MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 13

Find the difference of the arithmetic progression an if

$a_1 = 7$ and

$a_3 = 16$

0%
$4.5$
0%
$5$
0%
$5.5$
0%
$6$

Explanation

Given,

$a_1 = 7; a_3 = 16$
Common difference,

$d = ?$

$a_3 = a_2 + d = a_1+2d$
So,

$d = \dfrac{a_3 - a_1}{2}$

$= \dfrac{16 - 7}{2}$

$d = 4.5$
Hence, the common difference is

$4.5$ .

In a zoo, there are

$210$ visitors on first day,

$250$ visitors on second day and

$290$ visitors on third day, and so on in an arithmetic sequence. What is the total number of visitors on

$15^{th}$ day?

0%
$7,960$
0%
$7,250$
0%
$7,660$
0%
$7,350$

Explanation

The sequence is

$210, 250, 290 ..........$

$a = 210, d = 40$
First find the common difference:

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

$a_{15} = 210 + (15 - 1) 40$

$= 210 + 14 \times 40$

$a_{15} = 770$
We know that,

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

$+$ Last term]

$= \dfrac{15}{2} [210 + 770]$

$= 7.5 [980]$

$S_{15} = 7,350$
So, the total number of visitors on

$15^{th}$ day is

$7,350$ .

There are

$200$ students in IT department,

$210$ students in Computer science department and

$220$ students in Mechanic department in the college, and so on in an arithmetic sequence. What is the total number of students in

$8$ departments?

0%
$1,880$
0%
$1,760$
0%
$1,660$
0%
$1,560$

Explanation

The sequence is

$200, 210, 220 .....$

$a = 200, d = 10$
First find the common difference:

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

$a_8 = 200 + (8 - 1) 10$

$= 200 + 7 \times 10$

$a_8 = 270$
We know that,

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

$+$ Last term]

$= \dfrac{8}{2} [200 + 270]$

$= 4 [470]$

$S_8 = 1,880$
So, the total number of students in

$8$ departments is

$1,880$ .

What is the sum of the first

$n$ terms if the first term is

$2$ , the common difference is

$5$ and the

$n^{th}$ term is

$122$ in an arithmetic series?

0%
$1250$
0%
$1350$
0%
$1450$
0%
$1550$

Explanation

First term:

$a = 2$
last term,

$l = 122$

$d = 5$

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

$122=2+(n-1)5$

$122-2+5=5n$

$125=5n$

$n = 25$

$S_{n}=\dfrac{n}{2}[a+l]$

$S_{25}=\dfrac{25}{2}[2+122]$

$S_{25}=12.5[124]$

$S_{25}=1550$

There are

$100$ students in science department,

$150$ students in computer science department in a higher secondary school and so on in an arithmetic sequence. What is the total number of students in

$10$ departments?

0%
$3,960$
0%
$3,250$
0%
$3,660$
0%
$3,560$

Explanation

The sequence is

$100, 150 .....$

$a = 100, d= 50$
First find the common difference:

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

$a_{10} = 100 + (10 - 1) 50$

$= 100 + 9 \times 50$

$a_{10} = 550$
We know that,

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

$+$ Last term]

$= \dfrac{10}{2} [100 + 550]$

$= 5 [650]$

$S_{10} = 3,250$
So, the total number of students in

$10$ departments is

$3,250$ .

In a national treasure library, there are 10 visitors on first day, 50 visitors on second day and 90 visitors on third day, and so on in an arithmetic sequence. What is the total number of visitors on 5th day?

0%
460
0%
450
0%
440
0%
430

Explanation

The sequence is 10, 50, 90 ......

$a = 10, d = 40$
First find the common difference:

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

$a_5 = 10 + (5 - 1) 40$

$= 10 + 4 \times 40$

$a_5 = 170$
We know that,

$S_n = \dfrac{n}{2}$ [First term + Last term]

$= \dfrac{5}{2} [10 + 170]$

$= 2.5 [180]$

$S_5 = 450$
So, the total number of visitors on 5th day is 450

The first term of an A.P. is

$2$ and ninth term is

$58$ . What is the common difference?

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

We know the formula,

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

$a = 2; a_9 = 58$
Therefore,

$a_9 = a + (9 -1 )d$

$\Rightarrow 58 = 2 + (8) d$

$\Rightarrow 58 - 2 = 8d$

$\Rightarrow 56 = 8d$

$\Rightarrow d = 7$

What is the sum of the first forty terms of the arithmetic sequence 120, 115, 110, 105....?

0%
900
0%
700
0%
500
0%
300

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

$\therefore$ First Term:

$a = 120$

Common Difference:

$d = -5$

Total terms:

$n = 40$

Sum of

$40$ terms:

$S = \dfrac{40}2\times(2\times120 + (40-1)\times (-5))$

$S = 20\times (240 - 39\times 5)$

$S = 900$

What is the sum of the first nineteenth terms of the arithmetic sequence 2, 6, 10, 14.....?

0%
$522$
0%
$622$
0%
$722$
0%
$822$

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

$\therefore a = 2 d = 4 n = 19$

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

$S = \dfrac{19}{2}[2\times 2+(19 - 1)4]$

$S = \dfrac{19}{2}[4+72]$

$S = 9.5[76]$

$S =722$

In an arithmetic series,

$a_1 = 7$ and

$a_{12}=29$ . Find the sum of the first 12 terms.

0%
$116$
0%
$216$
0%
$316$
0%
$416$

Explanation

Given:

First term:

$a = 7$
Last term,

$l = 29$

Here,

$a_{n}=29$

and

$n = 12$

We know that

$S_{n}=\dfrac{n}{2}[a+l]$

$\therefore \ S_{12}=\dfrac{12}{2}[7+29]$

$=6\times 36$

$=216$

If the

$6^{th}$ term and

$12^{th}$ term of an A.P. are

$30$ and

$75$ respectively, find the sum of its

$20$ terms.

0%
$1275$
0%
$1150$
0%
$1140$
0%
$1130$

Explanation

The

$nth$ term and sum of n terms of an arithmetic sequence can be calculated by

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

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

$S_n$ = Sum of n terms

$\therefore$

$a+5d = 30$ (

$6^{th}$ term)

and

$a+11d = 75$ (

$11^{th}$ term)

Solving both, we get

$d=7.5$

$a=-7.5$

The sum of $20$ terms:

$=\dfrac{20}{2}(2(-7.5)+(20-1)7.5)$

$= 10(17 \times 7.5)$

$=1275$

Which of the term of the A.P. 11, 22, 33, 44.... is 550?

0%
$20$
0%
$30$
0%
$40$
0%
$50$

Explanation

Let the general form of A.P.,

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

$d = 11$
and

$a_{n}=550$

So,

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

$550 = 11 + (n-1)\times 11$

$550-11 = 11n-11$

$539+11= 11n$

$550 = 11n$

$n = 50$

Find which the term of the A.P. 20, 30, 40.... is 100.

0%
$11$
0%
$12$
0%
$13$
0%
$9$

Explanation

Let the general form of A.P.,

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

$d = 10$

$a_{n}=100$

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

$100 = 20 + (n-1)\times 10$

$100-20 = 10n-10$

$80+10 = 10n$

$90 = 10n$

$n = 9$

Find the sum of the following AP :

$2 , 4 ,6 , 8 ,...... (15$ terms).

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

Explanation

First term of the given arithmetic series = 2
Second term of the given arithmetic series = 4
Third term of the given arithmetic series = 6
Fourth term of the given arithmetic series = 8
Now, Second term - First term = 4 - 2 = 2
Third term - Second term = 6 - 4 = 2
Therefore, common difference of the given arithmetic series is 2.
The number of terms of the given A. P. series (n) = 15
We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

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

$S_{15}=\frac{15}{2}[2\times 2+(15-1)2]$

$S_{15}=7.5[4+28]$

$S_{15}=240$

Which term of the A.P. 18, 16, 14.... is -20?

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

Let the required term

$=n$

The general form of A.P.,

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

$d = -2$

$a_{n}=-20$

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

$-20 = 18 + (n-1)\times -2$

$-20-18 = -2n+2$

$-38-2 = -2n$

$-40 = -2n$

$n = 20$

Find the sum of the first 20 terms if the first term is -12 and the common difference is -5 in an arithmetic series.

0%
$-1190$
0%
$-1290$
0%
$-1390$
0%
$-1490$

Explanation

First term =

$a = -12$

$n = 20$

$d = -5$

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

$S_{20}=\frac{20}{2}[2\times -12+(20-1)(-5)]$

$S_{20}=10[-24-95]$

$S_{20}=-1190$

If the first term of an arithmetic sequence is

$-14$ and the common difference is

$2$ , find the sum of the first

$8$ terms.

0%
$-56$
0%
$-46$
0%
$-36$
0%
$-26$

Explanation

Let

$S_{n}$ be the sum of

$n$ terms of AP.

First term:

$a = -14$

$S_{8}=?$

$d = 2$

$n = 8$

We know that,

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

$S_{8}=\dfrac{8}{2}[2\times (-14)+(8-1)2]$

$S_{8}=4[-28+14]$

$S_{8}=-56$

$a_1=1.5, a_{20}=58.5$ in an arithmetic series, find

$S_{15}$ .

0%
$137.5$
0%
$237.5$
0%
$337.5$
0%
$437.5$

Explanation

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

$\Rightarrow a_{20} = 1.5+(20-1)d$

$\Rightarrow 58.5-1.5 = 19d$

$\Rightarrow 57= 19d$

$\Rightarrow d = 3$

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

$\Rightarrow S_{15} = \dfrac{15}{2}[2\times 1.5+(15-1)3]$

$\Rightarrow S_{15} = 7.5[3+42]$

$\Rightarrow S_{15} = 7.5[45]$

$\Rightarrow S_{15} = 337.5$

Given A.P is

$\dfrac{1}{3q},\dfrac{1-6q}{3q}, \dfrac{1-12q}{3q}$ . Find the common difference

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

Explanation

Given A.P is $\dfrac{1}{3q},\dfrac{1-6q}{3q}, \dfrac{1-12q}{3q}$

Then,

$\dfrac{1}{3q},\dfrac{1}{3q}-\dfrac{6q}{3q},\dfrac{1}{3q}-\dfrac{12q}{3q}$ will also be in AP.

Which results in,

$\dfrac{1}{3q},\dfrac{1}{3q}-2,\dfrac{1}{3q}-4$

It can be seen clearly that the series is decreasing by the quantity (2)

Hence, the common difference is

$-2$

In the arithmetic series, find the sum of the first

$n$ terms if the first term is

$-11$ , the common difference is

$4$ and the

$n^{th}$ term is

$137.$

0%
$1394$
0%
$2394$
0%
$3394$
0%
$4394$

Explanation

First term:

$a = -11$
Last term,

$l = 137$

$d = 4$

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

$137=-11+(n-1)4$

$137+11+4=4n$

$152=4n$

$n = 38$

$S_{n}=\dfrac{n}{2}[a+l]$

$S_{38}=\dfrac{38}{2}[-11+137]$

$S_{38}=19[126]$

$S_{38}=2394$

Find the common difference in an arithmetic seres, where

$S_{15}=322.5$ and the first is

$4$ .

0%
$1.5$
0%
$2.5$
0%
$3.5$
0%
$4.5$

Explanation

$S_{n}=\dfrac{n}{2}[t_1+t_n]$

$\Rightarrow 322.5=\dfrac{15}{2}[4+t_n]$

$\Rightarrow 322.5=7.5[4+t_n]$

$\Rightarrow 43=4+t_n$

$\Rightarrow t_n=39$

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

$\Rightarrow39 = 4+(15-1)d$

$\Rightarrow 35= 14d$

$\Rightarrow d = 2.5$

Find

${a}_{20}$ given that

${a}_{3}=9$ and

${a}_{8}=24$

0%
$52$
0%
$54$
0%
$56$
0%
$60$

Explanation

In a arithmetic series

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

$a$ is the first term of the sequence,

$d$ is the common difference, and

$n$ is the number of the term to find.

Since it is given that

${ a }_{ 3 }=9$ and

$a_8=24$

These two terms are

$8-3=5$ places apart, so, from the definition of a arithmetic sequence, we know that

$a_8=a_3+5d$ and we can use this to solve for the value of the common difference

$d$ :

$a_8=a_3+5d$

$24=9+5d$

$5d=15$

$d=3$

Since $a_3=a+2d$ ,

So,

$9=a+6$

$a=9-6=3$

To get ${a}_{20}$ ,

${a}_{20}=3+(20-1)(3)$

$=60$

If the

$14^{th}$ term of an arithmetic series is

$6$ and

$6^{th}$ term is

$14$ , then what is the

$95^{th}$ term?

0%
$-75$
0%
$75$
0%
$80$
0%
$-80$

Explanation

Given,

$14^{th}$ term

$=6$ ,

$6^{th}$ term

$=14$

We know

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

Therefore,

$6 = a + (14 - 1) d = a + 13d$ ....(1)
and

$14 = a + (6 - 1)d = a + 5d$ ....(2)

Subtracting equations (1) and (2),

$d=8d=-8\Rightarrow d=-1$
Putting this value in equation (1), we get

$6=a+13(-1)$

$\Rightarrow a=6+13$

$\Rightarrow a=19$
Thus

$T_{95} = a + (95 - 1)d$

$= a + 94d$

$= 19 - 94 = -75$

Find the sum of all natural numbers not exceeding

$1000$ , which are divisible by

$4$ but not by

$8$ .

0%
$62500$
0%
$62800$
0%
$64000$
0%
$65600$

Explanation

Natural Number divisible by

$4$ but not

$8$ are

$4,12,20,28,....996$

$996=4+(n-1)8$ (

$n^{th}$ term formula)

$249=1+(n-1)2$ (divide both sides by 4)

$248=(n-1)2 \\ 124+1=n \\ n=125$

$S_n=4+12+20+......till\ 125\ terms$

$\\ { S }_{ n }=4\left[ \cfrac { n }{ 2 } \left( 2a+(n-1)d \right) \right] \\ { S }_{ n }=4\times \cfrac { 125 }{ 2 } \left[ 2\times 1+124\times 2 \right] \\ { S }_{ n }=250(2+248)\\ { S }_{ n }=62500$

What is the sum of the first

$40$ even positive integers?

0%
$1600$
0%
$1560$
0%
$820$
0%
$1640$
0%
$400$

Explanation

According to problem:

$a=2,$ First term

$d=2,$ common difference

$n=40,$ number of terms

$\because~ S_{n}=\dfrac{n}{2} [2a+(n-1)d]=1640$

$\implies S_{n}=\dfrac{40}{2} [2\times 2+(40-1)2]=1640$

How many 3 digit numbers are divisible by

$6$ in all?

0%
$149$
0%
$150$
0%
$151$
0%
$166$

Explanation

${\textbf{Step -1: Write three digit multiple of 6 as A.P. having common difference as 6.}}$

${\text{Multiple of 6 between 100 and 999 are as follows:}}$

$102,315,324, \ldots ,996$

${\text{So, the formed A}}{\text{.P}}{\text{. is}}$ $102,315,324, \ldots ,996$

${\text{Where,}}$ $a = 102:$ ${\text{First term of A}}{\text{.P}}{\text{.}}$

$d = 6:$ ${\text{common difference of A}}{\text{.P}}{\text{.}}$

$l = 996:$ ${\text{Last term of A}}{\text{.P}}{\text{.}}$

${\text{Now we know that, formula for }}{{\text{n}}^{{\text{th}}}}{\text{ term of an A}}{\text{.P}}{\text{. is given by,}}$

${a_n} = a + \left( {n - 1} \right)d \ldots \left( 1 \right)$

${\text{Where,}}$ $a =$ ${\text{First term of A}}{\text{.P}}{\text{.}}$

$d =$ ${\text{common difference of A}}{\text{.P}}{\text{.}}$

${a_n} =$ ${{\text{n}}^{{\text{th}}}}{\text{term of an A}}{\text{.P}}{\text{.}}$

$n =$ ${\text{Total number of terms in A}}{\text{.P}}{\text{.}}$

${\textbf{Step -2: Substitute the known values in equation }}\left( \mathbf1 \right)$

$\Rightarrow 996 = 102 + \left( {n - 1} \right)6$

$\Rightarrow 996 - 102 = \left( {n - 1} \right)6$

$\Rightarrow 894 = \left( {n - 1} \right)6$

$\Rightarrow \left( {n - 1} \right) = \dfrac{{894}}{6}$

$\Rightarrow n - 1 = 149$

$\Rightarrow n = 149 + 1$

$\Rightarrow n = 150$

${\textbf{Therefore, option B. 150 is correct answer.}}$

What is the tenth term of the arithmetic sequence whose first term is

$x$ and whose third term is

$x + 6a$ ?

0%
$33a$
0%
$x + 24a$
0%
$x + 27a$
0%
$x + 30a$
0%
$x + 33a$

Explanation

Let

$a_1=x$
Given,

$a_3=x+6a$
Let

$n = 3$
We know,

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

$\therefore a_3=x+(3-1)d$

$\therefore x+6a-x=2d$

$\therefore \dfrac{6a}{2}=d$

$\therefore d=3a$
Now use

$a_1=x, d = 3a, n = 10$ in

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

$\therefore a_{10}=x+(10-1)3a$

$\therefore a_{10}=x+27a$

The sum of four numbers in AP is

$176$ . The product of 1st and last is

$1855$ . The mean of middle two is

0%
$42$
0%
$41$
0%
$44$
0%
$53$

Explanation

Let the numbers in A.P. be

$(a-3d), (a-d), (a+d), (a+3d)$

$\therefore (a-3d)+(a-d)+(a+d)+(a+3d)=176$

$\Rightarrow 4a=176$

$\therefore a=44$

$\therefore$ The mean of middle two is:

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

$=a=44$

The mean of five numbers in AP is

$89$ . The product of first and last terms is

$7885$ . The AM of first, third and fifth term is

0%
$83$
0%
$86$
0%
$89$
0%
$90$

Explanation

Let the numbers in A.P. be

$(a-2d), (a-d), a, (a+d), (a+2d)$

$\therefore \dfrac{(a-2d)+(a-d)+a+(a+d)+(a+2d)}{5}=89$

$\Rightarrow \dfrac{5a}{5}=89$

$\therefore a=89$

$\therefore$ The AM of 1st, 3rd & 5th term is:

$=\dfrac{(a-2d)+a+(a+2d)}{3}$

$=a=89$

The Sum of three numbers in AP is

$75$ , and product of extremities is

$609$ . The AM of first and second number is

0%
$\{21,25,29\}$ , AM $= 23$
0%
$\{13,17,21\}$ , AM $= 22$
0%
$\{21,25,29\}$ , AM $= 25$
0%
$\{21,22,29\}$ , AM $= 23$

Explanation

Let the numbers in A.P. be (a-d), a, (a+d)

$\therefore (a-d)+a+(a+d)=75$

$\Rightarrow a=25$

Also,

$(a-d)(a+d)=609$

$\Rightarrow a^{2}-d^{2}=609$

$\Rightarrow 25^{2}-d^{2}=609$

$\therefore d=\pm4$

The numbers are {21,25,29} or {29,25,21}

According to option, we take {21,25,29}

A.M. of first and second number

$=\dfrac{21+25}{2}=23$

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 13 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 13 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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