MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 15

Find the sum of

$2,4,6,8,,,........2n$

0%
$n(n+1)$
0%
$n^2$
0%
$n(n-1)$
0%
$n^2+n$

Explanation

The given sequence is

$2,4,6,8,..... 2n$

$\therefore a= 2, d = 2$ &

$a_k= 2n$

$\therefore 2n = a+(k-1)d$

$\Rightarrow 2n = 2+(k-1)2$

$\Rightarrow k = n$

Sum of

$n$ terms

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

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

$=n(n+1)$

The

$27^{th}$ term of AP

$7,9,11,13,15,17,19,\ldots$ is

0%
$59$
0%
$61$
0%
$63$
0%
None of these

Explanation

Given series is

$7,9,11,13,15,17,19,\dots$

$a=7$

$d=9-7$

$=2$

Formula for

$n^{th}$ term of an AP whose first term is

$a$ and common difference is

$d$ is given by,

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

So,

$27^{th}$ term is given as,

$a_{27}=7 + (27-1)2$

$=7+26\times2$

$=7+52$

$=59$

How many terms are there in the sequence 3, 6, 9, 12, ..., 111 ?

0%
$32$
0%
$37$
0%
$42$
0%
$49$

Explanation

The given sequence is an A.P. with first term

$a=3$ and common difference

$d=3$ . Let there be n terms in the given sequence. Then,

$nth \ term=111$

$a+(n-1)d=111$

$3+(n-1) \times 3=111$

$n=37$

Thus, the given sequence contains 37 terms.

Which of the following pair of terms do you prefer if the sum of two consecutive terms of an

$AP$ is given whose first term is

$a$ and common difference is

$d$ ?

0%
$a$ and $a+d$
0%
$a$ and $a-d$
0%
$a-d$ and $a+d$
0%
None of these

State true/false:

$3,4,5,6,7,......$ forms a progression. As they are formed by the rule

$n+1$

0%
True
0%
False

It is given that the sum of four terms of an

$AP$ is

$2$ and their product is zero, then what will be the terms?

0%
$-1, 0, 1, 2$
0%
$-2, 0, 2, 4$
0%
$-2, -1, 0, 2$
0%
None of these

Find the four consecutive terms of an

$AP$ whose sum is

$12$ and sum of

$2nd$ and

$4th$ term is

$8$ .

0%
$0, 2, 4, 6$
0%
$0, 3, 5, 7$
0%
$2, 4, 6, 7$
0%
None of these

In an arithmetic progression, if

${ S }_{ n }=n(5+3n)\quad and\quad { t }_{ n }=32$ , then the value of n is
[Note:

${ S }_{ n }$ and

${ t }_{ n }$ denote the sum of first n terms and

${ n }^{ th }$ term of arithmetic progression respectively.]

0%
4
0%
5
0%
6
0%
7

Find the sum of all odd integers between 2 and 100 divisible by 3.

0%
452
0%
754
0%
867
0%
765

Explanation

$\textbf{Step-1: Finding the value of n}$

$\text{The}$

$n^{th}$

$\text{term in an A.P. is given by}$

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

$\text{We know that the}$

$n^{th}$

$\text{term is 99}$

$99=3+(n-1)6$

$96=6(n-1)$

$n-1=16$

$n=17$

$\textbf{Step-2: Finding the required sum}$

$\text{Sum of n terms in an A.P. is}$

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

$S_n=\dfrac{17}{2}(3+99)=867$

$\textbf{Hence,The correct option is (C)}$

Choose the correct answer from the given four options in the following question:
In an A.P., if

$d = -4, n = 7, a_n = 4,$ then

$a$ is

0%
6
0%
7
0%
20
0%
28

Explanation

Hint :

$n^{th }$ term of an A.P. is

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

$a$ is first term and

$d$ is common difference.

Given:
common difference

$=d=-4$

$n=7$

$a_n=4$ [

$n^{th}$ term]

As we know, that
general term of an A.P. is given by

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

putting the given values in the above formula

$\Rightarrow 4=a+(7-1)(-4)$

$\Rightarrow 4=a+(6)(-4)$

$\Rightarrow 4=a-24\\ \Rightarrow 4+24=a$

$\Rightarrow 28=a$

Hence, value of first term

$,a$ is

$28$

Choose the correct answer from the followings for the sequence

$-10, -6, -2, 2, ...$

0%
an A.P. with $d = -16$
0%
an A.P. with $d = 4$
0%
an A.P. with $d = -4$
0%
not an A.P.

Explanation

Hint : In an A.P. difference between two consecutive terms is always same

Step 1 : find out the difference between two consecutive terms
for the given sequence

$-10,\,-6,\,-2,\,2,...$
second term - first term

$=-6-(-10)=4$
third term - second term

$=-2-(-6)=4$
fourth term - third term

$=2-(-2)=4$

Step 2 : Comparison of all the difference quantities and conclusion
Thus, difference between two consecutive terms is always

$4$

$\therefore \,$ common difference

$=d=4$

Final step : Hence the given sequence is an A.P. with

$d=4$ .

Choose the correct answer from the given four options in the following question:
The first four terms of an A.P., whose first term is

$-2$ and the common difference is

$-2$ , are

0%
$-2, 0, 2, 4$
0%
$-2, 4, -8, 16$
0%
$-2, -4, -6, -8$
0%
$-2, -4, -8, -16$

Explanation

Hint: use the formula of

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

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

Given:-
first term

$=a=a_1=-2$
common difference

$=d=-2$

Step 1 : find the second term by replacing $n$ with $2$ in the formula of $n^{th}$ term of the A.P

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

$\Rightarrow a_2=(-2)+(2-1)(-2)$

$\Rightarrow a_2=-2+(1)(-2)$

$\Rightarrow a_2=-2-2$

$\therefore a_2=-4$

Step 2 : find the third term by replacing $n$ with $3$ in the formula of $n^{th}$ term of the A.P

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

$\Rightarrow a_3=(-2)+(3-1)(-2)$

$\Rightarrow a_3=-2+(2)(-2)$

$\Rightarrow a_3=-2-4$

$\therefore a_3=-6$

Step 3 : find the fourth term by replacing $n$ with $4$ in the formula of $n^{th}$ term of the A.P

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

$\Rightarrow a_4=(-2)+(4-1)(-2)$

$\Rightarrow a_4=-2+(3)(-2)$

$\Rightarrow a_4=-2-6$

$\therefore a_4=-8$

Final step: write down the first four terms of A.P.
first four terms of A.P. are

$a_1,\,a_2,\,a_3,\,a_4$
i.e,

$-2,\,-4,\,-6,\,-8$

Next three consecutive numbers in the pattern are

11, 8, 5, 2...

0%
$0, – 3, – 6$
0%
$– 1, – 5, – 8$
0%
$– 2, – 5, – 8$
0%
$– 1, – 4, – 7$

Explanation

Option (d) is correct; any two numbers have the common difference of 3,

$\begin{array}{l}11-3=8,8-3=5, \ldots \\\text { Similarly, }(2-3)=-1,(-1-3)=-4,(-4-3)=-7\end{array}$
Hence, three consecutive terms will be -1, -4, and -7.

What is the common difference of an A.P. in which

$a_{18}-a_{14} = 32$ ?

0%
$8$
0%
$-8$
0%
$-4$
0%
$4$

Explanation

Hint: The general term formula of an

$A.P$ is

$\{a_n=a+(n-1)d\}$ will be used along with the related terms.

Given:
Let

$\text {a=first term (Let)}$

$\text {d= common difference (Let)}$

$a_{18}-a_{14}=32$

Step 1: Finding the general term expression of $a_{18}$
We know,

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

For

$n=18$

$a_{18}=a+(18-1)d=a+17d.....(i)$

Step 2: Finding the general term expression of $a_{14}$

We know,

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

For

$n=14$

$a_{14}=a+(14-1)d=a+13d.....(ii)$

Step 3: Finding the difference $a_{18}-a_{14}$ by subtracting $eq (i)$ from $eq. (ii)$

$eq.(ii)-eq. (i)$

$a_{18}-a_{14}=a+17d-(a+13d)=a+17d-a-13d=4d.....(iii)$

Step 4: Comparing and equating the value of $a_{18}-a_{14}$ from the question and $eq(iii)$

$a_{18}-a_{14}=4d=32$

$\Rightarrow 4d=32 \Rightarrow d=8$

Hence, the value of 'd' is

$8$

Final step: The value of

$\text {common difference, d=8}$

Two APs have the same common difference and the 1st term of one AP is -1 and first term of other AP is -Then the difference between their 4th terms is

0%
-1
0%
-8
0%
7
0%
-9

Explanation

Hint: use the formula of

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

Given:-
first term of one A.P.

$=a=-1$
first term of other A.P.

$=A=-8$
same common difference of both the A.Ps

$=d$

Step 1 : find the $4^{th}$ term of first A.P.

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

$\Rightarrow a_4=-1+(4-1)d$

$\therefore a_4=-1+3d$

Step 2 : find the $4^{th}$ term of other A.P.

$\Rightarrow A_n=A+(n-1)d$

$\Rightarrow A_4=-8+(4-1)d$

$\therefore A_4=-8+3d$

Step 3 : find the difference between $4^{th}$ terms of both A.P.s

$|a_4-A_4|=|(-1+3d)-(-8+3d)|=|-1+3d+8-3d|=|-1+8|=|7|=7$

Final step: Hence, difference between their fourth terms is

$7.$

In an A.P

$1^{st}$ term is

$1$ and the last term is

$20$ . The sum of all terms is

$=399$ then

$n=$ ....

0%
$42$
0%
$38$
0%
$21$
0%
$19$

Explanation

The is given that,

First term

$(a)=1$

Last term

$(t_n)=20$

Sum of terms

$(S_n)=399$

We know that,

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

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

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

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

$\Rightarrow 399=\dfrac{n}{2}(1+20)\\$

$\Rightarrow 399=\dfrac{21n}{2}\\$

$\Rightarrow 21n=399 \times 2\\$

$\Rightarrow n=\dfrac{798}{21}\\$

$\Rightarrow n=38$

$a$ is constant then

$a + 2a + 3a + ..... + na$ is

0%
$\dfrac{an(n + 1)}{4}$
0%
$\dfrac{an(n + 1)}{2}$
0%
$\dfrac{n(n + 1)}{2}$
0%
$\dfrac{an(n + 1)}{6}$

Explanation

For the terms

$a,\ 2a,\ 3a,\ 4a...,\ na$

${ a }_{ 2 }-{ a }_{ 1 }=2a-a$

$=a$

${ a }_{ 3 }-{ a }_{ 2 }=3a-2a$

$=a$

Thus, the difference between consecutive terms of A.P. is the same.

Sum of first n terms of A.P. is given by

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

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

$=\dfrac { n }{ 2 } \left[ 2a+na-a \right]$

$=\dfrac { n }{ 2 } \left[ a\left( n+2 \right) -a \right]$

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

$=\dfrac { n }{ 2 } \left[ a\left( n+1 \right) \right]$

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

In an A.P, if

$a_{4} = 8 \,\& \,a = 2,$ then its common difference is

0%
$6$
0%
$4$
0%
$2$
0%
$10$

Explanation

nth term of the A.P. is given by,

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

$\therefore { a }_{ 4 }=a+\left( 4-1 \right) d$

$\therefore 8=2+3d$ (Given)

$\therefore 3d=6$

$\therefore d=2$

Thus, common difference is 2.

If sum of

$n$ term of

$A.P.$ is

$S_{n}$ and

$S_{2n}=3S_{n}$ , then

$S_{3n}: S_{n}$ will be:

0%
$10$
0%
$11$
0%
$6$
0%
$4$

Explanation

$S_{2n}=3 S_{n}$

$\dfrac{2n}{2}[2a(2n-1)d]=\dfrac{3n}{2}[2a+(n-1)d]$

$\Rightarrow 4a+4nd-2d=6a+3nd-3d$

$\Rightarrow nd+d=2a$
Now,

$S_{3n}:S_{n}$

$\dfrac{S_{3n}}{S_{n}}=\dfrac{\dfrac{3n}{2}[2a+(3n-1)d]}{\dfrac{n}{2}[2a+(n-1)d]}$

$\dfrac{S_{3n}}{S_{n}}=\dfrac{3[nd+d+3nd-d]}{[nd+d+nd-d]}$

$[\because 2a=nd+d]$

$=\dfrac{3\times 4nd}{2\ nd}$

$=\dfrac{12}{2}$

$=6$
Hence, option

$(C)$ is correct.

If sum of

$n$ terms of

$A.P$ is

$3n^2+5n$ , then its which term is

$164$ :

0%
$12^{th}$
0%
$15^{th}$
0%
$27^{th}$
0%
$20^{th}$

Explanation

Given:

$S_{n}=3n^{2}+5n$

$S_{1}=3(1)^{2}+5(1)=8$

$S_{2}=3(2)^{2}+5(2)=22$

$S_{3}=3(3)^{2}+5(3)=42$

$S_{4}=3(4)^{2}+5(4)=68$

$\therefore a_{1}=S_{1}=8$

$a_{2}=S_{2}-S_{1}\Rightarrow 22-8\Rightarrow 14$

$a_{3}=S_{2}-S_{1}\Rightarrow 42-22\Rightarrow 20$

$a_{4}=S_{2}-S_{1}\Rightarrow 68-42\Rightarrow 26$
Thus

$A.P.$ will be

$8,14,20,26,......164$

$a=8$ ,

$d=14-8=6$ and

$a_{n}=164$

$\therefore 164=a+(n-1)d$

$164=8+(n-1)$

$(n-1)=156/6=26$

$\therefore n=26+1=27$
Hence, option

$(C)$ is correct.

A manufacture company made

$15000$ mobile phones in its

$17^{th}$ year. The production has been increasing by

$500$ every year since the first year.
How many mobile phones did the company make in second year?

0%
$7000$
0%
$7500$
0%
$8000$
0%
$8500$

In the A.P.

$2, -2, -6, -10........$ common difference (d) is:

0%
$-4$
0%
$2$
0%
$-2$
0%
$4$

If we divide

$18$ into three parts that are in

$AP$ and whose product is

$120,$ then formed

$AP$ is given by

0%
$2, 7, 9$
0%
$2, 5, 11$
0%
$2, 6, 9$
0%
$2, 4, 10$

Dishan started reading a novel which had

$240$ pages. At the end of the first day, she was on page number

$45$ . She slowed after that. She read

$39$ pages every day after the first day. How long did it take Dishan to complete that novel?

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

A roll of thread

$264$

$cm$ long is cut into four parts to make four circles. The radii of the circles increases by

$1$

$cm$ consecutively. Find the radius of smallest circle?

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

M.C.M college had enrollment of

$1625$ students in the year of

$2014$ which increases by

$100$ students per year. What was the enrollment in the year

$2020?$

0%
$2125$
0%
$2225$
0%
$2235$
0%
$2215$

Find an

$AP$ whose general term is given by

$3n+1$ .

0%
$4, 7, 10, 13, .....$
0%
$3, 6, 9, 12, .....$
0%
$3, 7, 12, 15, .....$
0%
$2, 5, 9, 13, .....$

In a quiz competition, if the first prize is worth

$Rs. 200$ . Each other prize is

$Rs.15$ less than its proceeding prize, then the worth of

$9th$ prize is ____.

0%
$60$
0%
$70$
0%
$80$
0%
$90$

Which of the following

$AP$ can be formed if the third and fifth terms are

$7$ and

$11$ respectively?

0%
$3, 10,7, 13, 11, .......$
0%
$3, 5, 7, 9, 11, ......$
0%
$1, 2, 7, 9, 11, .......$
0%
Nine of these

The arithmetic progression whose

$2nd$ and

$5th$ terms are

$13$ and

$19$ is

0%
$11, 13, 15, 17, ......$
0%
$11, 13, 14, 17, ......$
0%
$11, 13, 14, 16, ......$
0%
None of these

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 15 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 15 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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