MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 2

Strikers at a plant were ordered to return to work and were told they would be fined Rs.

$50$ the first day they failed to do so, Rs.

$75$ the second day, Rs.

$100$ the third day, and so on. If the strikers stayed out for

$6$ days, what was the fine for the sixth day?

0%
$200$
0%
$150$
0%
$125$
0%
$175$

Explanation

Strikers on plant fined Rs.

$50$ on first day,&

$75$ on second ,Rs

$100$ on third and so on,

difference in fine from day

$1$ to day

$2=Rs.25$

For

${ 6 }^{ th }$ day,

$n=6,a=50$ and

$d=25$ ,

$=a+(n-1)d\\ =50+(6-1)25\\ =50+(5\times 25)\\ =50+125\\ =175$

Answer is

$(D)$

Write the sum of first five terms of the following Arithmetic Progressions where, the common difference

$d$ and the first term

$a$ are given:

$a = 6, d = 6$

0%
$88$
0%
$89$
0%
$90$
0%
$91$

Explanation

Here the first term

$a=6$ and common difference

$d=6$ .
Now

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

$\therefore a_1=a=6$

$a_2=a+d=6+6=12$

$a_3=a+2d=6+2\times6=18$

$a_4=a+3d=6+3\times6=24$

$a_5=a+4d=6+4\times6=30$

$\therefore$ series in A.P is

$6,12,18,24,30,....$

Thus, the sum of first five terms

$=6+12+18+24+30=90$

Hence, the answer is

$90$ .

We can use the formula for sum of A.P

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

$\therefore S_5$ =

$\dfrac{5}{2} \times [2(6)+(5-1)(6)]=90$

Find the twenty fifth term of the A.P. : 12, 16, 20, 24, ..........

0%
108
0%
112
0%
116
0%
120

Explanation

Let

$a$ be the first term and

$d$ be the common difference of given AP.
Hence

$a=12$ and

$d=16-12=4$ .

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

$\Rightarrow t_{25}=12+24\times4$

$\Rightarrow t_{25}=12+96$

$\Rightarrow t_{25}=108$

Find the sum of first 20 natural numbers.

0%
210
0%
220
0%
230
0%
240

Explanation

We have to find the sum n terms of the AP 1,2,3,...
Here

$a=1,d=1$ .
Therefore, Required sum

$S_n=\frac n2[2a+(n-1)d]=\frac n2[2\times1+(n-1)\times1]$

$\Rightarrow S_n=\displaystyle \frac n2[2+n-1]$

$\Rightarrow S_n=\displaystyle \frac{n(n+1)}2$ ...(1)

Put

$n=20$ in (1), we get

$S_{20}=\displaystyle \frac{20(20+1)}2=210$

Find

$t_{11}$ from the given A.P. 4, 9, 14, ...

0%
54
0%
55
0%
51
0%
21

Explanation

Given series is

$4,9,14,....$

Clearly, the given sequence is an AP with first term

$a=4$ and the common difference

$d=5$ .

We know,

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

$t_{11}=a+10d$

$\therefore t_{11}=4+10\times5=54$

The sum of the first

$100$ natural numbers is

0%
$-5050$
0%
$5005$
0%
$5050$
0%
$-5005$

Explanation

We know, sum of a arithmetic progression, with first term as

$a$ and common difference

$d$ is

${S}_{n}=n[a+\frac{1}{2}(n-1)d]$

Here, given,

$a=1$ ,

$d=1$ and

$n=100$

${S}_{100}=100[1+\displaystyle\dfrac{1}{2}(100-1)1]$

$=100[1+\dfrac{99}{2}]$

$=100\times \displaystyle\frac{101}{2}$

$=5050$

Write the sum of the first five terms of the Arithmetic Progression for which the common difference

$d$ and the first term

$a$ are given below:

$a = 5, d = 2$

0%
$40$
0%
$45$
0%
$50$
0%
$55$

Explanation

Given: the first term

$a=5$ and common difference

$d=2$ .
Now

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

$\therefore a_1=a=5$

$a_2=a+d=5+2=7$

$a_3=a+2d=5+2\times2=9$

$a_4=a+3d=5+3\times2=11$

$a_5=a+4d=5+4\times2=13$

$\therefore$ The series in A.P. is

$5,7,9,11,13,.....$

Sum of the first five terms is

$=5+7+9+11+13=45$

Hence, the answer is

$45$ .

We can also use the formula for sum of A.P =>

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

$\therefore S_5$ =

$\dfrac{5}{2} \times [2(5)+(5-1)(2)]=45$

If the following sequence is an arithmetic progression, find its general term.
-5, 2, 9, 16, 23, 30, .......... is

0%
$7n-12$
0%
$6n-12$
0%
$5n-12$
0%
$4n-12$

Explanation

We can observe the difference between two consecutives terms is equal:

$2-(-5)=9-2= 16-9=23-16=7$

Hence, the sequence is an arithmetic progression with first term

$a=-5$ and common difference

$d= 7$ .
General term of an AP is

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

$t_n=-5+(n-1)\times 7$

$t_n=7n-12$

Find

$S_{10}$ , if a = 6 and d = 3.

0%
180
0%
185
0%
190
0%
195

Explanation

Given First term,

$a=6$ ,

Common difference,

$d=3$

Number of terms,

$n=10$

Sum of

$n$ terms is given as

$S_n$ =

$\displaystyle \frac n2[2a+(n-1)d]$

Substituting the values, we get

$S_{10} =\displaystyle \frac {10}2[2\times 6+(10-1)\times3]$

$=5(39)$

$=195$

How many terms of the sequence

$18, 16, 14,....$ should be taken so that their sum is zero?

0%
$19$
0%
$17$
0%
$18$
0%
$16$

Explanation

Here

$a=18,d=-2$ .
Let there are

$n$ terms so that the sum is zero.
Now,

$S_n=\dfrac n2[2a+(n-1)d]=0$

$\Rightarrow \dfrac n2[2\times18+(n-1)\times-2]=0$

$\Rightarrow 36+(n-1)\times-2=0$

$\Rightarrow n-1=18$

$\Rightarrow n=19$
Therefore

$19$ terms of the sequence has to be taken so that their sum is zero.

How many terms are there in the A.P.:

$-1,\displaystyle\frac{-5}{6},\displaystyle-\frac{2}{3},\displaystyle\frac{-1}{2}, ....., \displaystyle\frac{10}{3}$ ?

0%
$27$
0%
$18$
0%
$16$
0%
$9$

Explanation

Given A.P. is

$-1,\ \dfrac{-5}{6},\ \dfrac{-2}{3},\ \dfrac{-1}{2},\ .....,\ \dfrac{10}{3}$

Here, first term

$(a)=-1$ and the common difference

$(d)=\displaystyle \dfrac{-5}6-(-1)=\dfrac{1}6.$
Let

$n$ be the number of terms of the given A.P.

We know that the

$n_{th}$ term of an A.P. is:

$a_n=a+(n-1)d\\$
Here,

$a_n=\dfrac{10}3$

$\therefore \ a+(n-1)d=\dfrac{10}3$

$\Rightarrow -1+(n-1)\times\dfrac{1}6=\dfrac{10}3$

$\Rightarrow (n-1)\times\dfrac{1}6=\dfrac{13}3$

$\Rightarrow n-1=26$

$\Rightarrow n=27$

Therefore, there are

$27$ terms in the give A.P.

Find the sum of

$7 + 10\dfrac{1}{2} + 14+...+84$ .

0%
$\displaystyle\frac{2093}{2}$
0%
$\displaystyle\frac{1093}{2}$
0%
$\displaystyle\frac{2193}{2}$
0%
$\displaystyle\frac{3093}{2}$

Explanation

Clearly, terms of the given series form an AP with first term and last term

$a=7$ &

$l= a_n = 84$ respectively and common difference

$d=\displaystyle \frac{21}2-7=\frac72$ .

Let the given series has

$n$ terms.

Therefore,

$a_n=84=a+(n-1)d$

$84= 7+(n-1)\times\frac72$

$(n-1)\times\frac72=77$

$\therefore n=23$
Here we use the following formula to calculate the sum of terms:

$S_n =\dfrac {n}{2} (a+l)$

$\therefore$ Required sum

$=S_{23}=\displaystyle \dfrac{23}2(7+84)$

$\Rightarrow S_{23}=\displaystyle \frac{23\times91}2=\dfrac{2093}2$

The fourth term of an A.P. is

$4$ . Then the sum of the first

$7$ terms is ?

0%
$4$
0%
$28$
0%
$16$
0%
$40$

Explanation

Let a and d be the first term and common difference of A.P. Then,

$a_4=4\Rightarrow a+3d=4$ ...(1)
Now,

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

$\therefore S_7=\frac72[2a+(7-1)d]$

$\Rightarrow S_7=\frac72(2a+6d)$

$\Rightarrow S_7=7(a+3d)=7\times4=28$ (

$\because of (1)$ )
Therefore option B is correct

Write the arithmetic progression when first term

$a = -1.5$ and common difference

$d = -0.5$ .

0%
$-1.5, -2, -2.5, -3,.....$
0%
$-1.5, -1, -0.5, 0,.....$
0%
$-1.5, -2, -2.5, -5,.....$
0%
$-1.5, 2, 2.5, 3,.....$

Explanation

Here

$a=-1.5$ and

$d=-0.5$ .
The

$n^{th}$ term of the AP is

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

$\therefore a_1=a=-1.5$

$a_2=a+d\\=-1.5+(-0.5)\\=-2$

$a_3=a+2d\\=-1.5+2\times(-0.5)\\=-2.5$

$a_4=a+3d\\=-1.5+3\times(-0.5)\\=-3$ and so on.
Therefore, the required AP is

$-1.5,-2,-2.5,-3,...$

Find the nth term and 9th term of the sequence 3, 7, 11, 15.........

0%
$4n-1$ and $T_9=35$
0%
$4n+1$ and $T_9=37$
0%
$2n-1$ and $T_9=35$
0%
$4n-1$ and $T_9=39$

Explanation

We know,

$n^{th}$ term

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

$a =$ First term

$d =$ Common difference

$n =$ number of terms

$a_n = n^{th}$ term

Here,

$a=3\ and\ d=7-3\ or\ 11-7=4$

$a_n=a+(n-1)d=3+(n-1)4$

$a_n=4n-1$

For

$9$ th term put

$n=9$

$T_9=a_9=4\times9-1=35$

Find the

$18^{th}$ term of the A.P.,

$\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}$ .....

0%
$36\sqrt{2}$
0%
$35\sqrt{2}$
0%
$34\sqrt{2}$
0%
$33\sqrt{2}$

Explanation

Here

$a=\sqrt2$ and

$d=3\sqrt2-\sqrt2=2\sqrt2$ .
The nth term of the AP is

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

$\therefore a_{18}=a+17d$

$=\sqrt2+17\times2\sqrt2$

$=35\sqrt2$

$-150$ a term of the

$A.P$ :

$11, 8, 5, 2$ ?

0%
No
0%
Yes
0%
Data Insufficient
0%
Can't say

Explanation

Clearly the given sequence is an AP with first term

$a=11$ and the common difference

$d=8-11=-3$ .

Let

$-150$ be the

$n^{th}$ term of given AP. Then,

$a_n=-150\Rightarrow t_n= a+(n-1)d=-150$

$\Rightarrow 11+(n-1)\times-3=-150$

$\Rightarrow (n-1)\times-3=-161$

$\Rightarrow n-1=\dfrac{-161}{-3}$

$\Rightarrow n=\dfrac{161}{3}+1=\dfrac{164}3$
Clearly

$n$ is not an natural number, Hence,

$-150$ is not a term of given AP.

Write the first term

$a$ and the common difference

$d$ of the AP :

$-5, -1, 3, 7......$

0%
$a = -5, d = 4$
0%
$a =-5, d = -6$
0%
$a = -5, d = -4$
0%
$a = 5, d = -4$

Explanation

Given series is

$-5, -1, 3,7,......$

Clearly first term is

$a=-5$

And common difference

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

Find out whether the sequence

$1^2, 3^2, 5^2, 7^2$ ,... is an AP. If it is, find out the common difference.

0%
No
0%
Yes, $d=8$
0%
Yes, $d=-8$
0%
Yes, $d=9$

Explanation

Given series is

$1^2, 3^2, 5^2, 7^2,......$

We have,

$3^2-1^2=9-1=8$

$5^2-3^2=25-9=16$

$7^2-5^2=49-25=24$
This shows that the difference of a term and the preceding term is not always same.
Hence, the given sequence is not an AP.

Check if the series is an A.P. Find the common difference

$d$ and write three more terms of the the following series.

$2,\displaystyle\frac{5}{2}$ ,

$3,\displaystyle\frac{7}{2}, ...$

0%
$a=2$ ; $d=\dfrac{1}{2}$ ; other terms $4,$ $\dfrac{9}{2},5$
0%
$a=3$ ; $d=\dfrac{1}{2}$ ; other terms $6, 8, 10$
0%
$a=2$ ; $d=\dfrac{3}{2}$ ; other terms $7, 8, 15$
0%
$a=3$ ; $d=\dfrac{3}{2}$ ; other terms $4, 5, 15$

Explanation

It is an A.P. because

$a = 2$ ,

$d = \dfrac12$

$\because [(t_2 - t_1) = (t_3 - t_2) = (t_4 - t_3) = \dfrac12]$

$t_5 = \dfrac{7}{2} + \dfrac{1}{2} = 4$

$t_6 = 4 + \dfrac{1}{2} = \dfrac{9}{2}$

$t_7 = \dfrac{9}{2} + \dfrac{1}{2}= 5$

Check if the series is an AP. Find the common difference

$d$ and write three more terms of the the following series.

$2, 4, 8, 16,....$

0%
Common difference: $2$ and Other three terms are: $32, 64, 128$
0%
Common difference: $4$ and Other three terms are: $24, 32, 40$
0%
Common difference: $8$ and Other three terms are: $20, 24, 28$
0%
None of these

Explanation

Given series is

$2,4,8,16,.....$

In an AP series, the difference between the consecutive terms remain same:

$a_2-a_1=4-2=2$

$a_3-a_2=8-4=4$

$a_3-a_2\neq a_2-a_1$

This shows that the difference of a term and the preceding terms is not always same. Hence, the given sequence is not an AP.

For the following AP, write the first term and the common difference

$3, 1, -1, -3$

0%
First term: $-3$ and Common difference: $2$
0%
First term: $-1$ and Common difference: $-3$
0%
First term: $1$ and Common difference: $3$
0%
First term: $3$ and Common difference: $-2$

Explanation

Given series is

$3,1,-1,-3,......$

First term

$a=3$

And common difference

$d=a_2-a_1$

$=1-3$

$=-2$ .

Write first four terms of the AP, when the first term

$a$ and the common difference

$d$ are given as follows:

$a = 10, d =10$

0%
The $20, 40, 60, 80.$
0%
The first four terms of the AP are $10, 20, 30, 40.$
0%
The first four terms of the AP are $15, 30, 45, 60.$
0%
Data insufficient

Explanation

Here

$a=10$ and

$d=10$ .

Now,

$n^{th}$ term

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

$\therefore t_1=a=10$

$t_2=a+d=10+10=20$

$t_3=a+2d=10+2\times10=30$

$t_4=a+3d=10+3\times10=40$

Thus, the first four terms of given AP are

$10,20,30,40$ .

For the following AP, write the first term and the common difference

$-5, -1, 3, 7$

0%
First term: $7$ and Common difference: $7$
0%
First term: $3$ and Common difference: $-1$
0%
First term: $-5$ and Common difference: $4$
0%
First term: $3$ and Common difference: $-2$

Explanation

Given series is

$-5,-1,3,7,....$

First term

$a=-5$

And common difference

$d=a_2-a_1$

$=-1-(-5)$

$=-1+5$

$=4$

Write first four terms of the AP, when the first term

$a$ and the common difference

$d$ are given as follows:

$a = -2, d =0$

0%
$-2, -2, -2, -2$
0%
$-2, -4, -6, -2$
0%
$-2, -4, 4, -2$
0%
$-2, 4, 4, -2$

Explanation

Here

$a=-2$ and

$d=0$ .

Now,

$n^{th}$ term

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

$\therefore t_1=a=-2$

$t_2=a+d=-2+0=-2$

$t_3=a+2d=-2+2\times0=-2$

$t_4=a+3d=-2+3\times0=-2$

Thus, the first four terms of given AP are

$-2,-2,-2,-2$ .

For the following AP, write the first term and the common difference

$0.6, 1.7, 2.8, 3.9$

0%
First term: $0.6$ and Common difference: $1.1$
0%
First term: $1.7$ and Common difference: $2.8$
0%
First term: $1.7$ and Common difference: $-1.1$
0%
First term: $0.6$ and Common difference: $-2.8$

Explanation

Given series is

$0.6, 1.7, 2.8, 3.9,....$

First term

$a=0.6$

And common difference

$d=a_2-a_1$

$=1.7-0.6$

$=1.1$

Is it an AP?

$1, 4, 7, 10, 13, 16, 19, 22, 25, ...$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

Given series is

$1,4,7,10,13,16,19,22,25,....$

Since,

$4-1 = 7-4=10-7,.....$
i.e

$3=3=3,......$
Given the common difference is

$3$ .
Therefore, the given sequence is an AP.
Hence, option A is correct.

Check if the sequence is an AP

$1, 3, 9, 27,....$

0%
Yes, it is an AP
0%
No
0%
Data Insufficient
0%
Ambiguous

Explanation

Given series is

$1,3,9,27,....$

We have,

$3-1=2$

$9-3=6$

$27-9=18$
This shows that the difference of a term and the preceding term is now always same. Hence, the given sequence is not an AP.

$11^{th}$ term of the AP:

$-3,$

$-\dfrac{1}{2}, 2,...$ is

0%
$28$
0%
$22$
0%
$-38$
0%
$-48$

Explanation

Clearly, the given AP has first term

$a=-3$ and common difference

$d=\displaystyle \frac{-1}2-(-3)=2-\dfrac{(-1)}{2}\Rightarrow\frac52$ .

Also,

$n=11$

$\therefore a_{11}=a+10d$

$=-3+10\times\displaystyle \frac52$

$=-3+25$

$=22$
Hence, Option B is correct.

Find the sum of

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

$10$ terms

0%
$160$
0%
$245$
0%
$290$
0%
$300$

Explanation

Clearly, the given sequence is an AP with first term

$a=2$ , common difference

$d=5$ and

$n=10$ .
Now,

$S_{10}=\dfrac {10}2[2a+(10-1)d]$

$=5[2\times2+9\times5]$

$=245$

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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