MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 5

Choose the correct statement(s):

$A$ : Every sequence is a progression.

$B$ : Every progression is a sequence.

0%
Only $A$
0%
Only $B$
0%
Both $A$ and $B$
0%
None

Find the sum of first

$32$ multiples of

$4$ .

0%
$2,112$
0%
$2,712$
0%
$2,110$
0%
$2,111$

Explanation

The multiples of

$4$ are,

$4, 8, 12, 16,$ ............
Therefore,

$a= 4, d = 4, n=32$
We have

$S_{32} = \dfrac{n}{2} [2a + (n - 1)d]$

$= \dfrac{32}{2} [2 \times 4 + (32 - 1)4]$

$= 16 [8 + 31 \times 4]$

$= 16 [8 + 124]$
Therefore,

$S_{20} = 2,112$

Constant is subtracted from each term of an A.P. the resulting sequence is also an ______

0%
Geometric Sequence
0%
Arithmetic Progression
0%
Both A and B
0%
None of the above

Explanation

If a constant is subtracted from each term of an A.P. the resulting sequence is also an arithmetic progression.

For example : Let the sequence be

$\{ a_1, a_2, a_3 ...... a_4 \}$ be an arithmetic progression with common difference

$d$ .

Again subtract

$k$ to the above sequence.

Then the resulting sequence is

$a_1 -k, a_2 - k, a_3 - k ......$ be an A.P.

Find the difference of an A.P.

$a_n$ , if

$a_4 = 16$ and

$a_2 = 8$

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

Explanation

Given,

$a_4=16, a_2=8$

$a_4 = a + 3d=16$ ...... (1)

$a_2 = a + d=8$ ...... (2)
Multiplying equation (2) by

$3$ and then subtract the equation, we get

$16 = a + 3d$

$24 = 3a + 3d$
(-) (-) (-)
----------------

$-8 = - 2a$

$a = 4$
Put

$a= 4$ in equation (1), we get

$16 = 4 + 3d$

$\Rightarrow 16 - 4 = 3d$

$\Rightarrow d = 4$

The last term of the AP

$21, 18, 15$ ....... is

$-351$ . Find the nth term.

0%
$213$
0%
$123$
0%
$312$
0%
$-231$

A train can travel

$200$ m in the first hour,

$400$ m the next hour,

$600$ m the third hour and so on in an arithmetic sequence. What is the total distance the train travels in

$5$ hours?

0%
$2,000$
0%
$3,000$
0%
$4,000$
0%
$5,000$

Explanation

The sequence is

$200, 400, 600 .....$

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

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

$a_5 = 200 + (5- 1) 200$

$= 200 + 4 \times 200$

$a_5 = 1,000$
We know that,

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

$+$ Last term]

$= \dfrac{5}{2} [200 + 1,000]$

$= 2.5 [1,200]$

$S_5 = 3,000$
Hence, the train will travel

$3,000$ m in

$5$ hours.

Find the sum of first

$20$ multiples of

$13$ .

0%
$2,720$
0%
$2,730$
0%
$2,740$
0%
$2,750$

Explanation

The multiples of

$13$ are,

$13, 26, 39, .....$
Therefore,

$a = 13, d = 13,n=20$
We have

$S_{20} = \dfrac{n}{2} [2a + (n - 1)d]$

$= \dfrac{20}{2} [2 \times 13 + (20 - 1)13]$

$= 10 [26 + 19 \times 13]$

$= 10 [26 + 247]$
Therefore,

$S_{20} = 2,730$

Find the sum of the first 20 terms of the arithmetic series if

$a_{1} = 10$ and

$a_{20}=100$ .

0%
1000
0%
1100
0%
1200
0%
1300

Explanation

To find the sum of the first n terms of an arithmetic series use the formula,

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

$S_{20}=\dfrac{20(10+100)}{2}$

$S_{20}=\dfrac{2200}{2}$

$S_{20}=1100$

What is the sum of

$t_n= (2n-5)$ from n =10 to 150?

0%
$22650$
0%
$21855$
0%
$23250$
0%
$21250$

Explanation

$T_n = (2n-5)$

$T_{10} = 2\times10-5 = 15$

$T_{150} = 2\times15- 5 = 295$

Sum of AP, if first terms and last term is given:

$S = \dfrac n2(\text{first term } +\text{ last term})$

Here,

$n = 150-10+1 = 141$

$S = \dfrac{141}2(15+295)$

$S = 141\times155 = 21855$

Calculate the

$15^{\text{th}}$ term of the A.P.

$-3, -4, -5, -6, -7....$

0%
$-13$
0%
$-15$
0%
$-17$
0%
$-19$

Explanation

Let the general form of A.P.,

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

$d = -1$

$a_{15}=?$

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

$a_{15} = -3 + (15-1)\times -1$

$a_{15} = -3 -14$

$a_{15} = -17$

Find the

$8^{th}$ term of the A.P. :

$11, 14, 17, 20.....$

0%
$11$
0%
$-17$
0%
$17$
0%
$32$

Explanation

Given series is

$11,14,17,20,....$

Let the general form of A.P.,

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

$d = 3$ and first term

$a=11$

$a_{8}=?$

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

$a_{8} = 11 + (8-1)\times 3$

$a_{8} = 11 +21$

$a_{8} = 32$

3, 5, 7, 9, 11, 13, 15.... is an

0%
Geometric progression
0%
Harmonic progression
0%
Arithmetic progression
0%
None of above

A sequence of numbers in which each term is related to its predecessor by same law is called

0%
arithmetic series
0%
progression
0%
geometric series
0%
none of these

_______ is a series of successive events.

0%
Series
0%
Preceding sequence
0%
Progression
0%
Geometric progression

Which of the following is not in the form of A.P.?

0%
$1, -1, -3, -5, -7...$
0%
$0, 3, 6, 9, 12...$
0%
$4, 5, 7, 10, 14...$
0%
$-1, 2, 5, 8, 11..$

Explanation

For any series to be in Arithmetic Progression, the common difference (i.e., the difference between any two consecutive terms should be the same).

$4, 5, 7, 10, 14...$ is not in the form of A.P.
Here the common difference is not constant.

________ can be defined as arrangement of terms in which sequence of terms follow some conditions.

0%
Series
0%
Preceding sequence
0%
Progression
0%
Geometric progression

For given A.P.

$-\dfrac{1}{2}, -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{3}{2}, ..$ find the common difference.

0%
$-1$
0%
$-\dfrac{1}{2}$
0%
$\dfrac{3}{2}$
0%
$1$

Explanation

The general form of A.P. is

$a, a + d, a + 2d, a + 3d.....$

$-\dfrac{1}{2}, -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{3}{2}, ..$
Here the common difference is

$-1$ .

Which of the following is in the form of

$A.P$ ?

0%
$1, -1, -3, -5, -7,\dots$
0%
$0, 3, 2, 1, -2,\dots$
0%
$4, 5, 7, 10, 14,\dots$
0%
$-2, 2, -2, 2, -2,\dots$

Explanation

$1, -1, -3, -5, -7$ is in the form of

$A.P$ because the common difference between consecutive terms of this sequence is constant and that is equal to

$-2$ .

Find the sum of all the odd positive integers less than

$100$ .

0%
$2400$
0%
$2500$
0%
$2525$
0%
$2600$
0%
$2650$

Explanation

The possible odd positive integers:

$1,3,5,7,9,11..........99$
The first term,

$a$

$=1$
The difference between two consecutive terms

$=3-1=2$
The total no. of terms

$=50$
Applying sum of arithmetic progression formula,

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

$=\dfrac{50}{2}[2\times1+(50-1)2]$

$=25(2+98)=2500$
Hence, choice B is correct.

In an Arithmetic sequence,

$S_{n}$ represents the sum to

$n$ terms, what is

$S_{n} - S_{n - 1}$ ?

0%
$t_{1} + t_{2} + .... t_{n - 1}$
0%
$S_{n - 1}$
0%
$\displaystyle \sum_{n = 1}^{n - 2} t_{n}$
0%
$t_{n}$

Explanation

We know,

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

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

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

$=\dfrac{1}{2}[2an+n^{2}d-nd-2an+2a-n^{2}d+3nd-2d]$

$=\dfrac{1}{2}[2a+2nd-2d]$

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

$=t_{n}$

$(Ans \to D)$

The sum of first n terms of an A.P. is

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

Explanation

The sum of first n terms of an A.P. is

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

Find the sum of integers from

$1$ to

$35$ .

0%
$1160$
0%
$630$
0%
$1360$
0%
$1460$

Explanation

Given,

$n = 35$
We know

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

$S_{35}=\dfrac{35}{2}[35+1]$

$S_{35}=\dfrac {35[36]}{2}$

$S_{35}=\dfrac {1260}{2}=630$

$t_{n}$ be the

$n^{th}$ term of the A.P.

$-9, -14, -19, ....,$ what is the value of

$t_{30} - t_{20}$ ?

0%
$-35$
0%
$-50$
0%
$-55$
0%
$-65$

Explanation

Given series:

$-9,\ -14,\ -19,\ .\ .\ .$

Here

$a=-9$

and

$d=(-14)-(-9)=-5$

$t_{30}=(-9)+(30-1)(-5)$

$=-154$

$t_{20}=(-9)+(20-1)(-5)$

$=-104$

$\therefore t_{30}-t_{20}=(-154)-(-104)$

$=-50$

What is the sum of 12 odd numbers

$1, 3, 5, 7, 9.....?$

0%
$12$
0%
$144$
0%
$141$
0%
$124$

Explanation

First term of the given arithmetic series

$= 1$
Second term of the given arithmetic series

$= 3$
Third term of the given arithmetic series

$= 5$
Fourth term of the given arithmetic series

$= 7$
Now, Second term - First term

$= 3 - 1 = 2$
Third term - Second term

$= 5 - 3 = 2$
Therefore, common difference of the given arithmetic series is

$2.$
The number of terms of the given

$A. P.$ series

$(n) = 12$
We know that the sum of first n terms of the Arithmetic Progress, whose first term

$= a$ and common difference

$= d$ is

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

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

$S_{12}=6[2+22]$

$S_{12}=144$

What is the nth term of the arithmetic sequence

$1, 3, 5, 7, 9, 11...?$

0%
$n + 1$
0%
$n - 1$
0%
$2n + 1$
0%
$2n - 1$

Explanation

Clearly, the difference of successive terms of above sequence is constant which is 2

So given sequence is in AP with first term

$1$ and common difference

$2$

Hence general term is,

$a_n = a+(n-1)d=1+(n-1)2=2n-1$

Hence option (D) is correct choice

Find the nth term of an arithmetic sequence

$11, 22, 33, 44...$ .

0%
$n$
0%
$n - 1$
0%
$2n$
0%
$11n$

Explanation

Clearly, the difference of successive terms of above sequence is constant which is 11

So given sequence is in AP with first term 11 and common difference

$11$

Hence general term is,

$a_n = a+(n-1)d=11+(n-1)11=11n$

Hence option 'D' is correct choice

If the

$nth$ term of AP is

$2n+5$ . Then find the

$AM$ of first and last terms.

0%
$99$
0%
$98$
0%
$100$
0%
$44$

Explanation

Given:

$t_{n}=2n+5$

First term

$=2\times 1+5=7$

Last term

$=2\times 38+5=81$

$\therefore$ Arithmetic mean

$(A.M).=\dfrac{7+81}{2}=44$

Obtain the sum of the first

$56$ terms of an A.P. whose

$28^{th}$ and

$29^{th}$ terms are

$52$ and

$148$ respectively.

0%
$2100$
0%
$5600$
0%
$5200$
0%
$2600$

Explanation

We know that,

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

Here,

${t}_{28}=a+(28-1)d$

$\therefore$

$52=a+27d.......(i)$

Also,

${t}_{29}=a+(29-1)d$

$\therefore$

$148=a+28d........(ii)$

Adding equation

$(i)$ and

$(ii)$ , we get:

$a+27d=52$

$a+28d=148$
----------------------

$2a+55d=200.............(iii)$

Also,

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

${ S }_{ 56 }=\cfrac { 56 }{ 2 } \left[ 2\times a+(56-1)d \right]$

$\Rightarrow$

${S}_{56}=28(2a+55d)$

$\Rightarrow$

${S}_{56}=28\times 200$

$\Rightarrow$

${S}_{56}=5600$

$\therefore\$ The sum of the

$56$ terms of the given A.P. is

$5600$ .

$n^{th}$ term of AP is

$4n+1$ , then AM of

$11^{th}$ to

$20^{ th}$ terms is

0%
$61.5$
0%
$63$
0%
$63.5$
0%
$62$

Explanation

Given:

$t_{n}=4n+1$

$A.M.=\cfrac{t_{11}+t_{20}}{2}$

$=\cfrac{4\times 11+1+4\times 20+1}{2}$

$=63$

The common difference of A.P.:

$1, -1, -3, ....$ is _______

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

Explanation

The common difference d of A.P. is given by the difference between second term and first term.

$\therefore d=-1-1=-2$
Option

$C$ is correct.

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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