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 $5^{th}$ term of an AP is $20$ and the sum of its 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:
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 $(5n - n^2).$
$S_n= 5n - n^2$
$\Rightarrow S_{n-1}= 5(n - 1)-(n - 1)^2$
$= 5n - 5 - (n^2- 2n + 1)$
$= 5n - 5 - n^2 + 2n - 1$
$= -n^2 + 7n - 6$
$a_n = S_n - S_{n-1}$
$= 5n - n^2 - (-n^2 + 7n - 6)$
$= 5n - n^2 + n^2 - 7n + 6$
$= 6 - 2n$
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
- 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:
AnswerLet $a$ be frist term and d be the common difference.
$a = 5$
$d = 12 - 5 = 7$
$a_{50} = a + 49d$
$\Rightarrow a_{50} = 5 + 49(7)$
$\Rightarrow a_{50} = 348$
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 $(4n^2 + 2n).$
$S_n = 4n^2 + 2n$
$\Rightarrow S_{n-1}=4 n^2+2 n$
$= 4(n - 1)^2 + 2(n - 1)$
$= 4(n^2 - 2n + 1) + 2(n - 1)$
$= 4n^2 - 8n + 4 + 2n - 2$
$= 4n^2 - 6n + 2$
$a_n = S_n- S_{n-1}$
$= 4n^2 + 2n - (4n^2 - 6n + 2)$
$= 4n^2 + 2n - 4n^2 + 6n - 2$
$= 8n - 2$
View full question & answer→MCQ 171 Mark
Which term of the $AP\ 25, 20, 15,....$ is the first negative term?
- A
$10^{th}$
- B
$9^{th}$
- C
$8^{th}$
- ✓
$7^{th}$
AnswerCorrect option: D. $7^{th}$
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
$9^{th}$
- ✓
$10^{th}$
- C
$11^{th}$
- D
$12^{th}$
AnswerCorrect option: B. $10^{th}$
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 $10^{th}$ 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:
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 $AP\ 3, 8, 13, ..., 253$?
AnswerThe given $AP$ is $3, 8, 13, ....., 248, 253$
So, cinsider the $AP$ to be $253, 248,...., 13, 8, 3$
$a = 253$ 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 $5^{th}$ 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 291 Mark
The $7^{th}$ 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.
$a_8 = 17 \Rightarrow a + 7d = 17 .....(i)$
$a_{14} = 29 \Rightarrow a + 13d = 29 .....(ii)$
Subtracting $(i)$ from $(ii),$ we get
$6d = 12$
$\Rightarrow d = 2$
View full question & answer→