MCQ 511 Mark
The nth term of $a_n \ \ A.P.$ is $7 - 4n$, then its common difference is :
AnswerGiven : $a_n=7-4 n$
$\therefore a_1=7-4 \times 1=7-4=3$
$\therefore a_2=7-4 \times 2=7-8=-1$
$\therefore d = -1 - 3 = -4$
View full question & answer→MCQ 521 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.
$a_{13}= 4a_3$
$\Rightarrow a + 12d = 4(a + 2d)$
$\Rightarrow a + 12d = 4a + 8d$
$\Rightarrow 4d = 3a ....(i)$
$a_5= 16$
$\Rightarrow a + 4d = 16$
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 531 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 541 Mark
The sum of three terms of an $A.P$. is $72,$ then its middle term is :
AnswerLet the middle term be $a,$
then the first term is $a - d$ and next term is $a + d$
$\therefore a - d + a + a + d = 72$
$\Rightarrow 3a = 72$
$\Rightarrow a = 24$
View full question & answer→MCQ 551 Mark
Mark the correct alternative in the following : The common difference of the $A.P. \frac{1}{2\text{b}},\frac{1-6\text{b}}{2\text{b}},\frac{1-12\text{b}}{2\text{b}}, .....$ is
AnswerLet a be the first term and $d$ be the common difference.
The given $A.P$. is $\frac{1}{2\text{b}},\frac{1-6\text{b}}{2\text{b}},\frac{1-12\text{b}}{2\text{b}},\ .....$
Common difference $= d =$ Second $-$ First term
$=\frac{1-6\text{b}}{2\text{b}}-\frac{1}{2\text{b}}$
$=\frac{-6\text{b}}{2\text{b}}=-3$
Hence, the correct option is $(D)$.
View full question & answer→MCQ 561 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 $
$= 6 - 2n$
View full question & answer→MCQ 571 Mark
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 $30^{th}$ terms 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 581 Mark
Choose the correct answer from the given four options : What is the common difference of an $AP$ in which $a_{18}- a_{14}= 32?$
AnswerGiven, $a_{18}- a_{14}= 32$
$[\therefore a_n= a + (n - 1)d]$
$\Rightarrow a + (18 - 1)d - [a + (14 - 1)d] = 32$
$\Rightarrow a + 17d - a - 13d = 32$
$\Rightarrow 4d = 32$
$\Rightarrow d = 8$
which is the required common difference of an $AP$.
View full question & answer→MCQ 591 Mark
Choose the correct answer from the given four options : If the common difference of an $AP$ is $5,$ then what is $a_{18}– a_{13}?$
AnswerGiven, the common difference of $AP$
i.e.$,d = 5$
$ {\left[\therefore a_n=a+(n-1) d\right]} $
Now $,a_{18}-a_{13}=a+(18-1) d-[a+(13-1) d] $
$ a_{18}-a_{13}=a+17 \times 5-a 12 \times 5 $
$ a_{18}-a_{13}=85-60 $
$ a_{18}-a_{13}=25 $
View full question & answer→MCQ 601 Mark
Choose the correct answer from the given four options : If $7$ times the $7^{th}$ term of an $AP$ is equal to $11$ times its $11^{th}$ term, then its $18^{th}$ term will be :
AnswerAccording to the question,
$ 7 a_7=11 a_{11} $
$ {\left[\therefore a_n=a+(n-1) d\right]} $
$\Rightarrow 7[a + (7 - 1)d] = 11[a + (11 - 1)d]$
$\Rightarrow 7(a + 6d) = 11(a + 10d)$
$\Rightarrow 7a + 42d = 11a + 110d$
$\Rightarrow 4a + 68d = 0$
$\Rightarrow 2(2a + 34d) = 0$
$\Rightarrow 2a \ \ 34d = 0 \big[\therefore2\neq0\big]$
$\Rightarrow a + 17 d = 0 .......(i)$
$\Rightarrow 18^{th}$ term of an $AP, a_{18} = a + (18 - 1)d$
$= a + 17d = 0 \ [$from Eq. $(i)]$
View full question & answer→MCQ 611 Mark
Choose the correct answer from the given four options : The first four terms of an $AP,$ whose first term is $-2$ and common difference is $-2,$ are :
- A
$\{-2, 0, 2, 4\}$
- B
$\{-2, 4, – 8, 16\}$
- ✓
$\{-2, – 4, – 6, – 8\}$
- D
$\{-2, – 4, – 8, –16\}$
AnswerCorrect option: C. $\{-2, – 4, – 6, – 8\}$
Let the first four terms of an $AP$ are $a, a + d, a + 2d$ and $a + 3d$.
Given, that first term $, a = -2$ and common difference, $d = -2,$
then have an $AP$ as follows $-$
$= -2, -2 - 2, - 2 + 2(-2), -2 + 3(-2)$
$= \{-2, -4, -6, -8\}$
View full question & answer→MCQ 621 Mark
The first term of an $A.P$. is m and its common difference is $n,$ then its $10^{th}$ term is :
- A
$9m - n$
- B
$9m + n$
- C
$m - 9n$
- ✓
$m + 9n$
AnswerCorrect option: D. $m + 9n$
Given : $a = m, d = n$
$\therefore a_{10}= m + (10 - 1) n = m + 9n$
View full question & answer→MCQ 631 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 641 Mark
The first and last terms of an $A.P$. are $1$ and $11$. If their sum is $36,$ then the number of terms will be :
AnswerGiven : $a = 1, l = 11$ and $S_n= 36$
$\therefore\text{S}_{\text{n}} = \frac{\text{n}}{2}(\text{a =l}) $
$\therefore\text{S}_{\text{n}} = \frac{\text{n}}{2}(\text{1 + 11}) $
$\Rightarrow72 = \text{n}\times{12}$
$\Rightarrow\text{n}\times{6}$
View full question & answer→MCQ 651 Mark
If $a, b$ and $c$ are in $A.P$. then the relation between them is given by :
- A
$2c = a + b$
- B
$2a = b + c$
- C
$a = b + c$
- ✓
$2b = a + c$
AnswerCorrect option: D. $2b = a + c$
We know that if $a, b, c, ...$ are in $AP$ then $2^{nd}$ term $- 1^{st}$ term $= 3^{rd}$ term $- 2^{nd}$ term
$b - a = c - b$
$\Rightarrow 2b = a + c$
View full question & answer→MCQ 661 Mark
Choose the correct answer from the given four options : The $4^{th}$ term from the end of the $AP$ : $–11, –8, –5, ..., 49$ is :
AnswerWe know that, the $n^{th}$ term of an $AP$ from the end is
$a_n= l - (n - 1)d .....(i)$
Here $, l =$ Last term and $l = 49 \ [$given$]$
Common difference $, d = -8 - (-11)$
$d = -8 + 11$
$d = 3$
From Eq. $(i), a_4= 49 - (4 - 1)3$
$a_4= 49 - 9$
$a_4= 40$
View full question & answer→MCQ 671 Mark
From your pocket money, you save $Rs. 1$ on the first day, $Rs. 2$ on the second day, $Rs. 3$ on the third day and so on. The amount of money saved by you in the month of May $2013$ is :
- A
$Rs. 1000$
- B
$Rs. 500$
- ✓
$Rs. 496$
- D
$Rs. 498$
AnswerCorrect option: C. $Rs. 496$
Given : $a = 1, d = 2 - 1 = 1, n = 31$
$\therefore\text{s}_{\text{n}} = \frac{\text{n}}{2}[\text{2a}+(\text{n - 1})\text{d}]$
$\Rightarrow\text{s}_{31} = \frac{31}{2}[2\times1+(31 - 1)1]$
$\Rightarrow\text{s}_{31} = \frac{31}{2}[2+30]$
$=\frac{31}{2} \times32$
$= 31 \times 16$
$= \text{Rs}. 496$
$\therefore$ The amount saved by you in the month of may $2013$ is $Rs. 496$
View full question & answer→MCQ 681 Mark
The first term of an $A.P$. is $5,$ the last term is $45$ and the sum is $400$. The number of terms is :
AnswerGiven : $a = 5, l = 45, S_n= 400$
$\therefore\text{S}_{\text{n}} = \frac{\text{n}}{2}(\text{a + l})$
$\Rightarrow400 = \frac{\text{n}}{2} {(5 + 45)}$
$\Rightarrow 800=\text{n} \times50$
$\Rightarrow \text{n} = 16$
View full question & answer→MCQ 691 Mark
Mark the correct alternative in the following : If $7^{th}$ and $13^{th}$terms of an $A.P$. be $34$ and $64$ respectively, then its $18^{th}$ term is :
AnswerGiven,
$a_7=34$
and $a_{13}=64$
We know, $a_n=a+(n-1) d$
$7^{\text {th }}$ term, $a_7=a+(7-1) d$
$\Rightarrow 34 = a + 6d .....(i)$
$13^{\text {th }}$ term, $a_{13}=a+(13-1) d$
$\Rightarrow 64 = a + 12d .....(ii)$
By subracting eq. $(i)$ from $(ii)$ from eq. $(ii)$
$\Rightarrow 64 - 34 = a + 21d - (a + 6d)$
$\Rightarrow 30 = a + 12d - a - 6d$
$\Rightarrow 30 = 6d$
$\Rightarrow\ \text{d}=\frac{30}{6}$
Put the value of $d$ in eq. $(i)$
$\Rightarrow 34 = a + 6(5)$
$\Rightarrow 34 = a + 30$
$\Rightarrow a = 34 - 30 = 4$
Now we have to find $18^{\text {th }}$ term,
$ \Rightarrow a_{18}=4+(18-1) 5 $
$ \Rightarrow a_{18}=4+17 \times 5 $
$ \Rightarrow a_{18}=4+85 $
$ \Rightarrow a_{18}=89$
Hence, correct choice is $(C).$
View full question & answer→MCQ 701 Mark
If the second term of an $A.P$. is $13$ and its fifth term is $25,$ then its $7^{th}$ term is :
AnswerGiven : $a_2= 13$
$\Rightarrow a + (2 - 1) d = 13$
$\Rightarrow a + d = 13 ... (i)$
And $a_5= 25$
$\Rightarrow a + (5 - 1)d = 25$
$\Rightarrow a + 4d = 25 ... (ii)$
Solving eq. $(i)$ and $(ii),$
we get $a = 9$ and $d = 4$
$\therefore \ a_7= a + (7 - 1)d$
$= 9 + (7 - 1) \times 4$
$= 9 + 6 \times 4$
$= 9 + 24 = 33$
View full question & answer→MCQ 711 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 721 Mark
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
$\frac{\text{n+1}}{\text{2n}}$
- C
$\frac{\text{n}-1}{\text{n}}$
- ✓
$\frac{\text{n}+1}{\text{n}}$
AnswerCorrect option: D. $\frac{\text{n}+1}{\text{n}}$
Sum of $n$ even natural number $= n(n + 1)$
and sum of $n$ odd natural numbers $= n^2$
$\therefore\text{n}(\text{n}+1) = \text{kn}^{2}$
$\Rightarrow\text{k} = \frac{{\text{n(n + 1)}}}{\text{n}^2} = \frac{\text{n} + 1}{\text{n}}$
View full question & answer→MCQ 731 Mark
Mark the correct alternative in the following : The first three terms of an $A.P$. respectively are $3y - 1, 3y + 5$ and $5y + 1$. Then $, y$ equals :
AnswerSince $, 3y - 1, 3y + 5$ and $5y + 1$ are first three terms of an $A.P.$
Then, Second term $-$ First term $=$ Third term $-$ Second term $= d\ ($common difference$)$
$\Rightarrow 3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)$
$\Rightarrow 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5$
$\Rightarrow 6 = 2y - 4$
$\Rightarrow 2y = 6 + 4$
$\Rightarrow 2y = 10$
$\Rightarrow y = 5$
Hence, the correct option is $(C)$.
View full question & answer→MCQ 741 Mark
Choose the correct answer from the given four options : The sum of first five multiples of $3$ is :
AnswerThe first five multipies of $3$ are $3, 6, 9, 12$ and $15$.
Here, first term $, a = 3,$ common difference $, d = 6 - 3 = 3$
and number of terem $, n = 5$
$\bigg[\therefore\text{S}_{\text{n}}= \frac{\text{n}}{2}\left\{2\text{a}+\big(\text{n}-1\big)\text{d}\right\}\bigg] $
$\therefore\text{S}_{\text{5}}= \frac{\text{5}}{2}\big[2\text{a}+\big(\text{5}-1\big)\text{d}\big]$
$\text{S}_{5}=\frac{5}{2}\big[2\times3+4\times3\big]$
$\text{S}_{5}=\frac{5}{2}\big(6+12\big)=5\times9=45$
View full question & answer→MCQ 751 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 761 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 771 Mark
Mark the correct alternative in the following : The $n^{th}$ term of an $A.P,$ the sum of whose $n$ terms is $S_n$, is
- A
$S_n+S_{n-1} $
- ✓
$ S_n-S_{n-1} $
- C
$ S_n+S_{n+1} $
- D
$ S_n-S_{n+1} $
AnswerCorrect option: B. $ S_n-S_{n-1} $
$S_n$ is the sum of first $n$ terms
Last term $n^{th}$ term $= S_n-S_{n-1} $
View full question & answer→MCQ 781 Mark
Mark the correct alternative in the following : If the sum of $n$ terms of an $A.P.$ is $3n^2+ 5n$ then which of its terms is $164?$
- A
$ 26^{\text {th }} $
- ✓
$ 27^{\text {th }} $
- C
$ 28^{\text {th }} $
- D
AnswerCorrect option: B. $ 27^{\text {th }} $
Here, the sum of first $n$ terms is given by the expression,
$S_n= 3n^2+ 5n$
We need to find which term of the $A.P$. is $164$.
Let us take $164$ as the $n^{th}$ term
So we know that the $n^{th}$ term of an $A.P$. is given by,
$a_n= S_n- S_{n-1}$
So,
$ 164=S_n=S_{n-1} $
$ 164=3 n^2+5 n-\left[3(n-1)^2+5(n-1)\right]$
using the property,
$(a-b)^2+a^2+b^2-2 a b$
We get,
$ 164=3 n^2+5 n-\left[3\left(n^2+1-2 n\right)+5(n-1)\right]$
$ 164=3 n^2+5 n-\left[3 n^2+3-6 n+5 n-5\right] $
$164=3 n^2+5 n-\left(3 n^2+n-2\right) $
$ 164=3 n^2+5 n-3 n^2+n+2$
$ 164=6 n+2$
Further solving for $n,$ we get
$6n = 164 - 2$
$\text{n}=\frac{162}{6}$
$n = 27$
Therefore $,164$ is the $27^{th}$ term of the given $A.P.$
Hence the correct option is $(b)$.
View full question & answer→MCQ 791 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 801 Mark
The first four terms of the sequence $a_n= 2n + 3$ are :
- A
$\{3, 5, 7, 9\}$
- B
$\{1, 3, 5, 7\}$
- ✓
$\{5, 7, 9, 11\}$
- D
$\{5, 8, 11, 14\}$
AnswerCorrect option: C. $\{5, 7, 9, 11\}$
Given $an = 2n + 3$
$ \therefore a_1=2 \times 1+3=2+3=5 $
$ a_2=2 \times 2+3=4+3=7 $
$ a_3=2 \times 3+3=6+3=9 $
$ a_4=2 \times 4+3=11 $
$\therefore$ The first four terms are $\{5, 7, 9, 11\}.$
View full question & answer→MCQ 811 Mark
The next two terms of the $AP : k, 2k + 1, 3k + 2, 4k + 3, ... $ are :
- A
$5k + 5$ and $6k + 6$
- ✓
$5k + 4$ and $6k + 5$
- C
$4k + 4$ and $4k + 5$
- D
$5k$ and $6k$
AnswerCorrect option: B. $5k + 4$ and $6k + 5$
Given : $ k, 2k + 1, 3k + 2, 4k + 3, ...$
Here $, d = 2k + 1 - k = k + 1$
$\therefore$ The next two terms are
$4k + 3 + k + 1 = 5k + 4$ and $5k + 4 + k + 1 = 6k + 5$
View full question & answer→MCQ 821 Mark
The $n^{th}$ term of an $A.P$. the sum of whose $n$ terms is $S_n$ is :
- A
$ \mathrm{S}_{\mathrm{n}}+\mathrm{S}_{\mathrm{n}-1} $
- B
$ \mathrm{~S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}+1}$
- C
$ \mathrm{~S}_{\mathrm{n}}+\mathrm{S}_{\mathrm{n}+1}$
- ✓
$ \mathrm{~S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$
AnswerCorrect option: D. $ \mathrm{~S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$
$S_n$ is the sum of first $n$ terms
Last term nth term $= \mathrm{~S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$
View full question & answer→MCQ 831 Mark
Mark the correct alternative in the following : Let $S_n$ denote the sum of $n$ terms of an $A.P$. whose first term is a. If the common difference $d$ is given by $d=S_n-k S_{n-1}+S_{n-2},$ then $k =$
AnswerIn the given problem, we are given $d=S_n-k S_{n-1}+S_{n-1}$
We need to find the value of $k$
So here,
First term $= a$
Common difference $= d$
Sum of $n$ terms $= S_n$
Now, as we know
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\ ....\text{(i)}$
Also, for $n-1$ terms
$\text{S}_{\text{n}-1}=\frac{\text{n}-1}{2}[2\text{a}+[(\text{n}-1)-1]\text{d}]$
$=\frac{\text{n}-1}{2}[2\text{a}+[(\text{n}-1)-1]\text{d}]\ .....\text{(ii)}$
Further, for $n-2$ terms,
$\text{S}_{\text{n}-1}=\frac{\text{n}-1}{2}[2\text{a}+[(\text{a}-2)-1]\text{d}]$
$=\frac{\text{n}-1}{2}[2\text{a}+[(\text{a}-3)\text{d}]\ .....(\text{iii})$
Now, we are given
$\text{d}=\text{S}_\text{n}-\text{kS}_{\text{n}-1}+\text{S}_{\text{n}-2}$
$\text{d}+\text{kS}_{\text{n}-1} =\text{S}_{\text{n}}+\text{S}_{\text{n}-2}$
$\text{k}=\frac{\text{S}_\text{n}+\text{S}_{\text{n}-1}-\text{d}}{\text{S}_{\text{n}-1}}$
Using $(i), (ii)$ and $(iii)$ in the given equation, we get
$\text{k}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]+\frac{\text{n}-2}{2}[2\text{a}+(\text{n}-3)\text{d}]-\text{d}}{\frac{\text{n-1}}{2}[2\text{a}+(\text{n}-2)\text{d}]}$
Taking $\frac{1}{2}$ common, we get,
$\text{k}=\frac{\text{n}[2\text{a}+(\text{n}-1)\text{d}]+(\text{n}-2)[2\text{a}+(\text{n}-3)\text{d}]-2\text{d}}{(\text{n}-1)[2\text{a}+(\text{n}-2)\text{d}]}$
$=\frac{2\text{an}+\text{n}^2\text{d}-\text{nd}+2\text{an}+\text{n}^2\text{d}-3\text{nd}-4\text{a}-2\text{nd}+6\text{d}-2\text{d}}{2\text{an}+\text{n}^2\text{d}-2\text{nd}-2\text{d}-\text{nd}+2\text{d}}$
$=\frac{2\text{n}^2\text{d}+4\text{an}-6\text{nd}-4\text{a}+4\text{d}}{\text{n}^2\text{d}+2\text{an}-3\text{nd}+-2\text{a}+2\text{d}}$
Taking $2 $ common from the numerator, we get,
$\text{k}=\frac{2(\text{n}^2\text{d}+2\text{an}-3\text{nd}+-2\text{a}+2\text{d})}{\text{n}^2\text{d}+2\text{an}-3\text{nd}+-2\text{a}+2\text{d}}$
$=2$
Therefore $, k = 2$
Hence, the correct option is $(B)$.
View full question & answer→MCQ 841 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 $\left(4 n^2+2 n\right)$.
$ 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 851 Mark
The famous mathematician associated with finding the sum of the first $100$ natural numbers is :
AnswerThe famous mathematician associated with finding the sum of the first $100$ natural numbers is Gauss.
Gauss noticed that if he was to split the numbers into two groups $(1$ to $50$ and $51$ to $100),$ he could add them together vertically to get a sum of $101$.
Gauss realized then that his final total would be $50(101) = 5050$.
View full question & answer→MCQ 861 Mark
Which term of the $A.P. \ \{121, 117, 113, ...\}$ is its first negative term?
AnswerHere $, a = 121, d = 117 - 121 = -4$
for the first negative first $a_n < 0$
$\Rightarrow 121 + (n - 1) (-4) < 0$
$\Rightarrow 121 - 4n + 4 < 0$
$\Rightarrow 125 - 4n < 0$
$\Rightarrow\text{n}>\frac{125}{4}$
$\Rightarrow\text{n}>31\frac{1}{4}$
$\therefore 32^{nd}$ term is the first negative term.
View full question & answer→MCQ 871 Mark
The number of two digit numbers divisible by $3$ is :
AnswerThe two digit numbers divisible by $3$ are
$\{12, 15, 18, ... , 99\}$
Here $, a = 12, d = 15 - 12 = 3$ and $an = 99$
$\therefore an = a + (n - 1)d$
$\Rightarrow 99 = 12 + (n - 1) \times 3$
$\Rightarrow 87 = (n - 1) \times 387 = (n - 1) \times 3$
$\Rightarrow n - 1 = 29n - 1 = 29$
$\Rightarrow n = 30$
View full question & answer→MCQ 881 Mark
The $7^{th}$ term of an $A.P$. is $-1$ and its $16^{th}$ term is $17$. The nth term of the $A.P$. is :
- ✓
$(2n - 15)$
- B
$(15 - 2n)$
- C
$(3n + 8)$
- D
$(4n - 7)$
AnswerCorrect option: A. $(2n - 15)$
We have $,T_7=-1$
$\Rightarrow a + 6d = -1 ...(i)$
$T_{16}=17$
$\Rightarrow a + 15d = 17 ...(ii)$
On solving $(i)$ and $(ii),$ we get
$a = -13$ and $d = 2$.
$\therefore T_n=a+(n-1) d$
$= -13 + (n - 1) \times 2$
$= (2n - 15)$
View full question & answer→MCQ 891 Mark
In $a_n \ \ A.P$. if $d = -4, n = 7$ and $a_n= 4,$ then $'a\ '$ is :
AnswerGiven : $d = -4, n = 7$ and $a_n= 4$
$\therefore a_n= a + (n - 1)d$
$\Rightarrow 4 = a + (7 - 1) \times (-4)$
$\Rightarrow 4 = a + 6 \times -4$
$\Rightarrow 4 = a - 24$
$\Rightarrow a = 28$
View full question & answer→MCQ 901 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 $7^{\text {th }}$ term.
View full question & answer→MCQ 911 Mark
The $7^{th}$ term from the end of the $A.P.\ \{ -11, -8, -5,..., 49\}$ is :
AnswerReversing the given $A.P.$
we have $, 49, 46, 43 ,..., -11$
Here $, a = 49, d = 46 - 49 = -3$ and $n = 7$
$\therefore an = a + (n - 1)d$
$\Rightarrow a_7= 49 + (7 - 1) \times (-3)$
$= 49 + 6 \times (-3)$
$\Rightarrow a_7= 49 - 18 = 31$
View full question & answer→MCQ 921 Mark
Mark the correct alternative in the following : The sum of first $n$ odd natural numbers is :
- A
$2n - 1$
- B
$2n + 1$
- ✓
$ n^2 $
- D
$n^2-1 $
AnswerCorrect option: C. $ n^2 $
Let, odd numbers are,
$\{1, 3, 5, .....\}$
Here,
First term $, a = 1$
and Difference $, d = 3 - 1 = 2$
We know, sum of $n$ terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(1)+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2[1+\text{n}-1]$
$\Rightarrow\ \text{S}_\text{n}=\text{n}^2$
Hence, correct choice is $(C)$.
View full question & answer→MCQ 931 Mark
Mark the correct alternative in the following : If the sum of $n$ terms of an $A.P.$ be $3n^2+ n$ and its common difference is $6,$ then its first term is :
AnswerGiven,
$ S_n=3 n^2+n $
and $ d=6$
Putting $\mathrm{n}=0$ and $1$
$S_0=3(0)^2+0=3 \times 0+0=0+0=0$
And $S_1=3(1)^2+1=3 \times 1+1=3+1=4$
We know, $a_n+S_n=S_{n-1}$
First term $\mathrm{a}_1=\mathrm{S}_1-\mathrm{S}_0$
$ \Rightarrow a_1=4-0 $
$ \Rightarrow a_1=4$
Hence, correnct choice is $(d).$
View full question & answer→MCQ 941 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 $10^{\text {th}}$ term will be $210.$
View full question & answer→MCQ 951 Mark
The nth term of the $A.P.\ \{ 63, 65, 67, 69, ...\}$ and the $A.P. \ \{3, 10, 17, 24, ...\}$ are equal, then the value of $n$ is :
AnswerFor $1^{st}\ A.P. \ a = 63$ and $d = 2, T_n= 63 + (n - 1)^2$
$\Rightarrow T_n= 2n + 61 ... (i)$
for $2^{nd }\ AP\ a = 3, d = 7,$
hence $T_n= 3 + (n - 1)^7$
$T_n= 7n - 4 ... (ii)$
by condition, from $(i) $ and $(ii),$
$7n - 4 = 2n + 61$
$\Rightarrow 5n = 65$
$\therefore n = 13$
View full question & answer→MCQ 961 Mark
The common difference of the $A.P$ whose $S_n= 3n_2+ 7n$ is :
AnswerGiven : $S_n=3 n_2+7 n$
Putting $n = 1, 2, 3$ we get
$ S=a=3 \times(1)^2+7 \times 1=3+7=10 $
$ S_2=3 \times(2)^2+7 \times 2=12+14=26 $
$ S_3=3 \times(3)^3+7 \times 3=27+21=48$
Now, $a_2={S}_2-{S}_1=26-10=16$
$\therefore$ Common difference $(d)=a_2-a=16-10=6$
View full question & answer→MCQ 971 Mark
If the angles of a right angled triangle are in $A.P$. then the angles of that triangle will be :
- A
$45^\circ , 45^\circ , 90^\circ$
- B
$20^\circ , 70^\circ , 90^\circ$
- ✓
$30^\circ , 60^\circ , 90^\circ$
- D
$40^\circ , 50^\circ , 90^\circ$
AnswerCorrect option: C. $30^\circ , 60^\circ , 90^\circ$
Let the three angles of a triangle be $a - d, a$ and $a + d$
$\therefore a - d + a + a + d = 180^\circ$
$\Rightarrow 3a = 180^\circ$
$\Rightarrow a = 60^\circ$
$\therefore$ one angle is of $60^\circ$ and other is $90^\circ\ ($given$)$.
Let the third angle be $x^\circ $, then
$60^\circ + 90^\circ + x^\circ = 180^\circ$
$\Rightarrow 150^\circ + x^\circ = 180^\circ$
$\Rightarrow x^\circ = 180^\circ - 150^\circ = 30^\circ$
$\therefore$ The angles of the right $-$ angled triangle are $30^\circ , 60^\circ , 90^\circ$
View full question & answer→MCQ 981 Mark
The $17^{\text {th }}$ term of an $AP$ exceeds its $10^{\text {th }}$ 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 991 Mark
Choose the correct answer from the given four options : In the first term of an $AP$ is $-5$ and the common difference is $2,$ then the sum of the first $6$ terms is :
AnswerGiven $, a = -5$ and $d = 2$
$\bigg[\because\text{S}_{\text{n}}=\frac{\text{n}}{2}\left\{2\text{a}+\big(\text{n}-1\big)\text{d}\right\}\bigg]$
$\text{S}_{\text{6}}= \frac{6}{2}\big[2\text{a}+\big(6-1\big)\text{d}\big] $
$ S_6=3[2(-5)+5(2)] $
$ S_6=3(-10+10) $
$ S_6=0 $
View full question & answer→MCQ 1001 Mark
If $18, a, b, -3$ are in $A.P$. then $a + b =$
Answer$18, a, b, -3$ are in $A.P.$
then $a - 18 = -3 - b$
$\Rightarrow a + b = -3 + 18$
$\Rightarrow a + b = 15$
View full question & answer→