MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 3 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 3
Find the sums given below:
7 + 10$$\displaystyle\frac{1}{2}$$ + 14 +......+84
Report Question
0%
1066$$\displaystyle\frac{1}{2}$$
0%
1096$$\displaystyle\frac{1}{2}$$
0%
1046$$\displaystyle\frac{1}{2}$$
0%
1006$$\displaystyle\frac{1}{2}$$
Explanation
In given series of AP first term a=7, common difference d= $$10\frac 12-7=\frac {21}2-7=\frac 72$$ and last term l=84
Since, $$l=a+(n-1)d$$
$$\Rightarrow 84=7+(n-1)\frac 72 $$
$$\Rightarrow 84-7=(n-1)\frac 72 $$
$$\Rightarrow 77=(n-1)\frac 72 $$
$$\Rightarrow 77\times \frac 27=n-1 $$
$$\Rightarrow 22=n-1 $$
$$\Rightarrow n=23 $$
Since, $$ S_n=\frac n2(a+l) $$
$$\Rightarrow S_n=\frac {23}2(7+84) $$
$$\Rightarrow S_n=91 \times \frac {23}2 $$
$$\Rightarrow S_n=\frac {2093}2 $$
$$\Rightarrow S_n=1046\frac 12 $$
Find the sum of
$$-37, -33, -29,..$$ to $$12$$ terms
Report Question
0%
$$-180$$
0%
$$-210$$
0%
$$-360$$
0%
$$-390$$
Explanation
The given series is $$-37, -33, -29, .........$$ to $$12$$ terms
The given sequence is an AP with first term $$a=-37$$, common difference $$d=4$$ and $$n=12$$.
$$\therefore S_{n}=\displaystyle \frac {n}2[2a+(n-1)d]$$
$$\therefore S_{12}=\displaystyle \frac {12}2[2a+(12-1)d]$$
$$=6[2\times-37+11\times4]$$
$$=-180$$
Find the number of terms in each of the following AP's
$$7, 13, 19,....205$$
Report Question
0%
$$20$$ terms
0%
$$28$$ terms
0%
$$34$$ terms
0%
$$40$$ terms
Explanation
Clearly, the given Sequence is an AP with first term $$a=7$$ and common difference $$d=13-7=6$$.
Let $$205$$ be the $$nth$$ term of the given AP.
$$\Rightarrow a_n=205$$
$$\Rightarrow a+(n-1)d=205$$
$$\Rightarrow 7+(n-1)\times6=205$$
$$\Rightarrow (n-1)\times6=198$$
$$\Rightarrow n=34$$
Hence, there are $$34$$ terms in the given AP.
For an A.P. $$a = 7, d = 3, n = 8$$, find $$a_8$$.
Report Question
0%
$$a_8 = 18$$
0%
$$a_8 = 28$$
0%
$$a_8 = 16$$
0%
$$a_8 = 36$$
Explanation
Here, $$a=7, d=3$$ and $$n=8$$.
By using $$a_n=a+(n-1)d$$
Therefore, $$a_8=a+7d$$
$$=7+7\times3$$
$$=28$$
For an A.P. $$a = -18.9, d = 2.5 \ ,a_n = 3.6,$$ find $$n$$.
Report Question
0%
$$n=12$$
0%
$$n=10$$
0%
$$n=8$$
0%
$$n=6$$
Explanation
Here, $$a=-18.9, d=2.5$$ and $$a_n=3.6$$.
Therefore, $$a_n=a+(n-1)d$$
$$-18.9+(n-1)\times2.5=3.6$$
$$\Rightarrow (n-1)\times2.5=22.5$$
$$\Rightarrow n-1=9$$
$$\therefore n=10$$
Find common difference of the following
$$8, 15, 22, 29, .....$$
Report Question
0%
$$5$$
0%
$$6$$
0%
$$7$$
0%
$$8$$
Explanation
The given sequence is $$8, 15, 22, 29..... $$.
The common difference in this sequence is the difference between the consecutive terms.
So, the common difference would be $$2^{nd}$$ term - $$1^{st}$$ term $$= 15 - 8$$ i.e. $$7$$.
Difference between the $$3^{rd}$$ term and $$2^{nd}$$ term $$= 22 - 15$$ i.e. $$7$$.
Thus common difference between the terms in this sequence is $$7$$.
Check whether $$-150$$ is a term of the AP:
$$11, 8, 5, 2,...$$
Report Question
0%
No
0%
Yes
0%
Data Insufficient
0%
Ambiguous
Explanation
Clearly, the given Sequence is an AP with first term $$a=11$$ and common difference $$d=8-11=-3$$.
Let $$-150$$ be the $$nth$$ term of the given AP.
$$\Rightarrow a_n=-150$$
$$\Rightarrow a+(n-1)d=-150$$
$$\Rightarrow 11+(n-1)\times-3=-150$$
$$\Rightarrow (n-1)\times-3=-161$$
$$\Rightarrow -3n+3=-161$$
$$\Rightarrow -3n=-164\Rightarrow n=54\displaystyle \frac23$$
Since, $$n$$ is not a natural number.
Hence, $$-150$$ is not any term of given AP.
In AP
Given a =2, d = 8, S$$_n$$ = 90, find n and a$$_n$$
Report Question
0%
$$n = 5$$ and $$a_n = 34$$
0%
$$n = 10$$ and
$$a_n = 34$$
0%
$$n = 5$$ and
$$a_n = 54$$
0%
$$n = 10$$ and
$$a_n = 44$$
Explanation
In AP Given a =2, d = 8, S$$_n$$ = 90
Since, $$S_{n}=\frac n2 [2a+(n-1)d] $$
$$\Rightarrow 90=\frac n2[2\times 2+(n-1)8] $$
$$\Rightarrow 90=\frac n2[4+8n-8] $$
$$\Rightarrow 90=\frac n2[8n-4] $$
$$\Rightarrow 180= n[8n-4] $$
$$\Rightarrow 8n^2-4n-180=0 $$
$$\Rightarrow 2n^2-n-45=0 $$
$$\Rightarrow 2n^2-10n+9n-45=0 $$
$$\Rightarrow 2n(n-5)+9(n-5)=0 $$
$$\Rightarrow (2n+9)(n-5)=0 $$
but $$ 2n+9 \neq 0$$
$$\therefore n-5=0$$
$$\Rightarrow n=5 $$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow a_5=2+(5-1)8 $$
$$\Rightarrow a_5=2+32 $$
$$\Rightarrow a_5=34 $$
Find the $$20th$$ term from the last term of the AP $$3,8, 13,....253.$$
Report Question
0%
$$156$$
0%
$$180$$
0%
$$158$$
0%
$$188$$
Explanation
The AP is $$3, 8, 13, ....,253$$
Its first term $$= 3$$ and the common difference $$= 5$$
The AP in the reverse order will have the first term $$= 253$$ and the common difference $$= -5$$
The $$20th$$ term from the end of the given AP $$=$$ The $$20th$$ term of the AP in the reverse order$$= a + 19d$$
$$= 253 + 19 \times(-5) = 253 - 95 = 158$$
Which term of the sequence $$ 3, 8, 13, 18, ........$$ is $$498$$.
Report Question
0%
$$95th$$
0%
$$100th$$
0%
$$102th$$
0%
$$101th$$
Explanation
Given series is : $$3,8,13,18,........$$
$$a = 3, l =498, d = 5, n = ? $$
$$l =a + (n - 1)d $$
$$498=3+(n-1)5$$
$$498 - 3 = (n - 1)5$$
$$\dfrac{495}{5}=n-1$$
$$\therefore n =99+1 $$
$$\therefore n = 100 $$
So $$100^{th}$$ term is $$498$$
The $$16th$$ term of the AP: $$15,$$ $$\displaystyle \frac{25}{2},10,\frac{15}{2},........$$ is
Report Question
0%
$$\displaystyle \frac{45}{2}$$
0%
$$\displaystyle -\frac{45}{2}$$
0%
$$\displaystyle \frac{105}{2}$$
0%
$$\displaystyle -\frac{105}{2}$$
Explanation
Given AP: $$15,\dfrac{25}{2},10,\dfrac{15}{2},5........$$
First term $$a = 15$$
common difference $$ \displaystyle d=\dfrac{25}{2}-15=\dfrac{25-30}{2}=-\dfrac{5}{2}$$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow a_{16}=a+15d=15+15\times -\dfrac{5}{2}$$
$$ \Rightarrow\dfrac{30-75}{2}=-\dfrac{45}{2}$$
$$\therefore $$ Option B is correct.
If $$\displaystyle \frac{6}{5},$$ $$a,4$$ are in AP, the value of $$a$$ is
Report Question
0%
$$1$$
0%
$$13$$
0%
$$\displaystyle \frac{13}{5}$$
0%
$$\displaystyle \frac{26}{5}$$
Explanation
Given, $$\dfrac{6}{5},a,4$$ are in AP
If $$a_1,\ a_2,\ a_3$$ are in AP, the common difference will be the same
$$\Rightarrow d=a_2-a_1=a_3-a_2$$
For the given terms: $$\dfrac{6}{5},\ a,\ 4$$
$$\therefore$$ $$d= a-\dfrac{6}{5} = 4-a$$
$$\Rightarrow$$ $$2a=\dfrac{6}{5}+4$$
$$\Rightarrow$$ $$2a=\dfrac{6+20}{5}$$
$$\Rightarrow$$ $$a=\dfrac{26}{10}$$
$$=\dfrac{13}{5}$$
Is $$51$$ a term of the AP, $$5, 8, 11, 14,........?$$
Report Question
0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient
Explanation
Given series is $$5,8,11,14,....$$
Let $$51$$ be the $$ n^{th} $$ term.
$$\therefore a_n=51$$
$$a = 5, d = 3$$
We know, $$a_n=a+(n-1)d$$
$$\therefore 51 = a + (n - 1)d$$
$$51 = 5 + (n - 1)3$$
$$ \dfrac{46}{3} = n - 1$$
$$ \dfrac{46+3}{3} = n$$
$$ n=\dfrac{49}{3}$$
Since $$n$$ is not a natural no.
$$51$$ is not a term of AP.
The $$31st$$ term of the AP whose first two terms are respectively $$-2$$ and $$-7$$ is
Report Question
0%
$$-152$$
0%
$$150$$
0%
$$148$$
0%
$$-148$$
Explanation
Given, $$\displaystyle a_{1}=-2 , a_{2}=-7 $$
$$ \Rightarrow d=a_{2}-a_{1}= -7-(-2) = - 7 + 2 = - 5$$
Since, $$ a_n=a+(n-1)d $$
$$\displaystyle \Rightarrow a_{31}=a+30d=-2+30(-5)=-152$$
$$\therefore $$ Option A is correct.
The $$11th$$ term of AP whose $$3rd$$ term is $$11$$ and $$8th$$ term is $$26$$ is
Report Question
0%
$$25$$
0%
$$-2$$
0%
$$-8$$
0%
$$35$$
Explanation
Given, $$ a_3=11 , a_8=26 $$
if the terms are in $$A.P$$ then $$a_n = a+(n-1)d$$
$$\displaystyle \Rightarrow a_3=a+2d=11\ \ \ \ \ .......(1)$$
$$\displaystyle \Rightarrow a_8=a+7d=26\ \ \ \ \ .......(2)$$
Subtract equation $$(2)$$ from $$(1)$$
we get,
$$\Rightarrow - 5d = -15$$
$$\therefore d = 3$$
By putting $$d=3$$ in equation $$(1)$$ we get,
$$ \Rightarrow a +2(3) = 11 $$
$$\therefore a = 5$$
$$\displaystyle \Rightarrow a_{11}=a+10d$$ (by using formula)
$$=5+10(3)$$
$$=35$$
$$\therefore $$ Option D is correct.
If $$k, 2k -1$$ and $$2k + 1$$ are three consecutive terms as an $$A.P.$$ then find the value of $$k$$.
Report Question
0%
$$2$$
0%
$$3$$
0%
$$-3$$
0%
$$5$$
0%
$$4$$
Explanation
$$k, 2k - 1 , 2k + 1 $$ are three terms in A.P. Then,
Common difference between first two terms = $$2k - 1 - k $$ =$$k - 1$$
Common difference between next two terms = $$2k + 1 - 2k + 1 $$ =$$2$$
The common difference between consecutive terms should be equal.
Hence, $$k - 1 = 2$$
$$k = 3$$
The value of
$$\dfrac{1}{97}+\dfrac{2}{97}+.....+\dfrac{96}{97}$$
is:
Report Question
0%
$$48$$
0%
$$-48$$
0%
$$1$$
0%
None of the above
Explanation
We know,
Sum of $$n$$ natural numbers $$=\dfrac{n(n+1)}{2}$$
$$\therefore \dfrac{1}{97}+\dfrac{2}{97}+.....+\dfrac{96}{97}$$
$$= \dfrac{1+2+3+....+96}{97}$$
$$= \dfrac{96\times 97}{2\times 97}$$
$$=48$$
Hence, the answer is $$48$$.
If the $$n$$th term of an A.P is $$2n+3$$, then sum of first $$n$$ term of the A.P is
Report Question
0%
$$n(n+3)$$
0%
$$n(n+4)$$
0%
$$n(n-2)$$
0%
$$n(n+1)$$
Explanation
Here, $$a_n =2n+3 $$
$$\Rightarrow a_1 = 2(1)+3= 5$$
and $$a_2= 2(2)+3= 7$$
common difference $$d=7-5=2$$
sum of first n terms $${S}_{n}=\dfrac{n}{2}[2a+(n-1)d]$$
$$S_n=\dfrac{n}{2}[2\times 5+(n-1)2]$$
$$S_n=n[ 5+n-1]$$
$$S_n=n(n+4)$$
$$\therefore $$ Option B is correct.
If an A.P is given by $$7,12,17,22$$, then $$n$$th term is
Report Question
0%
$$2n+5$$
0%
$$4n+3$$
0%
$$5n+2$$
0%
$$3n+4$$
Explanation
Given AP: $$7,12,17,22.....$$
Here first term $$a=7$$ and common difference $$d=12-7=5$$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow a_n=7+(n-1)5 $$
$$\Rightarrow a_n=5n+2 $$
$$n$$th term is $$5n+2$$
$$\therefore$$ Option C is correct.
Find the sum of all odd natural numbers less than $$200$$.
Report Question
0%
$$1000$$
0%
$$10100$$
0%
$$9900$$
0%
$$10000$$
Explanation
The odd natural numbers less than 200 are $$1,\ 3,\ 5\, .\ .\ .\ 199$$
The numbers form an AP
Where, first term $$a=1$$
and common difference $$d= 2$$
and last term $$a_n = 199$$
We know that $$n^{th}$$ term of an AP $$=a+(n-1)d$$
$$\Rightarrow\;199= 1+(n-1)\times2$$
$$\Rightarrow\;\displaystyle\frac{198}{2}=n-1$$
$$\Rightarrow\;n=99+1$$
$$\Rightarrow\;n=100$$
We know
Sum of $$n$$ terms of an AP $$ = \dfrac{n}{2} (2a+(n-1)d)$$
Let the sum of odd natural numbers less than $$200$$ be $$S_{100}$$
$$\Rightarrow\;S_{100}=\displaystyle\frac{100}{2}[1+199]$$
$$\Rightarrow\;S_{100}=50\times200$$
$$=10000$$
Which of the following will not be a term of the sequence $$11,20,29,.....$$
Report Question
0%
$$326$$
0%
$$173$$
0%
$$388$$
0%
none
Explanation
As $${a}_{n}=a+(n-1)d$$
$$a=11, d=9$$
Take each option and see if $$(n-1)$$ is integer or not
Since, $$(n-1)=\cfrac{{a}_{n}-a}{d}$$, in other way if $${a}_{n}-a$$ is divisible by $$d$$ or not.
$${a}_{n}=326, \cfrac{326-11}{9}=\cfrac{315}{9}=35$$, divisible
$${a}_{n}=173, \cfrac{173-11}{9}=\cfrac{162}{9}=18$$, divisible
$${a}_{n}=388, \cfrac{388-11}{9}=\cfrac{377}{9}$$, not divisible
Hence option (3) is correct, as $$388$$ is not a term of the sequence.
Find the 12th term of the AP. whose first term is 9 and c
ommon difference is 10.
Report Question
0%
110
0%
119
0%
128
0%
130
Explanation
The nth term in an AP is given by $$\qquad \qquad { t }_{ n }\quad =\quad a\quad +\quad (n - 1)d$$, where 'a' is the first term, 'n' the number of terms and 'd' is the common difference.
So, the 12th term would be $$\qquad \qquad { t }_{ 12 }\quad =\quad 9\quad +\quad (12 - 1)10\\ \Longrightarrow \qquad { t }_{ 12 }\quad =\quad 9\quad +\quad 110\\ \Longrightarrow \qquad { t }_{ 12 }\quad =\quad 119$$
If $$\displaystyle t_{54}$$ an A.P is -61 and $$\displaystyle t_{4}=64$$ find $$\displaystyle t_{10}$$
Report Question
0%
59
0%
49
0%
36
0%
64
Explanation
$$t_{54}=a+53d=-64...........(1)$$
$$t_4=a+3d=61.........(2)$$
Subtracting eq 1 and 2
$$50d=-125$$
$$d=-2.5$$
Substitute the value of d in eq 2
$$a=64-(3\times -2.5)$$
$$a=64+7.5=71.5$$
$$t_{10}=a+9d$$
$$\Rightarrow 71.5+(9\times -2.5)$$
$$\Rightarrow 71.5-22.5=49$$
Find the number of terms in the sequence$$:\ 4,\ 12,\ 20,\ .\ .\ .\ 108$$
Report Question
0%
$$12$$
0%
$$19$$
0%
$$13$$
0%
$$14$$
Explanation
Given sequence: $$4,\ 12,\ 20,\ .\ .\ .\ 108$$
Here first term $$a=4$$,
common difference $$d=12-4=8$$
and last term $$ a_n=108$$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow 108=4+(n-1)8 $$
$$\Rightarrow 104=8n-8 $$
$$\Rightarrow 112=8n $$
$$\Rightarrow n=14 $$
So, the number of terms in the given sequence is $$14$$
The sum of the first $$40$$ natural numbers is
Report Question
0%
$$210$$
0%
$$820$$
0%
$$610$$
0%
$$710$$
Explanation
Sum of '$$n$$' natural numbers $$=\dfrac { n(n+ 1) }{ 2 } $$
$$\therefore$$ sum of $$40$$ natural numbers $$= \dfrac {40(40+ 1) }{ 2 } = 820$$.
hence, option B is correct.
If $$ 5, k, 11$$ are in $$AP,$$ then the value of $$k$$ is
Report Question
0%
$$6$$
0%
$$8$$
0%
$$7$$
0%
$$9$$
Explanation
If $$5, k$$ and $$11$$ are in AP, then
$$k - 5 = 11 - k$$
$$\therefore 2k= 16$$
$$\therefore k= 8$$
30th term of the A.P. : 10, 7, 4,.........., is :
Report Question
0%
97
0%
77
0%
-77
0%
-87
Explanation
nth term of an AP is given by $$\qquad \qquad \\ { t }_{ n }\quad =\quad a\quad +\quad (n\quad -\quad 1)d$$
In the given sequence, 'a' = 10, 'd' = -3, 'n' = 30.
So, 30th term = $$\qquad \qquad \\ \qquad { t }_{ 30 }\quad =\quad 10\quad +\quad (30\quad -\quad 1)(-3)\\ \Longrightarrow { t }_{ 30 }\quad =\quad 10\quad -\quad 87\quad =\quad -77$$.
The sum of the first $$22$$ terms of the A.P. $$8, 3, -2, ..........$$ is
Report Question
0%
$$-875$$
0%
$$-979$$
0%
$$-717$$
0%
$$1072$$
Explanation
Given, $$\displaystyle 8, 3, -2,.........$$ are in A.P.
Thus, first term and common difference of the A.P. is $$8$$ and $$-5$$ respectively.
$$\displaystyle a=8,d=3-8=-5,n=22$$
$$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} $$
$$\displaystyle =\frac { 22 }{ 2 } \left\{ 16+\left( 21 \right) \left( -5 \right) \right\} $$
$$\displaystyle =11\left\{ 16-105 \right\} $$
$$=-979$$
If $$\displaystyle { a }_{ n }=2n$$ of an A.P., then common difference is :
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Given, $$\displaystyle { a }_{ n }=2n$$
If, $$n=1$$, $$a_1=2$$,
$$n=2$$, $$a_2=4$$
$$n=3$$, $$a_3=6$$
$$\therefore \displaystyle { a }_{ 1 }=2,{ a }_{ 2 }=4,{ a }_{ 3 }=6$$
$$\displaystyle \therefore $$ Series is $$2, 4 ,6, .....$$
$$\therefore$$ common difference, $$\displaystyle $$ $$\displaystyle d=4-2=2$$
If 7 th term of AP is 34 and 13th term is 64 then its 18th term is :
Report Question
0%
87
0%
88
0%
89
0%
90
Explanation
Given, 7th term = 34 and 13th term = 64.
$$a\quad +\quad 6d\quad =\quad 34\\ a\quad +\quad 12d\quad =\quad 64$$.
From the above two equations, we can deduce, 'd' = 5.
$$\qquad -6d\quad =\quad -30\\ \Longrightarrow \quad d\quad =\quad 5$$. [subtracting both equations]
Substituting 'd' = 5 in either equation, will give 'a'.
$$\qquad \qquad a\quad +\quad 6d\quad =\quad 34\\ \Longrightarrow \qquad a\quad +\quad 6(5)\quad =\quad 34\\ \Longrightarrow \qquad a\quad +\quad 30\quad =\quad 34\quad or\quad a\quad =\quad 4$$.
First term 'a' = 4.
$$\qquad \qquad { t }_{ 18 }\quad =\quad a\quad +\quad 17d\\ \Longrightarrow \qquad { t }_{ 18 }\quad =\quad 4\quad +\quad 17(5)\\ \Longrightarrow \qquad { t }_{ 18 }\quad =\quad 89$$.
18th term of the sequence = 89.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 10 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
