MCQ 511 Mark
If the sum of first $m$ terms of an $A.P.$ is $2 m^2+3 m$, then what is its second term?
AnswerWe have, $S_m=2 m^2+3 m$
$\therefore S_1=2(1)^2+3(1)=5$ and
$S_2=2(2)^2+3(2)=8+6=14$
Now, $a_1=S_1=5$ and $a_2=S_2-S_1=14-5=9$
$\therefore$ Second term of the $A.P.$ is $9.$
View full question & answer→MCQ 521 Mark
If the sum of first $n$ terms of an $A.P.$ is given by $S_n=5 n^2+3 n$, then find its $n^{\text {th }}$ term.
- ✓
$10 n-2$
- B
$10 n+3$
- C
$2 n-10$
- D
$2 n+10$
AnswerCorrect option: A. $10 n-2$
It is given that $S_n=5 n^2+3 n$
We know that $S_n-S_{n-1}=a_n$
$\Rightarrow 5 n^2+3 n-\left\{5(n-1)^2+3(n-1)\right\}=a_n$
$\Rightarrow 5 n^2+3 n-\left\{5\left(n^2+1-2 n\right)+3 n-3\right\}=a_n$
$\Rightarrow 5 n^2+3 n-\left\{5 n^2-7 n+2\right\}=a_n$
$\Rightarrow 10 n-2=a_n$
View full question & answer→MCQ 531 Mark
If the $10^{\text {th }}$ term of an A.P. is 52 and $17^{\text {th }}$ term is 20 more than the $13^{\text {th }}$ term, then find the A.P.
AnswerCorrect option: D. $7,12,17, \ldots \ldots$
(d) : We have, $a_{10}=52$
$
\Rightarrow a+9 d=52\ldots(i)
$
Also, $a_{17}=20+a_{13}$
$
\Rightarrow a+16 d=20+a+12 d \Rightarrow 4 d=20 \Rightarrow d=5
$
From (i), $a+9 \times 5=52 \Rightarrow a+45=52 \Rightarrow a=7$
Thus, the A.P. is $7,12,17, \ldots$
View full question & answer→MCQ 541 Mark
Determine the A.P. whose $3^{\text {rd }}$ term is 5 and the $7^{\text {th }}$ term is 9 .
- A
$1,3,5,7,9, \ldots$
- ✓
$3,4,5,6,7, \ldots$
- C
$-1,2,5,8, \ldots$
- D
AnswerCorrect option: B. $3,4,5,6,7, \ldots$
(b) : We have,
$
\begin{aligned}
a_3 & =a+(3-1) d=a+2 d=5\ldots(i) \\
\text { and } a_7 & =a+(7-1) d=a+6 d=9\ldots(ii)
\end{aligned}
$
Solving the pair of linear equations (1) and (2), we get $a=3, d=1$
Hence, the required A.P. is $3,4,5,6,7, \ldots \ldots$
View full question & answer→MCQ 551 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 $a+8 d=11300$
$\therefore a+8 d-a-5 d=11300-8000$
$\Rightarrow 3 d=3300 \Rightarrow d=1100$
$\therefore a=8000-5 d=8000-5500=2500$
$a_8=a+7 d=2500+7(1100)=2500+7700=10200$
View full question & answer→MCQ 561 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 571 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 581 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 591 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 601 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 611 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 621 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 631 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 641 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 651 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 661 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 671 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 681 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 691 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 701 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 711 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 721 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 731 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 741 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 751 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 $=96$.
$\Rightarrow l=96$
$\Rightarrow a+(n-1) d=96$
$\Rightarrow 16+(n-1) 8=96$
$\Rightarrow (n-1) 8=80$
$\Rightarrow n-1=10$
$\Rightarrow n=11$
So there are $11$ terms in the $A.P.$
View full question & answer→MCQ 761 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 771 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 781 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 791 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 801 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 811 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 821 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 831 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 841 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 851 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 861 Mark
For an A.P., if d = - 4 a7 = 4 then its first term a=___________.