MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 3

Find the sums given below:
7 + 10

$\displaystyle\frac{1}{2}$ + 14 +......+84

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

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

$7, 13, 19,....205$

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

$a_8$ .

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

$n$ .

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

$8, 15, 22, 29, .....$

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:

$11, 8, 5, 2,...$

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

$_n$

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

$3,8, 13,....253.$

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

$498$ .

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

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.

$\displaystyle \frac{6}{5},$

$a,4$ are in AP, the value of

$a$ is

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:

$\dfrac{6}{5},\ a,\ 4$

$\therefore$

$d= a-\dfrac{6}{5} = 4-a$

$\Rightarrow$

$2a=\dfrac{6}{5}+4$

$\Rightarrow$

$2a=\dfrac{6+20}{5}$

$\Rightarrow$

$a=\dfrac{26}{10}$

$=\dfrac{13}{5}$

$51$ a term of the AP,

$5, 8, 11, 14,........?$

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

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

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.

$k, 2k -1$ and

$2k + 1$ are three consecutive terms as an

$A.P.$ then find the value of

$k$ .

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:

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

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

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

$200$ .

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

$11,20,29,.....$

0%
$326$
0%
$173$
0%
$388$
0%
none

Explanation

${a}_{n}=a+(n-1)d$

$a=11, d=9$
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 common difference is 10.

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

$\qquad \qquad { t }_{ 12 }\quad =\quad 9\quad +\quad (12 - 1)10\\ \Longrightarrow \qquad { t }_{ 12 }\quad =\quad 9\quad +\quad 110\\ \Longrightarrow \qquad { t }_{ 12 }\quad =\quad 119$

$\displaystyle t_{54}$ an A.P is -61 and

$\displaystyle t_{4}=64$ find

$\displaystyle t_{10}$

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

$:\ 4,\ 12,\ 20,\ .\ .\ .\ 108$

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

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.

$5, k, 11$ are in

$AP,$ then the value of

$k$ is

0%
$6$
0%
$8$
0%
$7$
0%
$9$

Explanation

$5, k$ 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 :

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

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.

$\displaystyle a=8,d=3-8=-5,n=22$

$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\}$

$\displaystyle =\frac { 22 }{ 2 } \left\{ 16+\left( 21 \right) \left( -5 \right) \right\}$

$\displaystyle =11\left\{ 16-105 \right\}$

$=-979$

$\displaystyle { a }_{ n }=2n$ of an A.P., then common difference is :

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 :

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.

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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