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
$8^{\text {th }}$
- ✓
$9^{\text {th }}$
- C
$10^{\text {th }}$
- D
$11^{\text {th }}$
AnswerCorrect option: B. $9^{\text {th }}$
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 $9^{\text {th }}$ term is $0$.
View full question & answer→MCQ 31 Mark
The $5^{th}$ term of an AP is $20$ and the sum of its $7^{th}$ and $11^{th}$ 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 $13^{th}$ term of an AP is 4 times its $3^{rd}$ term. If its $5^{th}$ term is $16$ then the sum of its first ten terms is:
- A
$150$
- ✓
$175$
- C
$160$
- D
$135$
AnswerLet a be frist term and d be the common difference.
$a_{13}= 4a_3$
$\Rightarrow a + 12d = 4(a + 2d)$
$\Rightarrow a + 12d = 4a + 8d$
$\Rightarrow 4d = 3a ....(i)$
$a_5= 16$
$\Rightarrow a + 4d = 16$
Substituting (i), we get
$\Rightarrow a + 3a = 16$
$\Rightarrow a = 4$
So, $d = 3$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_{10}=\frac{10}{2}\big[2(4)+9(3)\big]$
$\Rightarrow\text{S}_{10}=5\big[8+27\big]$
$\Rightarrow\text{S}_{10}=5[35]$
$\Rightarrow\text{S}_{10}=175$
View full question & answer→MCQ 81 Mark
The sum of first n term of an AP is $(3n^2+ 6n).$ 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=5 n-n^2 $
$ \Rightarrow S_{n-1}=5(n-1)-(n-1)^2 $
$ =5 n-5-\left(n^2-2 n+1\right) $
$ =5 n-5-n^2+2 n-1 $
$ =-n^2+7 n-6 $
$ 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 $
$ =6-2 n$
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:
- A
$6n + 3$
- ✓
$15$
- C
$12$
- D
$21$
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
- A
$320$
- ✓
$-320$
- C
$-352$
- D
$-400$
AnswerCorrect option: B. $-320$
Let 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:
- A
$343$
- B
$353$
- ✓
$348$
- D
$362$
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
$12, 15, 18, 21, ......., 99$
Here,
$a = 12$
$d = 3$
$a_n= a + (n - 1)d$
$\Rightarrow 99 = 12 + (n - 1)(3)$
$\Rightarrow 99 = 12 + 3n - 3$
$\Rightarrow 90 = 3n$
$\Rightarrow n = 30$
View full question & answer→MCQ 151 Mark
If an denotes the $n^{th}$ term of the AP $3, 8, 13, 18, .....$ then what is the value of $(a_{30}- a_{20})?$
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=4 n^2+2 n $
$ \Rightarrow S_{n-1}=4 n^2+2 n $
$ =4(n-1)^2+2(n-1) $
$ =4\left(n^2-2 n+1\right)+2(n-1) $
$ =4 n^2-8 n+4+2 n-2 $
$ =4 n^2-6 n+2 $
$ a_n=S_n-S_{n-1} $
$ =4 n^2+2 n-\left(4 n^2-6 n+2\right) $
$ =4 n^2+2 n-4 n^2+6 n-2 $
$ =8 n-2$
View full question & answer→MCQ 171 Mark
Which term of the AP $25, 20, 15,....$ is the first negative term?
- A
$10^{\text {th }}$
- B
$9^{\text {th }}$
- C
$8^{\text {th }}$
- ✓
$7^{\text {th }}$
AnswerCorrect option: D. $7^{\text {th }}$
The given AP is $25, 20, 15, .....$
$a = 25$ and $d = 20 - 25 = -5$
$a_n= a + (n - 1)d < 0$
$\Rightarrow 25 + (n - 1)(-5) < 0$
$\Rightarrow 5 - (n - 1) < 0$
$\Rightarrow 5 - n + 1 < 0$
$\Rightarrow 6 < 0$
So, the frist negative term will be the $7^{th}$ term.
View full question & answer→MCQ 181 Mark
Which term of the AP $21, 42, 63, 84, ....$ is the $210?$
- A
$9^{\text {th }}$
- ✓
$10^{\text {th }}$
- C
$11^{\text {th }}$
- D
$12^{\text {th }}$
AnswerCorrect option: B. $10^{\text {th }}$
The given AP is $21, 42, 63, 84, .....$
$a = 21$ and $d = 42 - 21 = 21$
$a_n= a + (n - 1)d$
$\Rightarrow 210 = 21 + (n -1)(21)$
$\Rightarrow 210 = 21 + 21n - 21$
$\Rightarrow 210 = 21n$
$\Rightarrow n = 10$
So, the $12^{\text {th }}$ term will be $210$.
View full question & answer→MCQ 191 Mark
The $17^{th}$ term of an AP exceeds its $10^{th}$ term by $21$. The common difference of the AP is:
AnswerLet a be the frist termm and d be the common difference.
$a_{17}= a_{10}+ 21$
$\Rightarrow a + 16d = a + 9d + 21$
$\Rightarrow 7d = 21$
$\Rightarrow d = 3$
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:
- A
$50$
- ✓
$-50$
- C
$30$
- D
$-30$
AnswerLet a be the frist term and d be the common difference.
$a_5= -3$
$\Rightarrow a + 4d = -3$
$\Rightarrow a + 4(-4) = -3$
$\Rightarrow a - 16 = -3$
$\Rightarrow a = 13$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_{10}=\frac{10}{2}\big[2(13) +9(-4)\big]$
$\Rightarrow\text{S}_{10}=5[-10]$
$\Rightarrow\text{S}_{10}=-50$
View full question & answer→MCQ 211 Mark
The sum of first $40$ positive integers divisible by $6$ is:
- A
$2460$
- B
$3640$
- ✓
$4920$
- D
$4860$
AnswerCorrect option: C. $4920$
The 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.
$5 + 13 + 21 +... + 181$
$a = 5$ and $d = 13 - 5 = 8$
$a_n= a + (n -1)d$
$\Rightarrow 181 = 5 + (n -1)8$
$\Rightarrow 176 = 8n - 8$
$\Rightarrow 8n = 184$
$\Rightarrow n = 23$
$\text{S}_\text{n}=\frac{23}{2}$ [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 $2^{nd}$ term of an AP is $13$ and its $5^{th}$ term is $25$. What is its $17{th}$ term?
AnswerLet a be the frist term and d be the common diffrerence.
$a_n= a + (n +1)d$
$a_2= 13$ and $a_5= 25$
$\Rightarrow a + d = 13$ and $a + 4d = 25$
Substracting the two equation we get
$3d = 12$
$\Rightarrow d = 4$
So, $a + 4 = 13 \Rightarrow a = 9$
$a_{17}= 9 + 16(4) = 9 + 64 = 73$
View full question & answer→MCQ 251 Mark
What is $20^{th}$ term from the end of the AP $3, 8, 13, ..., 253?$
- A
$163$
- ✓
$158$
- C
$153$
- D
$148$
AnswerThe given AP is $ 3, 8, 13, ....., 248, 253$
So, cinsider the AP to be $253, 248,...., 13, 8, 3$
$ 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 $20^{th}$ term will be $158$.
View full question & answer→MCQ 261 Mark
The $7^{th}$ term of an AP is $4$ and its common difference is $-4$. What is its first term?
AnswerLet a be the frist term.
$a_7= 4$
$\Rightarrow a + 6d = 4$
$\Rightarrow a + 6(-4) = 4$
$\Rightarrow a = 4 + 24$
$\Rightarrow a = 28$
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.
$a_{18}- a_{14}= 32$
$\Rightarrow a + 17d - (a + 13d) = 32$
$\Rightarrow a + 17d - a - 13d = 32$
$\Rightarrow 4d = 32$
$\Rightarrow d = 8$
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 $5^{\text {th }}$ term,
$28 = 4 + 4d$
$\Rightarrow 4d = 24$
$\Rightarrow d = 6$
$x_3=a+(3) d \ldots . .\left(x_3\right.$ is the fourth term$)$
$\Rightarrow x_3=4+3(6)$
$\Rightarrow x_3=22$
View full question & answer→MCQ 291 Mark
The $7th$ term of an AP is $-1$ its 16th 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, $a_n= a + (n - 1)d$
$\Rightarrow a_n= -13 + (n - 1)2$
$\Rightarrow a_n= -13 + 2n - 2$
$\Rightarrow a_n= 2n - 15$
View full question & answer→MCQ 301 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.
$a_8= 17 ⇒ a + 7d = 17 .....(i)$
$a_{14}= 29 ⇒ a + 13d = 29 .....(ii)$
Subtracting $(i)$ from $(ii)$, we get
$6d = 12$
$\Rightarrow d = 2$
View full question & answer→