MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 4

The sum of 2 + 4 + 6 + 8 + .......... + 80 is :

0%
$1540$
0%
$1640$
0%
$1740$
0%
$1440$

Explanation

Given sequence is:

$2+4+6+.......+80$

Here,

$\displaystyle a=2,d=4-2=2,$

$\displaystyle { a }_{ n }=80$

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

$\displaystyle \Rightarrow 80=2+\left( n-1 \right) 2$

$\displaystyle \Rightarrow 80=2n\Rightarrow n=40$

$\Rightarrow \displaystyle { S }_{ 40 }=\frac { 40 }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\}$

$\displaystyle =20\left\{ 4+78 \right\}$

$=20\times 82$

$=1640$

The sum of

$2, 7, 12, ..............$ to

$10$ terms is :

0%
$240$
0%
$248$
0%
$245$
0%
$250$

Explanation

Given series is:

$\displaystyle 2+7+12+........$ to

$10$ terms
Here,

$\displaystyle a=2,d=7-2=5,n=10$

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

$\displaystyle \Rightarrow { S }_{ 10}=\frac { 10 }{ 2 } \left\{ 4+45 \right\}$

$=5\times 49$

$=245$

Which term in the

$\text{A.P.}~~ 5,13,21,...$ is

$181$ ?

0%
${21}^{\text{st}}$
0%
${22}^{\text{nd}}$
0%
${23}^{\text{rd}}$
0%
${24}^{\text{th}}$

Explanation

Given sequence is

$5, 13, 21, ...$

Given:

$a = 5, ~d = 8, ~n^{th}\text{ term} = 181$

To find:

$n$

$\because~~ { t }_{ n }~ = a +(n- 1)d\\ \Longrightarrow 181 =5 + (n - 1)8\\ \Longrightarrow 181 = 5+ 8n - 8\\ \Longrightarrow 181= 8n - 3\\ \Longrightarrow 8n= 184\\\implies ~n =23$ .

$\therefore$

$181$ is the

$23^{\text{rd}}$ term.

If the sum of

$n$ terms of AP. is

$476, l = 20, a = 36$ , then

$n$ is equal to :

0%
$14$
0%
$15$
0%
$16$
0%
$17$

Explanation

Given,

$\displaystyle { S }_{ n }=476,l=20,a=36,n=?$
We know

$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left[ a+l \right]$

$\Rightarrow \displaystyle 476=\frac { n }{ 2 } \left[ 36+20 \right]$

$=\dfrac { 56n }{ 2 }$

$=28n$

$\Rightarrow \therefore n=\dfrac { 476 }{ 28 } =17$

$\displaystyle a=3,n=20$ and

$\displaystyle { S }_{ n }=300$ , then

$l$ is :

0%
$30$
0%
$25$
0%
$27$
0%
$14$

Explanation

Given,

$a=3, n=20, S_n=300$

We need to find

$l$

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

$\displaystyle \therefore 300=\frac { 20 }{ 2 } \left( 3+l \right)$

$\displaystyle \therefore l=27$

If numbers

$a, b$ and

$c$ are in AP, then

0%
$\displaystyle b-a=c-b$
0%
$\displaystyle b+a=c+b$
0%
$\displaystyle a-c=b-d$
0%
None of these

Explanation

Since, given three numbers are in AP, then the common difference will be same.

$\therefore$ Common difference =

$b - a = c - b$

The

$4^{th}$ term of an AP is

$14$ and its

$12^{th}$ term is

$70$ . What is its first term?

0%
$-10$
0%
$-7$
0%
$7$
0%
$10$

Explanation

As we know for an AP with first term

$a$ and common difference

$d$ , genral term is

$a_n = a+(n-1)d$

$\therefore \displaystyle a+3d=14$ and

$\displaystyle a+ 11d=70$

On subtracting both the above equations, we get

$\displaystyle 8d=56$
or

$\displaystyle d=7$

$\displaystyle \therefore \quad a=-7$

The sum of

$n$ natural number is :

0%
$\displaystyle \frac { n\left( n+1 \right) }{ 2 }$
0%
$\displaystyle \frac { n(n-1) }{ 2 }$
0%
$\displaystyle \left( n+2 \right)$
0%
$\displaystyle \frac { n+1 }{ 2 }$

Explanation

As the first

$n$ natural numbers are

$1,2,3,4,5,6,.........n$

They form an AP with

$a=1$ and

$d=1$

The sum of AP for

$n$ numbers

$=\dfrac{n}{2}(2a+(n-1)d)$

$=\dfrac{n}{2}(2+n-1)$

$=\dfrac{n(n+1)}{2}$

The

$3^{rd}$ term of an A.P. is

$-40$ and

$13^{th}$ term is zero, then

$d$ is equal to :

0%
$-4$
0%
$4$
0%
$0$
0%
$-1$

Explanation

Given

$3^{rd}$ term

$=-40$ ,

$13^{th}$ term

$=0$

Now using,

$a_n=a+(n-1)d$

$\therefore \displaystyle { a }_{ 3 }=a+2d=-40$ ......(i)
and

$\displaystyle { a }_{ 13 }=a+12d=0$ ......(ii)
Subtracting equations (i) and (ii), we get

$\displaystyle d=4$

$p, (p - 2)$ and

$3 p$ are in AP, then the value of

$p$ is

0%
$-3$
0%
$-2$
0%
$3$
0%
$2$

Explanation

Since, given three numbers are in AP, then the common difference between two consecutive terms will be same.

$\therefore$

$\displaystyle (p-2)-p=3p-(p-2)$

$\Rightarrow -2=2p+2$

$\displaystyle p=-2$

The common difference of the

$A.P:\;\;5, 3, 1, -1,\dots$ is

0%
$-2$
0%
$2$
0%
$-1$
0%
$3$

Explanation

Common difference of an

$A.P$ is equal to difference between the consecutive terms of the sequence,

$d= 3 - 5$

$= -2$

Hence, option

$A$ is the correct answer.

$\displaystyle l=20,d=-1,n=17$ , then the first term is :

0%
$30$
0%
$32$
0%
$34$
0%
$36$

Explanation

Given

$l = 20, d = -1, n = 17$

We need to find

$a$ .

By using

$l=a+(n-1)d$

$\therefore 20=a+(17-1)(-1)$

$20=a-16$

$a=20+16$

$\therefore a=36$

$50^{th}$ term of the AP.

$2, 5, 8, 11,.....$ is :

0%
$147$
0%
$149$
0%
$151$
0%
$153$

Explanation

Given series

$\displaystyle 2,5,8,11,.....$ is in A.P.

$\therefore \displaystyle a=2,d=3$

We need to find

$50^{th}$ term.

$\therefore n=50$

$\therefore \displaystyle { a }_{ 50 }=a+49d$

$=2+49\left( 3 \right)$

$=2+147$

$=149$

Find the sum of the following APs:

$-37, -33, -29, .....$ to

$12$ terms

0%
-180
0%
180
0%
200
0%
-200

Explanation

Here,

$a=-37, d=4$ and

$n=12$
Here we use the formula as given below:

$S_n = \dfrac{n}{2}[2a+(n-1)d]$

$\therefore S_{12}=\dfrac {12}{2}(2\times -37+4\times 11)=6\times -30=-180$

Find the sum of the following APs:

$2, 7, 12, ......,$ to

$10$ terms

0%
245
0%
345
0%
276
0%
250

Explanation

$S=\dfrac {n}{2}(2a+d(n-1))$

Here,

$a=2, d=5$ and

$n=10$

$Sum=\dfrac {10}{2}(2(2)+45)$

$=5\times 49=245$

The value obtained by subtracting the

$10^{th}$ term of an AP from the

$17^{th}$ term is

$56$ . Find the common difference.

0%
$7$
0%
$16$
0%
$9$
0%
$8$

Explanation

In an Arithmetic Progression the general term of an A.P. is given by

$T_n$ :

$\boxed {T_{n} = a+(n-1)d}$

According to the statement:

$T_{17}-T_{10} = 56$

$a+(17-1)d-[a+(10-1)d] = 56$

$a+16d-[a+9d] = 56$

$a+16d-a-9d = 56$

$7a = 56$

$\boxed {d = 8}$

The common difference of the sequence

$5,8,11,14,$ is

0%
$3$
0%
$-3$
0%
$0$
0%
$1$

Explanation

Given,

$5,8,11,14$

$\therefore 8-5=3$

$11-8=3$

$14-11=3$

The common difference of the sequence

$5,8,11,14$ is

$3$

In an AP, the

$9th$ term is

$-72$ . The

$10th$ term is

$60$ less than the

$4th$ term. Find the first term of the AP

0%
$8$
0%
$-8$
0%
$-152$
0%
$10$

Explanation

Given that:

$a_9=-72$ and

$a_{10}=a_4-60$

Let

$a$ be the first term and

$d$ be the common difference then

$a+8d=-72$ ......(1)

And

$a+9d=a+3d-60$

$\Rightarrow 6d=-60\Rightarrow d=-10.$

Substituting this value in equation (1), we get

$a+8\times (-10)=-72$

$a=80-72$

$a=8$

Hence,

The first term is

$8.$

The first term of an AP is

$-50$ and the

$50th$ term is

$48$ . Find the common difference

0%
$4$
0%
$1$
0%
$-3$
0%
$2$

Explanation

Given,

First term

$a = -50$

$50^{th}$ term,

$T_{50} = 48$

$d = ?$

$\boxed {T_{n} = a+(n-1)d}$

$T_{50} = 48$

$48 = -50+(50-1)d$

$48 = -50+49d$

$49d = 98$

$\boxed {d = 2}$

Find the sum of the following APs:

$0.6, 1.7, 2.8, ......$ to

$100$ terms

0%
5475
0%
5505
0%
6589
0%
3844

Explanation

Here

$a=0.6, d=1.1$ and

$n=100$
As we know,

$S_n = \dfrac {n}{2}[2a+(n-1)d]$

$\therefore S_{100}=\dfrac {100}{2} (2(0.6)+1.1\times 99)=50\times 110.1=5505$

In an AP, the

$4th$ term is

$36$ . The

$21st$ term is

$108$ more than the

$9th$ term. Find the common difference.

0%
$12$
0%
$9$
0%
$4$
0%
$-3$

Explanation

Given:

$T_{4} = 36$

$T_{21} = 108+T_{9}$

$d = ?$

General term of an A.P.

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

$T_{21} = 108+T_{9}$

$a+(21-1)d = 108+a+(9-1)d$

$20d = 108 +8d$

$12d = 108$

$\boxed {d = 9}$

Find the common difference in the series: 0.2, 0.9, 1.6 ..............

0%
0.5
0%
0.1
0%
0.7
0%
0.6

Explanation

First term,

$a_1 = 0.2$

Second term,

$a_2 = 0.9$

Common difference,

$d = a_2 - a_1$

$d= 0.9 - 0.2$

$\therefore d = 0.7$

Find the sum of the arithmetic series

$6 + 12 + 18 + ....... 96$ .

0%
$815$
0%
$816$
0%
$817$
0%
$819$

Explanation

Given series is

$6+12+18+....96$ .

Here

$a = 6$ ; last term,

$l = 96, d = 12-6 =6$
We know that, last term,

$l = a + (n - 1) d$

$\Rightarrow 96 = 6 + (n - 1) 6$

$\Rightarrow 96 = 6 + 6n - 6$

$\Rightarrow 96 = 6n$

$\Rightarrow n = \dfrac{96}{6}$

$\Rightarrow n = 16$ = total number of terms in the series

We know that,

Sum

$(S_n) = \dfrac{n}{2}$ [First term

$+$ Last term]

$= \dfrac{16}{2} [6 + 96]$

$= 8 [102]$

Therefore,

$S_{16} = 816$

Find the common difference, if the sum of first n terms will be (n - 2) (n - 1).

0%
1
0%
2
0%
3
0%
4

Explanation

The sum of first n terms,

$S_n = \dfrac{n}{2} (2a + (n - 1)d)$
On comparing the equations, we get

$\dfrac{n}{2} (2a + (n - 1) d) = (n - 2)(n + 1)$

$(2a - d) n + dn^2 = n^2 - n - 2$
On comparing the factors of

$n^2$ , we find the value of

$d= 1$

$\therefore$ The common difference is 1.

Find the common difference in the sequence 4, 8, 12, 16, .......... 20

0%
8
0%
3
0%
4
0%
2

Explanation

First term,

$a_1 = 4$
Second term,

$a_2 = 8$
Common difference,

$d = a_2 = a_1$

$d = 8 - 4 = 4$

$\therefore$ The common difference is 4.

Find the number of terms in an A.P:

$-1,\ -5,\ -9,\ ..,\ - 197$

0%
48
0%
49
0%
50
0%
47

Explanation

Given: AP:

$-1, \ -5,\ -9,\ ...,\ -197$

Here, First term

$a = -1$

Common difference

$d = a_2 - a_1 = -5 - (-1) = - 4$

And last term

$a_n = - 197$

We know that,

$a_n = a + (n - 1) d$

$\Rightarrow -197 = -1 + (n - 1) - 4$

$\Rightarrow -197 = -1 - 4n + 4$

$\Rightarrow -197 + 1 - 4 = -4n$

$\Rightarrow n = 50$

$\therefore$ There are 50 terms in the A.P

Which term of the

$A.P. 2, 9, 16, 23 .............$ is

$100$ ?

0%
$15$
0%
$10$
0%
$11$
0%
$12$

Explanation

$a = 2$
common difference,

$d = a_2 - a_1, a - 2 = 7$
we know that,

$a_n = a + (n - 1) d$

$a_n = 100$

$100 = a + (n - 1) 7$

$100 = 2 + 7n - 7$

$100 - 2 + 7 = 7n$

$\dfrac{105}{7} = n$

$n = 15$
Hence,

$15th$ term of

$A.P$ . is

$100$ .

Find 11th term of the A.P. : -3, 1, 5, ........

0%
36
0%
37
0%
38
0%
39

Explanation

First term,

$a_1 = -3$
Common difference,

$d = a_2 - a_1 = 1 - (-3) = 4$

We know that,

$a_n = a_1 + (n - 1) d$

$a_{11} = - 3 + (11- 1) 4$

$= -3 + (10) 4$

$= - 3 + 40$

$a_{11} = 37$

_____ is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

0%
Geometric value
0%
Geometric series
0%
Arithmetic progression
0%
Arithmetic mean

Explanation

$\text {Arithmetic progression}$ is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

Ex:

$1, 2, 3, 4$ ..... In this sequence, the first number is

$1$ and the next number is

$2$ . The difference between the two numbers is

$1$ and so on.

By adding

$1$ to the preceding number, we are forming the next term and so on, and the series is in

$AP$ .

The sum of n terms of an arithmetic sequence can be calculated by .........................

0%
$S_n = \dfrac{n}{2} [2a+ (n - 1) d]$
0%
$S_n = \dfrac{n}{2} [2a+ (n + 1) d]$
0%
$S_n = \dfrac{2n}{2} [2a+ (n - 1) d]$
0%
$S_n = \dfrac{2n}{2} [2a- (n - 1) d]$

Explanation

The sum of n terms of an arithmetic sequence can be calculated by

$S_n = \dfrac{n}{2} [2a+ (n - 1) d]$
Where, n denotes the total number of terms
a = first term
d = common difference

$S_n$ = Sum of n terms

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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