be the first term and $d$ be the common difference of the given $A.P$.
According to question, $9 a_9=15 a_{15}$
$\Rightarrow 9(a+8 d)=15(a+14 d)$
$\Rightarrow 9 a+72 d=15 a+210 d$
$\Rightarrow 6 a+138 d=0 $
$\Rightarrow a+23 d=0$
$\Rightarrow a+(24-1) d=0$
$ \Rightarrow a_{24}=0$
View full question & answer→MCQ 1841 Mark
If $m^{\text {th }}$ term of an A.P. is $1 / n$ and $n^{\text {th }}$ term is $1 / m$, then the sum of first $m n$ terms is
- A
$m n+1$
- ✓
$\frac{m n+1}{2}$
- C
$\frac{m n-1}{2}$
- D
$\frac{m n-1}{3}$
AnswerCorrect option: B. $\frac{m n+1}{2}$
(b): We have, $a+(m-1) d=1 / n$ and $a+(n-1) d=1 / m$
Solving (1) and (2), we get $a=\frac{1}{m n}, d=\frac{1}{m n}$
$
\therefore \quad S_{m n}=\frac{m n}{2}\left[\frac{2}{m n}+(m n-1)\left(\frac{1}{m n}\right)\right]=\frac{m n+1}{2} .
$
View full question & answer→MCQ 1851 Mark
If $a,(a-2)$ and $3 a$ are in A.P, then the value of $a$ is
Answer(b) : Since $a,(a-2), 3 a$ are in A.P., we have $(a-2)-a=3 a-(a-2) \Rightarrow-2=2 a+2 \Rightarrow a=-2$
View full question & answer→MCQ 1861 Mark
The sum $(-6)+(0)+(6)+\ldots .$. upto $13^{\text {th }}$ term is given below :
Here, $a=-6$ and $d=0-(-6)=6$ (Step- 1$)$
$
\begin{array}{ll}
\therefore \quad & S_{13}=\frac{13}{2}[2 \times(-6)-(13-1) 6](\text { Step-2) } \\
& =\frac{13}{2}[-12+72]=\frac{13}{2} \times(60)=390(\text { Step-3) }
\end{array}
$
In which step is there an error in solving?
View full question & answer→MCQ 1871 Mark
If the seventh term of an $A.P.$ is $1 / 9$ and its ninth term is $1 / 7,$ find common difference.
- A
$1$
- B
$2 / 63$
- C
$3 / 64$
- ✓
$1 / 63$
AnswerCorrect option: D. $1 / 63$
Given, $a_7=1 / 9$ and $a_9=1 / 7$
$\Rightarrow a+6 d=1 / 9$
and $a+8 d=1 / 7$
Subtracting $(i)$ from $(ii),$ we get
$2 d=\frac{1}{7}-\frac{1}{9}=\frac{2}{63}$
$\Rightarrow d=\frac{1}{63}$
View full question & answer→MCQ 1881 Mark
What is the common difference of four terms in an A.P. such that the ratio of the product of the first and fourth terms to that of the second and third is $2: 3$ and the sum of all four terms is 20 ?
Answer(d) : Take the four terms as $a-3 d, a-d, a+d, a+3 d$ with common difference $2 d$.
Sum $=4 a=20 \Rightarrow a=5$
Also, $3\left(a^2-(3 d)^2\right)=2\left(a^2-d^2\right) \Rightarrow d=1,-1$
$\therefore \quad$ Common difference, $2 d=2$ or -2
View full question & answer→MCQ 1891 Mark
If the common difference of an A.P. is 5, then what is $a_{18}-a_{13}$ ?
Answer(c) : Given, the common difference of A.P. i.e, $d=5$
Now, $a_{18}-a_{13}=[a+(18-1) d]-[a+(13-1) d]$
$=a+(17 \times 5)-a-(12 \times 5)=85-60=25$
View full question & answer→MCQ 1901 Mark
The famous mathematician associated with finding the sum of the first 100 natural numbers is
Answer(c) : Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., $1,2,3, \ldots ., 100$.
View full question & answer→MCQ 1911 Mark
Find how many two$-$digit numbers are divisible by $6.$
AnswerThe two$-$digit numbers that are divisible by $6$ are $12,18,24, \ldots, 96$.
It forms an $A.P.$ with $a=12, d=6$
Since, $n^{\text {th }}$ term of an $A.P.$ is $a_n=a+(n-1) d$
$\therefore 12+(n-1) \times 6=96$
$\Rightarrow 2+(n-1)=16$
$\Rightarrow n=14+1=15$
Thus, there are $15$ two$-$digit numbers that are divisible by $6.$
View full question & answer→MCQ 1921 Mark
For what value of $n$, are the $n^{\text {th }}$ terms of two A.P.'s $52,54,56, \ldots \ldots$ and $4,12,20, \ldots .$. equal ?
Answer(d) : For the A.P., 52, 54, 56, ...
We have, $a=52, d=54-52=2$
$
\Rightarrow a_n=52+(n-1) 2=2 n+50
$
For the A.P., $4,12,20$,....
We have, $a=4, d=12-4=8$
$
\therefore a_n=4+(n-1) 8=8 n-4
$
Now, it is given that the $n^{\text {th }}$ terms of the two A.P.'s are equal.
$
\therefore \quad 2 n+50=8 n-4 \Rightarrow 6 n=54 \Rightarrow n=9
$
View full question & answer→MCQ 1931 Mark
The common difference of the A.P. $\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}, \ldots$. is
Answer(c) : Here, $a_1=\frac{1}{p}$ and $a_2=\frac{1-p}{p}$
$\therefore \quad$ Common difference $=a_2-a_1$
$
=\frac{1-p}{p}-\frac{1}{p}=\frac{1-p-1}{p}=\frac{-p}{p}=-1
$
View full question & answer→MCQ 1941 Mark
If $p^{\text {th }}$ term of an A.P. is $\frac{3 p-1}{6}$, then sum of first $n$ terms of the A.P. is
- ✓
$\frac{n}{12}[3 n+1]$
- B
$\frac{n}{12}[3 n-1]$
- C
$\frac{n}{6}[3 n+1]$
- D
$\frac{n}{6}[3 n-1]$
AnswerCorrect option: A. $\frac{n}{12}[3 n+1]$
(a) : Given, $a_p=\frac{3 p-1}{6}$
When $p=1, a_1=\frac{3-1}{6}=\frac{1}{3}$; When $p=n, a_n=\frac{3 n-1}{6}$
We know that, $S_n=\frac{n}{2}\left[a_1+a_n\right]$
$
\Rightarrow S_n=\frac{n}{2}\left[\frac{1}{3}+\frac{3 n-1}{6}\right] \Rightarrow S_n=\frac{n}{2}\left[\frac{2+3 n-1}{6}\right]=\frac{n}{12}[3 n+1]
$
View full question & answer→MCQ 1951 Mark
The sum of the first $n$ terms of an A.P. is $5 n-n^2$. Find the $n^{\text {th }}$ term of this A.P.
- A
$3-2 n$
- B
$6+2 n$
- ✓
$6-2 n$
- D
$3+2 n$
AnswerCorrect option: C. $6-2 n$
(c): We have, $S_n=5 n-n^2$
$
\begin{aligned}
\therefore \quad S_{n-1} & =5(n-1)-(n-1)^2 \\
& =5 n-5-\left(n^2+1-2 n\right)=-n^2+7 n-6 \\
& n^{\text {th }} \text { term of A.P., } a_n=S_n-S_{n-1} \\
& =5 n-n^2-\left(-n^2+7 n-6\right) \\
& =5 n-n^2+n^2-7 n+6=-2 n+6
\end{aligned}
$
View full question & answer→MCQ 1961 Mark
The numbers $-11,-7,-3,1,5, \ldots \ldots$ are
- A
in A.P. with $d=-18$
- B
in A.P. with $d=-4$
- ✓
in A.P. with $d=4$
- D
AnswerCorrect option: C. in A.P. with $d=4$
(c) : We have, $(-7+11)=(-3+7)=(1+3)$$=(5-1)=4
$
So, given numbers are in A.P. with $d=4$.
View full question & answer→MCQ 1971 Mark
The value of $x$ for which $(8 x+4),(6 x-2)$ and $(2 x+7)$ are in $A.P.,$ is
- ✓
$\frac{15}{2}$
- B
$\frac{2}{15}$
- C
$-\frac{15}{2}$
- D
$-\frac{2}{15}$
AnswerCorrect option: A. $\frac{15}{2}$
Since $(8 x+4),(6 x-2)$ and $(2 x+7)$ are in $A.P.$
$\therefore (6 x-2)-(8 x+4)=(2 x+7)-(6 x-2)$
$\Rightarrow 2(6 x-2)=(8 x+4)+(2 x+7)$
$\Rightarrow 12 x-4=10 x+11 $
$\Rightarrow 2 x=15 $
$\Rightarrow x=15 / 2$
View full question & answer→MCQ 1981 Mark
A man starts repaying a loan with first monthly installment of $₹ 1000 .$ If he increases the installment by $₹ 50$ every month, what amount will he pay in the $30^{\text {th }}$ installment?
- A
$₹ 1450$
- ✓
$₹ 2450$
- C
$₹ 2050$
- D
$₹ 2040$
AnswerCorrect option: B. $₹ 2450$
First monthly installment $=\text {₹} 1000$
Second monthly installment $=\text {₹}(1000+50)=\text {₹} 1050$
Third monthly installment $=\text {₹}(1050+50)=\text {₹}1100$ and so on.
$1000, 1050, 1100, ......$ forms an $A.P.,$
where $a=1000, d=50, n=30$
Now, $a_n=a+(n-1) d$
$\Rightarrow a_{30}=1000+(30-1) 50=1000+1450=2450$
$\therefore 30^{\text {th }}$ installment $=\text {₹} 2450$
View full question & answer→MCQ 1991 Mark
The number of terms in the $A.P.\ 3, 6, 9, 12, \ldots, 111$ is
Answer$A.P.$ is $3,6,9, \ldots ., 111$
Here, $a=3, d=6-3=3$
$a_n=a+(n-1) d$
$\Rightarrow 111=3+(n-1) 3$
$\Rightarrow 108=3(n-1)$
$\Rightarrow n-1=36$
$\Rightarrow n=37$
View full question & answer→MCQ 2001 Mark
The production of TV in a factory increases uniformly by a fixed number every year. It produced 8000 TV's in $6^{\text {th }}$ year $\& 11300$ in $9^{\text {th }}$ year, find the production in $8^{\text {th }}$ year.
Answer(d) : Let number of TV produced in first year be $a$
Let increment in every year $=d$
Given, $a_6=8000, a_9=11300$
$\Rightarrow a+5 d=8000$ and $a+8 d=11300$
$\therefore a+8 d-a-5 d=11300-8000$
$\Rightarrow 3 d=3300 \Rightarrow d=1100$
$\therefore a=8000-5 d=8000-5500=2500$
$a_8=a+7 d=2500+7(1100)=2500+7700=10200$
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