MCQ 11 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 21 Mark
Which term of the $AP 72, 63 54, ....$ is $0?$
- A
$8^{th}$
- ✓
$9^{th}$
- C
$10^{th}$
- D
$11^{th}$
AnswerCorrect option: B. $9^{th}$
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 31 Mark
The $5^{\text {th }}$ term of an $AP$ is $20$ and the sum of its $7^{\text {th }}$ and $11^{\text {th }}$ 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 41 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 51 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 61 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 71 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 ten terms is:
AnswerLet $a$ be frist term and $d$ be the common difference.
$a_{13}=4 a_3$
$\Rightarrow a+12 d=4(a+2 d)$
$\Rightarrow a+12 d=4 a+8 d$
$\Rightarrow 4 d=3 a \ldots .(i)$
$a_5=16$
$\Rightarrow a+4 d=16$
Substituting $(i),$ we get
$\Rightarrow a+3 a=16$
$\Rightarrow a=4$
So, $d=3$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{10}=\frac{10}{2}[2(4)+9(3)]$
$\Rightarrow S_{10}=5[8+27]$
$\Rightarrow S_{10}=5[35]$
$\Rightarrow S_{10}=175$
View full question & answer→MCQ 81 Mark
The sum of first $n$ term of an $AP$ is $\left(3 n^2+6 n\right)$. The common difference of the $AP$ is:
AnswerThe sum of frist n term of an AP is $\big(3\text{n}^2+6\text{n}\big).$
$\Rightarrow\text{S}_\text{n-1}=3(\text{n}-1)^2+6(\text{n}-1)$
$=3\big(\text{n}^2-2\text{n}+1\big)+6(\text{n}-1)$
$=3\text{n}^2-6\text{n}+3+6\text{n}-6$
$=3\text{n}^2-3$
$\text{a}_\text{n}=\text{S}_\text{n}-\text{S}_\text{n-1}$
$=3\text{n}^2+6\text{n}-3\text{n}^2-3$
$=6\text{n}+3$
Let $d$ be the common difference of thew $AP.$
$\text{d}=\text{a}_\text{n}-\text{a}_\text{n-1}$
$=(6\text{n}+3)-\big[6(\text{n}-1)+3\big]$
$=(6\text{n}+3)-6(\text{n}-1)-3$
$=6$
View full question & answer→MCQ 91 Mark
The sum of first n term of an $AP$ is $(5n^2 - n^2)$. The common difference of the $AP$ is:
- A
$(5 - 2n)$
- ✓
$(6 - 2n)$
- C
$(2n - 5)$
- D
$(2n - 6)$
AnswerCorrect option: B. $(6 - 2n)$
The sum of frist n term of an $AP$ is $(5n - n^2).$
$S_n= 5n - n^2$
$\Rightarrow S_{n-1}= 5(n - 1)-(n - 1)^2$
$= 5n - 5 - (n^2- 2n + 1)$
$= 5n - 5 - n^2 + 2n - 1$
$= -n^2 + 7n - 6$
$a_n = S_n - S_{n-1}$
$= 5n - n^2 - (-n^2 + 7n - 6$
$= 5n - n^2 + n^2 - 7n + 6$
$= 6 - 2n$
View full question & answer→MCQ 101 Mark
If the $n^{th}$ term of an $AP$ is $(2n + 1)$ then the sum of its first three terms is:
Answer$n^{th}$ term is given to be $(2n + 1).$
Let the frist term be $a$ and the common diffrerance be $d.$
$\Rightarrow a = 2(1) + 1 = 3$
The second term $= 2(2) + 1 = 5$
$d = 5 - 3 = 2$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\ \text{S}_3=\frac{3}{2}\big[2((3)+(3-1)2\big]$
$\therefore\ \text{S}_3=\frac{3}{2}\big[6+4\big]$
$\therefore\ \text{S}_3=15$
View full question & answer→MCQ 111 Mark
The common difference of the AP $\frac{1}{\text{p}},\frac{1-\text{p}}{\text{p}},\frac{1-2\text{p}}{\text{p}},....$ is:
AnswerCommon difference
$=\frac{1-\text{p}}{\text{p}}-\frac{1}{\text{p}}$
$=\frac{1-\text{p}-1}{\text{p}}$
$=\frac{-\text{p}}{\text{p}}$
$= - 1$
View full question & answer→MCQ 121 Mark
The sum of first 16 terms of the AP 10, 6, 2, .... is
AnswerLet a be the term and d be the common difference.
AP is 10, 6, 2, ......
a = 10 and d = 6 - 10 = -4
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n-1)}\text{d}\big]$
$\Rightarrow\text{S}_{16}=\frac{16}{2}\big[2(10)+15(-4)\big]$
$\Rightarrow\text{S}_{16}=8[20-60]$
$\Rightarrow\text{S}_{16}=8[-40]$
$\Rightarrow\text{S}_{16}=-320$
So, the sum is -320.
View full question & answer→MCQ 131 Mark
An $AP 5, 12, 19, ....$ has $50$ term. Its last term is:
AnswerLet a be frist term and $d$ be the common difference.
$a=5$
$d=12-5=7$
$a_{50}=a+49 d$
$\Rightarrow a_{50}=5+49(7)$
$\Rightarrow a_{50}=348$
So, its last term is $348 .$
View full question & answer→MCQ 141 Mark
How many two$-$digit numbers are divisible by $3?$
AnswerThe two-digit numbers divisible by $3$ start from
$12,15,18,21, \ldots \ldots ., 99$
Here,
$a=12$
$d=3$
$a_n=a+(n-1) d$
$\Rightarrow 99=12+(n-1)(3)$
$\Rightarrow 99=12+3 n-3$
$\Rightarrow 90=3 n$
$\Rightarrow n=30$
View full question & answer→MCQ 151 Mark
If an denotes the $n ^{\text {th }}$ term of the $AP 3,8,13,18, \ldots .$. then what is the value of $\left( a _{30}- a _{20}\right)$ ?
AnswerThe given $AP$ is $3, 8, 13, 18, .....$
$a = 3$ and $d = 8 - 3 = 5$
$a_{30} - a_{20}$$= a + 29d - (a +19d)$
$= a + 29d - a - 19d$
$= 10d$
$= 10(5)$
$= 50$
View full question & answer→MCQ 161 Mark
The sum of first n terms of an AP is $(4n^2 + 2n)$. The nth term of this $AP$ is:
- A
$(6n - 2)$
- B
$(7n - 3)$
- ✓
$(8n - 2)$
- D
$(8n + 2)$
AnswerCorrect option: C. $(8n - 2)$
The sum of frist n terms of an $AP$ is $(4n^2 + 2n).$
$S_n = 4n^2 + 2n$
$\Rightarrow Sn_{-1} =4n^2 + 2n$
$= 4(n - 1)^2 + 2(n - 1)$
$= 4(n^2 - 2n + 1) + 2(n - 1)$
$= 4n^2 - 8n + 4 + 2n - 2$
$= 4n^2 - 6n + 2$
$a_n = S_n- S_{n-1}$
$= 4n^2 + 2n - (4n^2 - 6n + 2)$
$= 4n^2 + 2n - 4n^2 + 6n - 2$
$= 8n - 2$
View full question & answer→MCQ 171 Mark
Which term of the $AP 25, 20, 15,....$ is the first negative term?
- A
$10^{th}$
- ✓
$9^{th}$
- C
$8^{th}$
- D
$7^{th}$
AnswerCorrect option: B. $9^{th}$
$9^{th}$
View full question & answer→MCQ 181 Mark
Which term of the $AP 21, 42, 63, 84, ....$ is the $210?$
- A
$9^{th}$
- ✓
$10^{th}$
- C
$11^{th}$
- D
$12^{th}$
AnswerCorrect option: B. $10^{th}$
The given $AP$ is $21,42,63,84, \ldots .$.
$a=21 \text { and } d=42-21=21$
$a_n=a+(n-1) d$
$\Rightarrow 210=21+(n-1)(21)$
$\Rightarrow 210=21+21 n-21$
$\Rightarrow 210=21 n$
$\Rightarrow n=10$
So, the $10^{\text {th }}$ term will be $210 .$
View full question & answer→MCQ 191 Mark
The $17^{\text {th }}$ term of an $A P$ exceeds its $10^{\text {th }}$ term by $21$ . The common difference of the $A P$ is:
AnswerLet a be the frist termm and d be the common difference.
$a_{17}=a_{10}+21$
$\Rightarrow a+16 d=a+9 d+21$
$\Rightarrow 7 d=21$
$\Rightarrow d=3$
View full question & answer→MCQ 201 Mark
The$ 5^{th}$ term of an $AP$ is $-3$ and its common difference is $-4$. The sum of its first $10$ term is:
AnswerLet $a$ be the frist term and $d$ be the common difference.
$a_5=-3$
$\Rightarrow a+4 d=-3$
$\Rightarrow a+4(-4)=-3$
$\Rightarrow a-16=-3$
$\Rightarrow a=13$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{10}=\frac{10}{2}[2(13)+9(-4)]$
$\Rightarrow S_{10}=5[-10]$
$\Rightarrow S_{10}=-50$
View full question & answer→MCQ 211 Mark
The sum of first 40 positive integers divisible by 6 is:
AnswerThe frist 40 positive numbers divisible by 6 will be
6, 12, 18, 24, ......
Here,
a = 6
d = 6
n = 40
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_{40}=\frac{40}{2}\big[2(6)+39(6)\big]$
$\Rightarrow\text{S}_{40}=20[12+234]$
$\Rightarrow\text{S}_{40}=20[246]$
$\Rightarrow\text{S}_{40}=4920$
View full question & answer→MCQ 221 Mark
How many terms of the AP 3, 7, 11, 15, .... will make the sum 406?
AnswerLet a be the frist term and d be the common diffrerence.
AP is 3, 7, 11, 15, .....
a = 3 and d = 7 - 3 = 4
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow406=\frac{\text{n}}{2}\big[2(3)+(\text{n}-1)(4)\big]$
$\Rightarrow812=\text{n}[6+4\text{n}-4]$
$\Rightarrow812=\text{n}[2+4\text{n}]$
$\Rightarrow812=2\text{n}+4\text{n}^2$
$\Rightarrow4\text{n}^4+2\text{n}-812=0$
$\Rightarrow2\text{n}^2+\text{n}-406=0$
$\Rightarrow2\text{n}^2-28\text{n}+29-406=0$
$\Rightarrow2\text{n}(\text{n}-14)+29(\text{n}-14)=0$
$\Rightarrow(\text{n}-14)(2\text{n}+29)=0$
$\Rightarrow\text{n}=14\text{ or }\text{n}=-\frac{29}{2}$
So, clearly, n = 14 since n cannot be negative n or a fraction.
Hence, 14 terms will make the sum 406.
View full question & answer→MCQ 231 Mark
$(5 + 13 + 21 + .... + 181) = ?$
- A
$2476$
- B
$2337$
- C
$2219$
- ✓
$2139$
AnswerCorrect option: D. $2139$
Let a be the term and d be the common difference.
$5+13+21+\ldots+181$
$a=5 \text { and } d=13-5=8$
$a_n=a+(n-1) d$
$\Rightarrow 181=5+(n-1) 8$
$\Rightarrow 176=8 n-8$
$\Rightarrow 8 n=184$
$\Rightarrow n=23$
$\text{S}_\text{n}=\frac{23}{2}$ [frist term + last term]
$\Rightarrow\text{S}_\text{n}=\frac{23}{2}[5+181]$
$\Rightarrow\text{S}_\text{n}=\frac{23}{2}[186]$
$\Rightarrow\text{S}_\text{n}=2139$
So, the sum is $2139.$
View full question & answer→MCQ 241 Mark
The $2^{\text {nd }}$ term of an $AP$ is $13$ and its $5^{\text {th }}$ term is $25$ . What is its $17^{\text {th }}$ term?
AnswerLet a be the frist term and $d$ be the common diffrerence.
$a_n=a+(n+1) d$
$a_2=13 \text { and } a_5=25$
$\Rightarrow a+d=13 \text { and } a+4 d=25$
Substracting the two equation we get
$3 d=12$
$\Rightarrow d=4$
So, $a+4=13 \Rightarrow a=9$
$a_{17}=9+16(4)=9+64=73$
View full question & answer→MCQ 251 Mark
What is $20^{\text {th }}$ term from the end of the $AP 3,8,13, \ldots, 253$ ?
AnswerThe given $AP$ is $3,8,13, \ldots . ., 248,253$
So, cinsider the AP to be $253,248, \ldots ., 13,8,3$
$a=253 \text { and } d=248-253=-5$
$a_n=a+(n-1) d$
$\Rightarrow a_{20}=253+19(-5)$
$\Rightarrow a_{20}=253-95$
$\Rightarrow a_{20}=158$
So, the $20^{\text {th }}$ term will be $158$
View full question & answer→MCQ 261 Mark
The $7^{\text {th }}$ term of an $AP$ is $4$ and its common difference is $-4$ . What is its first term?
AnswerLet a be the frist term.
$a_7=4$
$\Rightarrow a+6 d=4$
$\Rightarrow a+6(-4)=4$
$\Rightarrow a=4+24$
$\Rightarrow a=28$
View full question & answer→MCQ 271 Mark
What is the common difference of an $AP$ in which $a_{18}-a_{14}=32 ?$
AnswerLet a be the frist term and d be the common difference.
$a_{18}-a_{14}=32$
$\Rightarrow a+17 d-(a+13 d)=32$
$\Rightarrow a+17 d-a-13 d=32$
$\Rightarrow 4 d=32$
$\Rightarrow d=8$
View full question & answer→MCQ 281 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,
$28=4+4 d$
$\Rightarrow 4 d=24$
$\Rightarrow d=6$
$x_3=a+(3) d \ldots . .\left(x_3 \text { is the fourth term }\right)$
$\Rightarrow x_3=4+3(6)$
$\Rightarrow x_3=22$
View full question & answer→MCQ 291 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.
$a_7=-1$
$\Rightarrow a+6 d=-1 \ldots$
$a_{16}=17$
$\Rightarrow a+15 d=17$
Subtracting $(i)$ from $(ii),$ we get
$9 d=18$
$\Rightarrow d=2$
Substituting in $(i),$ we get $a=-13$
$\text { So, } a_n=a+(n-1) d$
$\Rightarrow a_n=-13+(n-1) 2$
$\Rightarrow a_n=-13+2 n-2$
$\Rightarrow a_n=2 n-15$
View full question & answer→MCQ 301 Mark
The $8^{\text {th }}$ term of an $AP$ is $17$ and its $14^{\text {th }}$ term is $29$ . The common difference of the $AP$ is:
AnswerLet a be the frist term and d be the common difference.
$a_8=17 \Rightarrow a+7 d=17 \ldots . .(i)$
$a_{14}=29 \Rightarrow a+13 d=29 \ldots \ldots$
Subtracting $(i)$ from $(ii),$ we get
$6 d=12$
$\Rightarrow d=2$
View full question & answer→