MCQ 1011 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : The sum of the series with the nth term. $t_n= (9 - 5n)$ is $(465),$ when no. of terms $n = 15$.
Reason : Given series is in $A.P$. and sum of $n$ terms of an $A.P$. is $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n-1})\text{d]}$
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ isthe correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A).$
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true
Assertion $(A)$ is false but reason $(R)$ is true
View full question & answer→MCQ 1021 Mark
What is the probability of a sure event ?
- A
$0$
- B
$\frac{1}{2}$
- ✓
$1$
- D
Less than $1$
AnswerThe probability of a sure event is always $1.$
View full question & answer→MCQ 1031 Mark
The $A.P$. whose third term is $16$ and the difference of $5^{th}$ term from the $7^{th}$ term is $12,$ then the $A.P$. is :
- A
$\{4, 6, 8, 10,\}$
- B
$\{4, 14, 24, 34,\}$
- ✓
$\{4, 10, 16, 22,\}$
- D
$\{4, 11, 18, 25,\}$
AnswerCorrect option: C. $\{4, 10, 16, 22,\}$
Given : $a_3= 16$
$\Rightarrow a + 2d = 16 ... (i)$
And $a_7- a5 = 12$
$\Rightarrow a + 6d - a - 4d = 12$
$\Rightarrow 2d = 12$
$\Rightarrow d = 6$
Putting the value of d in eq. $(i),$ we get
$a + 2 \times 6 = 16$
$\Rightarrow a = 4$
$\therefore A.P$. is $\{4, 10, 16, 22,\}$
View full question & answer→MCQ 1041 Mark
Mark the correct alternative in the following : If the sum of $n$ terms of an $A.P.$ is $2n^2+ 5n,$ then its $n^{th}$ term is :
- A
$4n - 3$
- B
$3n - 4$
- ✓
$4n + 3$
- D
$3n + 4$
AnswerCorrect option: C. $4n + 3$
Given,
$S_n=2 n^2-5 n$
We know, $a_n=S_n-S_{n-1}$
$ S_n=2 n^2+5 n $
$ S_{n-1}=2(n-1)^2+5(n-1) $
$ \Rightarrow S_{n-1}=2\left(n^2+1-2 n\right)+5 n-5 $
$ {\left[\because(a-b)^2=a^2+b^2-2 a b\right]} $
$ \Rightarrow S_{n-1}=2 n^2+2-4 n+5 n-5 $
$ \Rightarrow S_{n-1}=2 n^2+n-3$
$a_n=2 n^2+5 n-\left(2 n^2+n-3\right)$
$ \Rightarrow a_n=2 n^2+5 n-2 n^2-n+3 $
$ \Rightarrow a_n=4 n+3$
Hence, correct choice is $(C).$
View full question & answer→MCQ 1051 Mark
If $9$ times the $9^{th}$ term of an $A.P$. is equal to $11$ times the $11^{th}$ term, then its $20^{th}$ term is :
AnswerAccording to question,
$9 \times a_9=11 \times a_{11}$
$\Rightarrow 9[a + (9 - 1)d] = 11[a + (11 - 1)d]$
$\Rightarrow 9[a + 8d] = 11[a + 10d]$
$\Rightarrow 9a + 72d = 11a + 110d$
$\Rightarrow 2a + 38d = 0$
$\Rightarrow a + 19d = 0$
$\Rightarrow a + (20 - 1) d = 0$
$\Rightarrow a_{20}= 0$
View full question & answer→MCQ 1061 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 1071 Mark
In an $A.P.$ if $a = 4, n = 7$ and $a_n= 4,$ then the value of $d$ is :
AnswerGiven : $a = 4, n = 7 $ and $an = 4,$ then
$a_n= a + (n - 1)d$
$\Rightarrow 4 = 4 + (7 - 1)d$
$\Rightarrow 4 - 4 = 6d$
$\Rightarrow 6d = 0$
$\Rightarrow d = 0$
View full question & answer→MCQ 1081 Mark
Mark the correct alternative in the following : The $9^{th}$ term of an $A.P$. is $449$ and $449^{th}$ term is $9$. The term which is equal to zero is :
- A
$501^{th}$
- B
$502^{th}$
- C
$508^{th}$
- ✓
AnswerIn the given problem, let us take the first term as a and the common difference as $d$.
Here, we are given that,
$a_9=449 .....(i)$
$a_{449}=9 .....(ii)$
We need to find $n$
Also, we know,
$a_n=a+(n-1) d$
For the $9^{th}$ term $(n = 9),$
$a_9=a+(9-1) d$
$449 = a + 8d \ ($Using $i)$
$a = 449 - 8d .....(iii)$
Similarly, for the $449^{\text {th }}$ term $(n = 449),$
$a449 = a + (449 - 1)d$
$9 = a + 448d \ ($Using $ii)$
$a = 9 - 448d .....(iv)$
Subtracting $(3)$ from $(4),$ we get,
$a - a = (9 - 448d) - (449 - 8d)$
$0 = 9 - 448d - 449 + 8d$
$0 = -440 - 440d$
$440d = -440$
$d = -1$
Now, to find $a,$ we substitute the value of $d$ in $(3)$
$a = 449 - 8(-1)$
$a = 449 + 8$
$a = 457$
So, for the given $A.P \ \ d = -1$ and $a = 457$
So, let us take the term equal to zero as the $n^{th}$ term.
So $, a_n= 457 + (n - 1)(-1)$
$0 = 457 - n + 1$
$n = 458$
So $, n = 458$
Therefore, the correct option is $(D).$
View full question & answer→MCQ 1091 Mark
Choose the correct answer from the given four options : Which term of the $AP : \{21, 42, 63, 84,........\}$ is $210?$
- A
$9^{\text {th }}$
- ✓
$10^{\text {th }}$
- C
$11^{\text {th }}$
- D
$12^{\text {th }}$
AnswerCorrect option: B. $10^{\text {th }}$
Let nth term of the given $AP$ be $210$
Here, first term $, a = 21$
and common difference $, d = 42 - 21 = 21$ and $a_n= 210$
$\therefore a_n= a + (n - 1)d$
$\Rightarrow 210 = 21 + (n - 1)21$
$\Rightarrow 210 = 21 + 21n -21$
$\Rightarrow 210 = 21n$
$\Rightarrow n = 10$
View full question & answer→MCQ 1101 Mark
Progressions with equal common difference are known as :
AnswerProgressions with an equal common difference are known as Arithmetic Progression.
i.e the difference between any two consecutive terms is constant throughout the series,
this constant difference is called common difference usually denoted by the letter $d$
if a is the $1^{st}$ term, $d$ is a common difference,
then the $AP$ is represented by $a, a + d, a + 2d, a + 3d ... a + (n - 1)d$
View full question & answer→MCQ 1111 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 1121 Mark
Choose the correct answer from the given four options : The sum of first $16$ terms of the $AP \ \{10, 6, 2, ......\}$ is:
- ✓
$-320$
- B
$320$
- C
$-325$
- D
$-400$
AnswerCorrect option: A. $-320$
Given : $AP$ is $\{0, 6, 2, .......\}$
Here, first term $a = 10,$ common difference $, d = -4$
$\bigg[\therefore\text{S}_{\text{n}}= \frac{\text{n}}{2}\left\{2\text{a}+\big(\text{n}-1\big)\text{d}\right\}\bigg] $
$\text{S}_{16}= \frac{\text{16}}{2}\big[2\text{a}+\big(\text{16}-1\big)\text{d}\bigg] $
$ S_{16}=8[2 \times 10+15(-4)] $
$ S_{16}=8(20-60)=8(-40) $
$ S_{16}=-320 $
View full question & answer→MCQ 1131 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 $d = 7 - 3 = 4$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow406=\frac{\text{n}}{2}\big[2(3)+(\text{n}-1)(4)\big]$
$\Rightarrow812=\text{n}[6+4\text{n}-4]$
$\Rightarrow812=\text{n}[2+4\text{n}]$
$\Rightarrow812=2\text{n}+4\text{n}^2$
$\Rightarrow4\text{n}^4+2\text{n}-812=0$
$\Rightarrow2\text{n}^2+\text{n}-406=0$
$\Rightarrow2\text{n}^2-28\text{n}+29-406=0$
$\Rightarrow2\text{n}(\text{n}-14)+29(\text{n}-14)=0$
$\Rightarrow(\text{n}-14)(2\text{n}+29)=0$
$\Rightarrow\text{n}=14$ 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 1141 Mark
Mark the correct alternative in the following : If $S_1$ is the sum of an arithnetic progression of $'n\ '$ odd number of terms and $S_2$ the sum of the terms of the series in odd places, then $\frac{\text{S}_1}{\text{S}_2}=$
- ✓
$\frac{2\text{n}}{\text{n}+1}$
- B
$\frac{\text{n}}{\text{n}+1}$
- C
$\frac{\text{n}+1}{2\text{n}}$
- D
$\frac{\text{n}+1}{\text{n}}$
AnswerCorrect option: A. $\frac{2\text{n}}{\text{n}+1}$
Given,
$S_1$ is the sum of an $A.P$. of n odd number of terms,
$S_2$ is the sum of the terms if the series in odd plaves,
Let, $A.P$. is $a_2, a_2, a_3,....., a_n$
Here,
$n$ is odd
$d$ is difference
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}$
then, $\text{S}_1=\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}]$
Now, $S^2$ be the sum if the terms of the series in odd places.
Number of terms, $\frac{\text{n}+1}{2}$
Difference $, d = 2d$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_2=\frac{\frac{\text{n}+1}{2}}{2}\bigg[2\text{a}_1+\Big(\frac{\text{n}+1}{2}-1\Big)2\text{d}\bigg]$
$\Rightarrow\ \text{S}_2=\frac{\text{n}+1}{2}\Big[2\text{a}_1+\frac{\text{n}-1}{2}\times2\text{d}\Big]$
$\Rightarrow\ \text{S}_2=\frac{\text{n}+1}{4}[2\text{a}_1+(\text{n}-1)\text{d}]$
Now, we have to find $\frac{\text{S}_1}{\text{S}_2}$,
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}+1}{4}[2\text{a}_1+(\text{n}-1)\text{d}]}$
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{4\text{n}}{2(\text{n}+1)}$
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{2\text{n}}{(\text{n}+1)}$
Hence, correct chice is $(A)$.
View full question & answer→MCQ 1151 Mark
If the angles of a triangle are in $A.P$. and the greatest angle is twice the least then one of its angles is :
- ✓
$60^\circ$
- B
$45^\circ$
- C
$70^\circ$
- D
$50^\circ$
AnswerCorrect option: A. $60^\circ$
Let $a - d, a, a + d$ be the three angles of the triangle that form $A.P.$
Given : that the greatest angle is twice the least.
Now $, 2(a - d) = a + d$
$2a - 2d = a + d$
$a = 3d ... (i)$
Now by angle sum property,
($a - d) + a + (a + d) = 180^\circ $
$3a = 180^\circ $
$a = 60^\circ ... (ii)$
from $(i)$ and $(ii),$
$3d = 60^\circ $
$d = 20^\circ $
Now, the angles are,
$a - d = 60^\circ - 20^\circ = 40^\circ $
$a = 60^\circ $
$a + d = 60^\circ + 20^\circ = 80^\circ $
View full question & answer→MCQ 1161 Mark
If the sum of three consecutive terms of an increasing $A.P$. is $51$ and the product of the first and third of these terms is $273,$ then the third term is:
AnswerLet three consecutive terms of an increasing $A.P$. be $(a - d), d, (a + d)$ where a is the first term and $d$ be the common difference.
$\therefore a - d + a + a + d = 51$
$\Rightarrow 3a + 51$
$\Rightarrow \text{a}= \frac{51}{3} = 17$
and product of the first and third terms
$= (a - d) (a + d) = 273$
$ \Rightarrow a^2-d^2=273 $
$ \Rightarrow(17)^2-d^2=273 $
$ \Rightarrow 289-d^2=273 $
$\Rightarrow \text{d}^2 = 289 - 273 = 16 = (\underline{+}4)^2$
$\therefore\text{d} = \underline{+} 4$
$\therefore$ The $A.P$. is increasing
$\therefore d = 4$
Now third term $= a + d$
$= 17 + 4 = 21$
View full question & answer→MCQ 1171 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 1181 Mark
Mark the correct alternative in the following : In an $AP. \ S_p=q, S_q=p$ and $S_r$ denotes the sum of first $r$ terms. Then, $S_{p+q}$ is equal to :
- A
$0$
- ✓
$-(p + q)$
- C
$p + q$
- D
$pq.$
AnswerCorrect option: B. $-(p + q)$
Given,
$S_p= q$
and $S_q= p$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
then, $\text{S}_\text{p}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$\Rightarrow\ \text{q}=\frac{\text{P}}{2}[2\text{a}+\text{pd}-\text{d}]$
$\Rightarrow\ 2\text{q}=2\text{ap}+\text{p}^2\text{d}-\text{pd}\ .....\text{(i)}$
Again, $\text{Sq}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$\Rightarrow\ \text{p}=\frac{\text{q}}{2}[2\text{a}+\text{qd}-\text{d}]$
$\Rightarrow\ 2\text{p}=2\text{aq}+\text{q}^2\text{d}-\text{qd}\ .....\text{(ii)}$
By subtracting eq. $(i)$ from eq. $(ii)$
$ \Rightarrow 2 p-2 q=2 a q+q^2 d-q d-\left(2 a p+p^2 d-p d\right) $
$ \Rightarrow 2(p-q)=2 a q+q^2 d-q d-2 a p-p^2 d+p d $
$\Rightarrow 2(p-q)=2 a q-2 a p+q^2 d-p^2 d-q d+p d $
$ \Rightarrow-2(q-p)=2 a(q-p)+d\left(q^2-p^2\right)-d(q-p) $
$\Rightarrow -2(q - p) = 2a(q - p) + d(q + p)(q - p) -d(q - p)$
$\Rightarrow -2(q - p) = (q - p)(2a + d(q + p)-d)$
$\Rightarrow\ \frac{-2(\text{q}-\text{p})}{\text{q}-\text{p}}=2\text{a}+\text{dq}+\text{dp}-\text{d}$
$\Rightarrow -2 = 2a + (q + p - 1)d$
Now, we have to find sum of $p + q$
$\Rightarrow\ \text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$\Rightarrow\ \text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}\times-2\ [\because $From eq. $(iii)]$
$\Rightarrow\ \text{S}_{\text{p}+\text{q}}=-(\text{p}+\text{q})$
Hence, correct choice is $(B).$
View full question & answer→MCQ 1191 Mark
Mark the correct alternative in the following: If $18^{\text {th }}$ and $11^{\text {th }}$term of an $A.P.$ are in the ratio $3 : 2,$ then its $21^{\text {st}}$and $5^{\text {th }}$terms are in the ratio :
- A
$3 : 2$
- ✓
$3 : 1$
- C
$1 : 3$
- D
$2 : 3$
AnswerCorrect option: B. $3 : 1$
$18^{th}$ term $: 11^{th}$ term $= 3 : 2$
$\Rightarrow\ \frac{\text{a}_{18}}{\text{a}_{11}}=\frac{3}{2}$
$\Rightarrow\ \frac{\text{a}+17\text{d}}{\text{a}+10\text{d}}=\frac{3}{2}$
$\Rightarrow\ 2\text{a}+34\text{d}=3\text{a}+30\text{d}$
$\Rightarrow\ 34\text{d}-30\text{d}=3\text{a}-2\text{a}$
$\Rightarrow\ \text{a}=4\text{d}$
Now $\frac{\text{a}_{21}}{\text{a}_5}=\frac{\text{a}+20\text{d}}{\text{a}+4\text{d}}$
$=\frac{4\text{d}+20\text{d}}{4\text{d}+4\text{d}}$
$=\frac{24\text{d}}{8\text{d}}=\frac{3}{1}$
$\text{a}_{21}:\text{a}_5=3:1$
View full question & answer→MCQ 1201 Mark
Mark the correct alternative in the following : If the $n^{th}$ term of an $A.P.$ is $2n + 1,$ then the sum of first $n$ terms of the $A.P.$ is :
- A
$n(n - 2)$
- ✓
$n(n + 2)$
- C
$n(n + 1)$
- D
$n(n - 1)$
AnswerCorrect option: B. $n(n + 2)$
Given,
$a_n= 2n + 1 = l$
Putting $n = 1, 2, 3, .....$
$a_1= 2(1) + 1 = 2 + 1 = 3,$
$a_2= 2(2) + 1 = 4 +1 = 5,$
and $a_3= 2(3) = 1 = 6 + 1 = 7$
Now, $A.P.$ is $3, 5, 7, .... (2n + 1)$
Here,
First term $, a = 3$
and Difference $, d = 5 - 3 = 2$
We know, sum of $n$ terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(3+2\text{n}+1)$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(2\text{n}+4)$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2(\text{n}+2)$
$\Rightarrow\ \text{S}_\text{n}=\text{n}(\text{n}+2)$
Hence, correct choice is $(B).$
View full question & answer→MCQ 1211 Mark
Mark the correct alternative in the following : If $S_n$ denote the sum of the first $n$ terms of an $A.P.$ If $S_{2n}= 3S_n,$ then $S_{3n}: S_n$ is equal to :
AnswerHere, we are given an $A.P.$
whose sum of $n$ terms is $S_n$ and $S_{2n}= 3S_n$.
We need to find $\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}$.
Here we use the following formula for the sum of $n$ terms of an $A.P.$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; $a =$ first term for the given $A.P.$
$d =$ common difference of the given $A.P.$
$n = $ number of terms
So, first we find $S_{3n}$
$\text{S}_{3\text{n}}=\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]$
$=\frac{3\text{n}}{2}[2\text{a}+3\text{nd}-\text{d}] ..... (\text{i})$
Similarly,
$\text{S}_{2\text{n}}=\frac{2\text{n}}{\text{2}}[2\text{a}+(2\text{n}-1)\text{d}]$
$=\frac{2\text{n}}{2}[2\text{a}+2\text{nd}-\text{d}]\ .....(\text{ii})$
Also,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2\text{a}+\text{nd}-\text{d}]\ .....(\text{iii})$
Now, $S_{2n}= 3S_{n}$
So, using $(ii)$ and $(iii),$ we get,
$\frac{2\text{n}}{2}(2\text{a}+2\text{nd}-\text{d})=3\Big[\frac{\text{n}}{2}(2\text{a}+\text{nd}-\text{d})\Big]$
$\frac{2\text{n}}{2}(2\text{a}+2\text{nd}-\text{d})=\frac{3\text{n}}{2}(2\text{a}+\text{nd}-\text{d})$
On further solving, we get,
$2(2\text{a}+2\text{nd}-\text{d})=3(2\text{a}+\text{nd}-\text{d})$
$4\text{a}+4\text{nd}-2\text{d}=6\text{a}+3\text{nd}-3\text{d}$
$2\text{a}=\text{nd}+\text{d}\ .....\text{iv}$
So,
$\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=\frac{\frac{3\text{n}}{2}[2\text{a}+3\text{nd}-\text{d}]}{\frac{\text{n}}{(2)}[2\text{a}+\text{nd}-\text{d}]}$
Taking $\frac{\text{n}}{2}$ common, we get,
$\frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=\frac{3(2\text{a}+3\text{nd}-\text{d})}{(2\text{a}+\text{nd}-\text{d})}$
$=\frac{3(\text{nd}+\text{d}+3\text{nd}-\text{d})}{(\text{nd}+\text{d}+\text{nd}-\text{d})}$
$=\frac{3(4\text{nd})}{2\text{nd}}$
$=6$
$\therefore\ \frac{\text{S}_{3\text{n}}}{\text{S}_\text{n}}=6$
Hence, the correct option is $(b).$
View full question & answer→MCQ 1221 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 1231 Mark
$S_n- S_n- 1 =4$
Answerformula is $a(n) = S(n) - S(n - 1)$
In general $, S(n) = a(1) + a(2) + ... a(n-1) + a(n)$
In general $S(n-1) = a(1) + a(2) + ... a (n-1)$
If you subtract one from the other, you are left with $a(n)$
because all of the $A's$ from $a(0)$ to $a(n - 1)$ cancel out.
$a(1) - a(1) = 0$
$a(2) - a(2) = 0$
$a(n-1) - a(n-1) = 0$
$S(n) - S(n - 1) = a(n)$
View full question & answer→MCQ 1241 Mark
Mark the correct alternative in the following : The first and last terms of an $A.P$. are $1$ and $11$. If the sum of its terms is $36,$ then the number of terms will be :
AnswerGiven,
First term $, a = 1$
Last term, $a_n= 11 = l$
and Sum of $11$ term, $S_{11}= 36$
We know, Sum of $n$ terms,
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\Rightarrow\ 36=\frac{\text{n}}{2}(1+11)$
$\Rightarrow\ 36\times2=\text{n}\times12$
$\Rightarrow\ \text{n}=\frac{36\times2}{12}$
$\Rightarrow\ \text{n}=6$
Hence, correct choice is $(b)$.
View full question & answer→MCQ 1251 Mark
Mark the correct alternative in the following : If $\frac{1}{\text{x}+2},\frac{1}{\text{x}+3},\frac{1}{\text{x}+5}$ are in $A.P$. then $, x =$
AnswerGiven,
$\frac{1}{\text{x}+2},\frac{1}{\text{x}+3},\frac{1}{\text{x}+5}$ are in $A.P.$
Then difference between terms is equal
$\Rightarrow\ \frac{1}{\text{x}+3}-\frac{1}{\text{x}+2}=\frac{1}{\text{x}+5}-\frac{1}{\text{x}+3}$
$\Rightarrow\ \frac{(\text{x}+2)-(\text{x}+3)}{(\text{x}+3)(\text{x}+2)}=\frac{(\text{x}+3)-(\text{x}+5)}{(\text{x}+5)(\text{x}+3)}$
$\Rightarrow\ \frac{\text{x}+2-\text{x}-3}{\text{x}+2}=\frac{\text{x}+3-\text{x}-5}{\text{x}+5}$
$\Rightarrow\ \frac{-1}{\text{x}+2}=\frac{-2}{\text{x}+5}$
$\Rightarrow\ -1(\text{x}+5)=-2(\text{x}+2)$
$\Rightarrow\ -\text{x}-5=-2\text{x}-4$
$\Rightarrow\ -\text{x}+2\text{x}=-4+5$
$\Rightarrow\ \text{x}=1$
Hence, carrect chice is $(C).$
View full question & answer→MCQ 1261 Mark
Choose the correct answer from the given four options : In an $AP,$ if $d = –4, n = 7, a_n= 4, $ then $a$ is :
AnswerIn an $ AP, \ a_n= a + (n - 1)d$
$\Rightarrow 4 = a + (7 - 1)(- 4)\ ($by given conditions$)$
$\Rightarrow 4 = a + 6(-4)$
$\Rightarrow 4 + 24 = a$
$\Rightarrow a = 28$
View full question & answer→MCQ 1271 Mark
If $k, 2k - 1$ and $2k + 1$ are three consecutive terms of an $A.P$. the value of $k$ is :
Answer$(2k - 1) - k = (2k + 1) - (2k - 1)$
$k - 1 - = 2$
$\Rightarrow k = 3$
View full question & answer→MCQ 1281 Mark
What is $20^{\text {th }}$ 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 $20^{\text {th}}$ term will be $158 .$
View full question & answer→MCQ 1291 Mark
In an $A.P$. if $S_n= 3n^2+ 2n,$ then the value of $d$ is :
AnswerGiven : $ S_n=3 n^2+2 n $
$ S_1=3(1)^2+2 \times 1=3+2=5 $
$\Rightarrow a=5 $
$ S_2=3(2)^2+2 \times 2=3 \times 4+4$
$=12+4=16 $
$ \Rightarrow a_1+a_2=16 $
$ \Rightarrow a_2=11 $
$ \therefore d=a_2-a_1=11-5=6 $
View full question & answer→MCQ 1301 Mark
The sum of first $n$ terms of an $A.P$. is $(4n^2+ 2n).$ The nth term of this $A.P$. is :
- ✓
$(8n - 2)$
- B
$(6n - 2)$
- C
$(7n - 3)$
- D
$(8n + 2)$
AnswerCorrect option: A. $(8n - 2)$
$T_n=(S_n-S_n-1)$
$=\left(4 n^2+2 n\right)-\left\{4(n-1)^2+2(n-1)\right\}$
$=\left(4 n^2+2 n\right)-\left(4 n^2-6 n+2\right)$
$= (8n - 2)$
View full question & answer→MCQ 1311 Mark
Mark the correct alternative in the following : The sum of $n$ terms of an $A.P$. is $3n^2+ 5n,$ then $164$ is its :
- A
$24^{\text {th }}$ term
- ✓
$27^{\text {th }}$ term
- C
$26^{\text {th }}$ term
- D
$25^{\text {th }}$ term
AnswerCorrect option: B. $27^{\text {th }}$ term
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 $,64$ is the $27^{th}$ term of the given $A.P.$
Hence the correct option is $(B).$
View full question & answer→MCQ 1321 Mark
If the common difference of an $A.P$. is $5,$ then the value of $a_{20}- a_{13}$ is :
AnswerGiven : $a_{20}-a_{13} $ and $d=5 $
$ \Rightarrow a_{20}-a_{13}=a+(20-1) d-[a+(13-1) d] $
$ =a+(20-1) \times 5-[a+(13-1) \times 5] $
$ \Rightarrow a_{20}-a_{13}=a+95-[a+60] $
$= a + 95 - a - 60 = 35$
View full question & answer→MCQ 1331 Mark
In an $A.P.$ if $a = 3.5, d = 0$ and $n = 101,$ then $a_1=$
AnswerGiven : $a = 3.5, d = 0$ and $n = 101,$
then $a_n= a + (n - 1)d$
$= 3.5 + (101 - 1) \times 0 $
$= 3.5 + 0 = 3.5$
View full question & answer→MCQ 1341 Mark
The 5th term of an $A.P$. is $20$ and the sum of its $7^{th}$ and $11^{th}$ terms is $64$. The common difference of the $A.P$. is :
Answer$T_5= 20$
$ \Rightarrow a + 4d = 20 ...(i)$
$(T_7+ T_{11}) = 64$
$\Rightarrow (a + 6d) + (a + 10d) = 64$
$\Rightarrow a + 8d = 32 ...(ii)$
Subtracting $(i)$ from $(ii),$ we get
$4d = 12$
$\Rightarrow d = 3$
View full question & answer→MCQ 1351 Mark
Mark the correct alternative in the following : The sum of n terms of two $A.P.'s$ are in the ratio $5n + 9 : 9n + 6.$ Then, the ratio of their $18^{\text {th }}$ term is
- A
$\frac{179}{321}$
- B
$\frac{178}{321}$
- C
$\frac{175}{321}$
- ✓
$\frac{184}{321}$
AnswerCorrect option: D. $\frac{184}{321}$
Given,
$\frac{\text{Sum of A.P.}_1}{\text{Sum of A.P.}_2}=\frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{5\text{n}+9}{9\text{n}+6}\ .....{\text{(i)}}$
We know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P._1$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P._2$
$\text{S}'_\text{n}=\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}]$
Put the value in Eq. $(i)$
$\Rightarrow\ \frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{5\text{n}+9}{9\text{n}+6}=\frac{[2\text{a}+(\text{n}-1)\text{d}]}{[2\text{a}'+(\text{n}-1)\text{d}']}$
Now, put the $n = 2n - 1$
$\Rightarrow\ \frac{5(2\text{n}-1)+9}{9(2\text{n}-1)+6}=\frac{[2\text{a}+(2\text{n}-1-1)\text{d}]}{[2\text{a}'+(2\text{n}-1-1)\text{d}']}$
$\Rightarrow\ \frac{10\text{n}-5+9}{18\text{n}-9+6}=\frac{[2\text{a}+(2\text{n}-2)\text{d}]}{[2\text{a}'+(2\text{n}-2)\text{d}']}$
$\Rightarrow\ \frac{10\text{n}+4}{18\text{n}-3}=\frac{2\text{a}+2(\text{n}-1)\text{d}}{2\text{a}'+2(\text{n}-1)\text{d}'}$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{2[\text{a}+(\text{n}-1)\text{d}]}{2[\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n}-1)\text{d}']}$
We kmow, $a_n= a + (n - 1)d$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{\text{a}_\text{n}}{\text{a}'_\text{n}}$
Now put $n = 18$
$\Rightarrow\ \frac{2[5(18)+2]}{3[6(18)-1]}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{2[90+2]}{3[108-1]}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{2\times92}{3\times107}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{184}{321}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
Hence, the correct option is $(D).$
View full question & answer→MCQ 1361 Mark
Mark the correct alternative in the following : If $k, 2k - 1$ and $2k + 1$ are three consecutive terms of an $A.P,$ the value of $k$ is
AnswerSince $, k, 2k - 1$ and $2k + 1$ are three consecutive terms of an $A.P$.
Then, Second term $-$ First term $=$ Third term $-$ Second term $= d\ ($common difference$)$
$\Rightarrow 2k -1 - k = 2k + 1 - (2k - 1)$
$\Rightarrow k - 1 = 2k + 1 - 2k + 1$
$\Rightarrow k - 1 = 2$
$\Rightarrow k = 2 + 1$
$\Rightarrow k = 3$
Hence, the correct option is $(B)$.
View full question & answer→MCQ 1371 Mark
If the first term of an $A.P$. is $2$ and common difference is $4,$ then the sum of its first $40$ terms is :
- A
$200$
- B
$1600$
- C
$2800$
- ✓
$3200$
AnswerCorrect option: D. $3200$
In an $A.P. a = 2$ and $d = 4, n = 40$
$\therefore\text{s}_{\text{n}} = \frac{\text{n}}{2}[\text{2a}+(\text{n - 1})\text{d}]$
$=\frac{40}{2}[2\times2+(40-1)\times4]$
$=20[4+39\times4] = 20\times(4+156)$
$= 20\times160 = 3200$
View full question & answer→MCQ 1381 Mark
The $10^{th}$ term of an $A.P. \ \{2, 7, 12, ...\}$ is :
AnswerHere $, a = 2, d = 7 - 2 = 5$ and $n = 10$
$a_n= a + (n - 1)d$
$\Rightarrow a_{10}= 2 + (10 - 1) \times 5$
$= 2 + 9 \times 5$
$\Rightarrow a_{10}= 2 + 45 = 47$
View full question & answer→MCQ 1391 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.
$a_7= 4$
$\Rightarrow a + 6d = 4$
$\Rightarrow a + 6(-4) = 4$
$\Rightarrow a = 4 + 24$
$\Rightarrow a = 28$
View full question & answer→MCQ 1401 Mark
Mark the correct alternative in the following : $\text{If }\frac{5+9+13+, ....\text{ to n terms}}{7+9+11+, .....\text{to }(\text{n}+1)\text{ term}}=\frac{17}{16},\text{then n }=$
AnswerSum of $\{5 + 9 + 13 + .....\}$ to $n$ term
$=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Here $, a = 5, d = 9 - 5 = 4$
$\therefore\ \text{Sum}=\frac{\text{n}}{2}[2\times5+(\text{n}-1)\times4]$
$=\frac{\text{n}}{2}[10+4\text{n}-4]$
$=\frac{\text{n}}{2}[6+4\text{n}=\text{n}(3+2\text{n})]$
and sum of $7 + 9 + 11 + ..... to (n + 1)$ terms
$=\frac{\text{n}+1}{2}[2\times7+(\text{n}+1-1)2]$
$=\frac{\text{n}+1}{2}[14+2\text{n}]=(\text{n}+1)(7+\text{n})$
$\therefore\ \frac{5+9+13+\ .....\text{to n terms}}{7+9+11+\ .....\text{ to }(\text{n}+1)\text{ terms}}=\frac{17}{16}$
$\Rightarrow\ \frac{\text{n}(3+2\text{n})}{(\text{n}+1)(7+\text{n})}=\frac{17}{16}$
$ \Rightarrow 16 n(3+2 n)=17(n+1)(7+n) $
$ \Rightarrow 48 n+32 n^2=17\left(n^2+8 n+7\right) $
$ \Rightarrow 48 n+32 n^2=17 n^2+136 n+119 $
$ \Rightarrow 48 n+32 n^2-17 n^2-136 n-119=0 $
$ \Rightarrow 15 n^2-88 n-119=0 $
$ \Rightarrow 15 n^2-105 n+17 n-119=0 $
$\begin{Bmatrix}\because\ 15\times(-119)=1785 \\ -1785 = 17\times(105) \\ -88 = 17 - 105 \end{Bmatrix}$
$\Rightarrow 15n(n - 7) + 17(n - 7) = 0$
$\Rightarrow (n - 7)(15n + 17) = 0$
Either $, n - 7 = 0,$ then $n = 7$ or $15n + 17 = 0,$
then $\text{n}=\frac{-17}{15}$ which is not possible being fraction.
$\therefore\ \text{n}=7$
View full question & answer→MCQ 1411 Mark
Choose the correct answer from the given four options : In an $AP,$ if $a = 3.5, d = 0, n = 101,$ then $a_n$ will be :
- A
$0$
- ✓
$3.5$
- C
$103.5$
- D
$104.5$
AnswerFor an $AP\ a_n = a + (n - 1)d $
$= 3.5 + (101 - 1) \times 0$
$[$by given conditions$]$
$\therefore = 3.5$
View full question & answer→MCQ 1421 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 1431 Mark
Choose the correct answer from the given four options : Two $\text{APs}$ have the same common difference. The first term of one of these is $–1$ and that of the other is $– 8$. Then the difference between their $4^{th}$ terms is :
AnswerLet the common diference of two $AP$ s are $d_1$ and $d_2,$ respectively.
Bycondition, $d_1= d_2= d$
Let the first term of first $AP (a_1) = -1$
and the first term of second $AP (a_2) -8$
we know that, the $n^{th}$ term of an $AP, T_1= a + (n - 1)d$
$\therefore 4^{th}$ term of first $AP, T_4= a + (4 - 1)d = -1 + 3d$
and $4^{th}$ term of second $AP, T_4= a_2+ (4 - 1)d = -8 + 3d$
Now, the difference between there $4^{th}$ term is i.e.,
$[T_4- T'_4] = (-1 + 3d) - (-8 + 3d)$
$= -1 + 3d + 8 - 3d = 7$
View full question & answer→MCQ 1441 Mark
The first three terms of an $A.P$. respectively are $3y - 1, 3y + 5$ and $5y + 1$. Then, $y$ equals :
Answer$2 (3y + 5) = 3y - 1 + 5y + 1$
$\Rightarrow 6y + 10 = 8y$
$\Rightarrow 10 = 2y$
$\Rightarrow y = 5$
View full question & answer→MCQ 1451 Mark
If $7^{th}$ and $13^{th}$ terms of an $A.P$. be $34$ and $64$ respectively, then its $18^{th}$ term is :
Answer$7^{th}$ term $(a_7) = a + 6d = 34 ..... (i)$
$13^{th}$ term $(a_{13}) = a + 12d = 64 ..... (ii)$
Subtracting $(i)$ from $(ii)$ we have,
$6d = 30$
$\Rightarrow d = 5$
$\therefore a + 12 \times 5 = 64$
$\Rightarrow a + 60 = 64$
$\Rightarrow a = 64 - 60 = 4$
$\therefore 18^{th}$ term $(a_{18}) = a + 17d $
$= 4 + 17 \times 5 = 4 + 85 = 89$
View full question & answer→MCQ 1461 Mark
Mark the correct alternative in the following : If $18, a, b, -3$ are in $A.P.,$ the $a + b =$
AnswerHere, we are given four terms which are in $A.P.,$
First term $\left(a_1\right)=$
Second term $\left(a_2\right)=$
Third term $\left(a_3\right)=$
Fourth term $\left(a_4\right)=$
So, in an $A.P$. the difference of two adjacent terms is always constant.
So, we get $,d = a_2- a_1$
$d = a - 18 .....(i)$
Also,
$d = a_4- a_3$
$d = -3 - b .....(ii)$
Now, on equating $(1)$ and $(2),$ we get,
$a - 18 = -3 - b$
$a + b = 18 - 3$
$a + b = 15$
Therefore $, a + b = 15$
Hence the correct option is $(D).$
View full question & answer→MCQ 1471 Mark
Mark the correct alternative in the following : If $Sn$ denote the sum of $n$ terms of an $A.P$. with first term a and common difference $d$ such that $\frac{\text{S}_\text{x}}{\text{Sk}_\text{x}}$ is independent of $x,$ then
- A
$d = a$
- ✓
$d = 2a$
- C
$a = 2d$
- D
$d = -a$
AnswerCorrect option: B. $d = 2a$
$Sn$ is the sum of first $n$ terms a is the first term add $d$ is the common difference
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}$
$\frac{\text{S}_\text{x}}{\text{S}_\text{kx}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}}{\frac{\text{kx}}{2}[2\text{a}+(\text{kx}-1)\text{d]}}$
$\because\ \frac{\text{S}_\text{x}}{\text{S}_\text{kx}}$ is independent of $x$
$\therefore\ \frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}}{\frac{\text{kx}}{2}[2\text{a}+(\text{kx}-1)\text{d]}}$ is independent of $x$
$\therefore\ \frac{\frac{\text{n}}{2}[2\text{a}+\text{xd}-\text{d]}}{\frac{\text{kx}}{2}[2\text{a}+\text{kdx}-\text{d]}}$
$\Rightarrow\ \frac{2\text{a}-\text{d}}{\text{k}(2\text{a}-\text{d})}$ is in dependent of $x$ if $2\text{a}-\text{d}\neq0$
If $2a - d = 0,$ then $d = 2a.$
View full question & answer→MCQ 1481 Mark
Mark the correct alternative in the following : If the sum of $P$ terms of an $A.P$. is $q$ and the sum of $q$ terms is $p,$ then the sum of $p + q$ terms will be :
- A
$0$
- B
$p - q$
- C
$p + q$
- ✓
$-(p + q)$
AnswerCorrect option: D. $-(p + q)$
In the given problem, we are given $S_p=q$ and $S_q=p$
We need to find $S_{p+q}$
Now, as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
So
$\text{S}_\text{p}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$\text{q}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$2\text{q}=2\text{ap}+\text{p}(\text{q}-1)\text{d}\ .....(\text{i})$
Similarly,
$\text{S}_\text{q}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$\text{p}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$2\text{p}=2\text{aq}+\text{q}(\text{q}-1)\text{d}\ .....\text{(ii)}$
Subtracting $(ii)$ from $(i),$ we get
$2q - 2p = 2ap + [p(p - 1)d] - 2aq - [q(q - 1)d]$
$2q - 2p = 2a(p - q) + [p(p - 1) -1(q - 1)]d$
$-2(p - q) = 2a(p - q) + [(p^2- q^2) - (p - q)]$
$-2 = 2a + (p + q - 1)d .....(iii)$
Now,
$\text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$\text{S}_{\text{p}+\text{q}}=\frac{(\text{p}+\text{d})}{2}(-2)$
$\text{S}_{\text{p}+\text{q}}= -(\text{p}+\text{d})$
Thus, $S_{p+q}= -(p + q)$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 1491 Mark
Mark the correct alternative in the following : If the first term of an $A.P$. is $2$ and common difference is $4,$ then the sum of its $40$ terms is :
- ✓
$3200$
- B
$1600$
- C
$200$
- D
$2800$
AnswerCorrect option: A. $3200$
In the given problem, we need to find the sum of $40$ terms of an arithmetic progression,
where we are given the first term and the common difference.
So, here we use the following formula for the sum of $n$ terms of an $A.P.,$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; $a =$ first term for the given $A.P.$
$d =$ common difference of the given $A.P.$
$n =$ number of terms
Given,
First term $(a) = 2$
Common difference $(d) = 4$
Number of terms $(n) = 40$
So, using the formula we get,
$\text{S}_{40}=\frac{40}{2}[2(2)+(40-1)(4)]$
$= (20)[4 + (39)(4)]$
$= (20)[4 + 156]$
$= (20)(160)$
$= 3200$
Therefore, the sum of first $40$ terms for the given $A.P$. is $S_{40}= 3200$.
So, the correct option is $(a).$
View full question & answer→MCQ 1501 Mark
Mark the correct alternative in the following : Sum of $n$ term of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ .....$ is
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
The series is given
$\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ ....$
$\Rightarrow\ \sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+\ .....$
Here $\text{a}=\sqrt{2}$ and ${d} = 2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2\sqrt{2}+(\text{n}-1)\sqrt{2}]$
$=\frac{\text{n}}{2}[2\sqrt{2}+\sqrt{2}\text{n}-\sqrt{2}]$
$=\frac{\text{n}}{2}(\sqrt{2}\text{n}+\sqrt{2})$
$=\frac{\text{n}\sqrt{2}}{2}(\text{n}+1)=\frac{\text{n}(\text{n}+1)}{\sqrt{2}}$
View full question & answer→