MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 10

Which term of the AP :

$3,8, 13, 18,..$ is

$78?$

0%
$t_{16}$
0%
$t_{18}$
0%
$t_{14}$
0%
None of these

Explanation

Given :- A.P. is

$3, 8 , 13, 18, ....$

To get :- which term is

$78$ .

solution :-

First term

$= 3 = a_1$

common diff.

$= d = (8 - 3) = (13 - 8) = (18 - 13) = 5$

$\therefore$ Let the required term be n

$\therefore a_n = 78$

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

$\Rightarrow 78 = 3 + (n - 1) d$

$\Rightarrow 78 = 3 + (n - 1) 5$

$\Rightarrow 78 = 3 + 5n - 5$

$\Rightarrow 78 = -2 + 5n$

$\Rightarrow 5n = 78 + 2 = 80$

$\Rightarrow 5n = 80$

$\Rightarrow n = 16$

$\therefore$ so the required term is

$16^{th}$ term.

i.e.

$t_{16}$

Find the common difference of an AP. whose first term is

$100$ and the sum of whose first six terms is five times the sum of the next six terms.

0%
$10$
0%
$-10$
0%
$5$
0%
$-5$

Explanation

Here

$a=100$

Let difference is

$d$ .

$\Rightarrow a_1+a_2+a_3+a_4+a_5+a_6=5(a_7+a_8+a_9+a_{10}+a_{11}+a_{12})$

So by the formula,

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

$a$ &

$l$ are the first and last term of an AP, we have

$6\left(\dfrac{a_1+a_6}{2} \right)=5\times 6\left(\dfrac{a_7+a_{12}}{2}\right)$

$\Rightarrow a_1+a_6=5(a_7+a_{12})$

$\Rightarrow a+a+5d=5(a+6d+a+11d)$

$\Rightarrow 2a+5d=10a+85d$

$\Rightarrow 80d=-8a$

$\Rightarrow d=\dfrac{-a}{10}\Rightarrow \dfrac{-100}{10}\Rightarrow -10$

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

$\dfrac {1}{3}(2n+1)$ , then the sum of its

$19$ term is

0%
$131$
0%
$132$
0%
$133$
0%
$134$

Explanation

Let

$T_n$ be the

$n^{th}$ term of an AP

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

On comparing with the given expression it is clear that

$d=\dfrac{2}{3}$ and

$a-d=\dfrac{1}{3}$

$\Rightarrow a-\dfrac{2}{3}=\dfrac{1}{3}$

$\Rightarrow a=1$

We know that sum of first

$n$ terms,

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

$a$ &

$d$ are the first term amd common difference of an AP.

$\therefore S_{19}=\dfrac {19}{2}\left [ 2+(18)\dfrac{2}{3} \right ]\\=133$

Hence, option

$(C)$ is correct.

Which term of AP :

$3, 15, 27, 39,..$ will be

$132$ more than its

$54th$ term?

0%
t $_{65}$
0%
t $_{64}$
0%
t $_{60}$
0%
None of these

Explanation

Clearly, the given Sequence is an AP with first term

$a=3$ and common difference

$d=15-3=12$ .
Let

$nth$ term be the

$132$ more than

$54th$ term of the given AP.

$\Rightarrow a_n=a_{54}+132$

$\Rightarrow a+(n-1)d=a+53d+132\Rightarrow (n-1)\times12=53\times12+132$

$\Rightarrow n-1=53+11$

$\Rightarrow n=65$
Hence,

$a_{65}$ is

$132$ more than

$a_{54}$

The sum of the first nineteen terms of an A.P.

$a_1,\, a_2,\, a_3$ .......... if it is known that

$a_4\, +\, a_8\, +\, a_{12}\, +\, a_{16}\, =\, 224$ is .......... .

0%
$1064$
0%
$896$
0%
$532$
0%
$448$

Explanation

We know that

$n^{th}$ term of an AP is given as,

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

And sum of first

$n$ terms,

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

$a$ and

$d$ are the first term amd common difference of an AP.

Since,

${ a }_{ 4 }+{ a }_{ 8 }+{ a }_{ 12 }+{ a }_{ 16 }=224$

$\therefore a+3d+a+7d+a+11d+a+15d=224\\ \Rightarrow 4a+36d=224\\\Rightarrow a+9d=56$

Now

$\displaystyle { S }_{ 19 }=\frac { 19 }{ 2 } \left[ 2a+\left( 19-1 \right) d \right] \\ =19\left[ a+9d \right] \\=19\left[ 56 \right] \\ =1064$

$9^{th}$ and

$19^{th}$ terms of an AP are

$35$ and

$75$ , then its

$20^{th}$ term is

0%
$78$
0%
$79$
0%
$80$
0%
$81$

Explanation

${ T }_{ 9 }=35\Rightarrow a+\left( 9-1 \right) d=35\Rightarrow a+8d=35$

${ T }_{ 19 }=75\Rightarrow a+\left( 19-1 \right) d=75\Rightarrow a+18d=75$

Solving these two we get

$d=4,a=3$

Therefore,

${ T }_{ 20 }=a+\left( 20-1 \right) d=3+19\times 4=79$

The sum of all integers between

$81$ and

$719$ which are divisible by

$5$ is

0%
$51800$
0%
$50800$
0%
$52800$
0%
None of these

Explanation

All integers between

$81$ and

$719$ which are divisible by

$5$ form an AP with common difference

$5$ .

First term

$a=85$ , last term

$l=715$ and common difference is

$d=5$

Then, apply the formula for

$n^{th}$ term of an AP

$715=85+\left( n-1 \right) 5$

$126=n-1$

$\Rightarrow n=127$

To get the required sum, apply the formula for sum of

$n$ terms of an AP

$\displaystyle { S }_{ n }=\dfrac { n }{ 2 } \left( a+l \right) \\=\dfrac { 127 }{ 2 } \left( 85+715 \right)\\ =50800$

The value of

$1 + 3 + 5 + 7 + 9 + ............. + 25$ is:

0%
$196$
0%
$625$
0%
$225$
0%
$169$

Explanation

We take the first term

$a=1$ , last term

$l=25$ , common difference

$d=2$ and the number of terms

$=n$ .

Then, since it is an A.P. series,

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

$n=\dfrac { l-a }{ d } +1\\ =\dfrac { 25-1 }{ 2 } +1\\$

$=\dfrac{25-1+2}{2}$

$=\dfrac{25+1}{2}$

$=\dfrac{26}{2}$

$=13$

Let

${ S }_{ n }$ be the sum of

$n=13$ terms

${ \therefore \quad S }_{ n }=\dfrac { n }{ 2 } \times \left( a+l \right) \\ =\dfrac { 13 }{ 2 } \times \left( 1+25 \right) \\ =13\times 13=169.$

So the required sum

$=169$ .

Ramkali saved

$Rs. 5$ in the first week of a year and then increased her weekly savings by

$Rs. 1.75.$ If in the nth week, her weekly savings become

$Rs. 20.75,$ find

$n.$

0%
$n=10$
0%
$n=12$
0%
$n=16$
0%
$n=20$

Explanation

Given that

$a =5$

$d = 1.75$

$a_{n} =20.75$

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

$20.75=5+\left(n-1\right)1.75$

$15.75=\left(n-1\right)1.75$

$\left(n-1\right)=\displaystyle \frac{15.75}{1.75}=\frac{1575}{175}=\frac{63}{7}=9$

$\therefore n-1 = 9$

$\therefore n=10$

The sum of

$3^{rd}$ and

$15^{th}$ elements of an arithmetic progression is equal to the sum of

$6^{th},\, 11^{th}\, and\, 13^{th}$ elements of the same progression. Then which element of the series should necessarily be equal to zero?

0%
$1st$
0%
$9th$
0%
$12th$
0%
None of the above

Explanation

Let the first term of AP be

$a$ and difference be

$d$

We know that

$n$ th term,

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

$a$ &

$d$ are the first term amd common difference of an AP.

$\therefore$ Then third term will be

$=a+2d$

$15^{th}$ will be

$=a+14d$

$6^{th}$ will be

$=a+5d$

$11^{th}$ will be

$=a+10d$

$13^{th}$ will be

$=a+12d$
Then the equation will be

$a+2d+a+14d = a+5d+a+10d+a+12d$

$2a+16d = 3a+27d$

$a+11d=0$

We understand

$a+11d$ will be the

$12^{th}$ term of arithmetic progression.
So, the correct answer is

$12$ .

The sum of first

$n$ terms of the

$A.P$ .

$\sqrt{2}\, +\, \sqrt{8}\, +\, \sqrt{18}\, +\, \sqrt{32}\, +\, ........$ is ......

0%
$\displaystyle \frac{n(n\, +\, 1)}{2}$
0%
$2n (n + 1)$
0%
$\displaystyle \frac{n(n\, +\, 1)}{\sqrt{2}}$
0%
$1$

Explanation

The series can be rewritten as

$\sqrt{2} + 2 \sqrt{2} + 3\sqrt{2} + 4 \sqrt{2} + .....$
So the nth sum

$S_{n} = \sqrt{2} \times \left(1 + 2 + 3 + ..... + n \right)$

$= \displaystyle \frac{\sqrt{2}\times n\times \left(n+1\right)}{2}$

$=\displaystyle \frac{n\left(n+1\right)}{\sqrt{2}}$

The

$9th$ term of an AP is

$499$ and

$499th$ terms is

$9.$ The term which is equal to zero is

0%
$501 th$
0%
$502 th$
0%
$508 th$
0%
None of these

Explanation

$a_{9}=a+8d$

$a_9=499 (given)$

$\therefore a+8d=499.........(i)$

$a_{499}=a+498d$

$a_{499}=9$

$\therefore a+498d=9.........(ii)$
Subtract (i) from (ii)

$\Rightarrow 490 d = - 490$

$\Rightarrow d = -1$
substitute the value of

$d$

$\Rightarrow a + 8(-1) = 499$

$\Rightarrow a = 507$

$\therefore$ For

$a_{n}=0$

$\Rightarrow a+(n-1)d=0$

$\Rightarrow 507+(n-1)(-1)=0$

$\Rightarrow 507=n-1$

$\Rightarrow n=508$
So,

$508^{th}$ term is equal to zero.

$8^{th}$ term of series

$\displaystyle 2\sqrt{2}+\sqrt{2}+0+......$ will be

0%
$\displaystyle -5\sqrt{2}$
0%
$\displaystyle 5\sqrt{2}$
0%
$\displaystyle 10\sqrt{2}$
0%
$\displaystyle -10\sqrt{2}$

Explanation

Given series is :

$2\sqrt2+ \sqrt2+0+.......$

Then,

$a=2\sqrt{2},d=(\sqrt 2-2\sqrt 2)=-\sqrt{2}$

$a_{8}=a+7d$

$\therefore 2\sqrt{2}+7\times (-\sqrt{2})$

$\therefore a_{8}=-5\sqrt{2}$

Find the sum of all whole numbers divisible by

$5$ but less than

$100$ .

0%
$950$
0%
$925$
0%
$880$
0%
$1050$

Explanation

All whole numbers divisible by

$5$ but less then

$100$ are

$0,5,10,15,20........95$
It is an A.P. with first term, common difference and last term as

$0, 5$ and

$95$ respectively.

$a=0,d=5,a_n=95$

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

$95=0+(n-1)5$

$5n=100$

$n=20$
Hence total numbers are

$=20$
Sum of numbers are

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

$S_{20}=\dfrac{20}{2}[0+(20-1)5]$

$S_{20}=10\times 95\\\ \ \ \ \ =950$
The sum of all natural number divisible by

$5$ but less than

$100$ is

$950$ .

The

$4th$ term from the end of the AP

$-11, -8, -5, ....................49$ is

0%
$37$
0%
$40$
0%
$43$
0%
$58$

Explanation

Given series of AP:

$- 11, -8, -5,.......49$
New series :

$49,........-5,-8,-11$

In new series of AP first term

$a = 49,$ common difference

$d = (-8) - (-5) = - 3$

$\displaystyle a_{4}=a+3d=49+3(-3)=49-9=40$

$\therefore$ Option B is correct.

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

0%
$530, 820$
0%
$179, 587$
0%
$560, 758$
0%
$164, 850$

Explanation

$\textbf{Step1-Find sequence of an A.P according to question}$

$\text{The lowest 3 digit number is }$

$101$

$\text{ and the highest number is}$

$998$ .

$\text{Therefore, the AP is }$

$101,104,107, ....... , 998$

$\text{ where, the first term is }$

$a=101$ ,

$\text{common difference is $$d=104 - 101 = 3$$ and }$

$\text{n$^{th}$ term is }$

$a_n=998$

$\textbf{Step2-Find the total number of terms of sequence}$

$\text{Let us find the number of terms is n}$

$\text{We know that the n$^{th}$}$

$\text{term of AP is given by}$

$a_{n} =$

$a+(n-1)d$

$\text{Now, substituting the values, we get:}$

$998=101+(n-1)3\\ \Rightarrow (n-1)3=998-101\\ \Rightarrow (n-1)3=897\\ \Rightarrow n-1=\dfrac { 897 }{ 3 } \\ \Rightarrow n-1=299\\ \Rightarrow n=299+1\\ \Rightarrow n=300$

$\textbf{Step3-Find sum of 300 terms of an A.P}$

$\text{We know }$

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

$\text{(1st term + last term) }$

$\text{Therefore, we have:}$

$S=\dfrac { 300 }{ 2 } (101+998)$

$S=150\times 1099$

$S=164850$

$\textbf{Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.}$

The sum of all odd integers between

$2$ and

$50$ divisible by

$3$ is

0%
$192$
0%
$216$
0%
$168$
0%
$240$

Explanation

The odd intergers between 2 and 50 divisible by 3 are

$3, 9, 15,......45$
This sequence forms an AP with first term

$a=3$ and common difference

$d=9-3=6$
Since,

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

$\Rightarrow 45=3+(n-1)6$

$\Rightarrow 42=(n-1)6$

$\Rightarrow 7=n-1$

$\Rightarrow n=8$
Number of terms in given sequence are 8.

$S_n=\dfrac n2(a+l)$

$\Rightarrow S_n=\dfrac{8}{2}(3+45)$

$\Rightarrow S_n=4(48)$

$\Rightarrow S_n= 192$

$\therefore$ Option A is correct.

If the

$5$ th term of an A.P is eight times the first term and

$8$ th term exceeds twice the

$4$ th term by

$3$ , then the common difference is

0%
$7$
0%
$5$
0%
$6$
0%
$3$

Explanation

$n^{th}$ term of an AP is given as,

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

Where

$a$ and

$d$ are first term and common difference respectively.

According to the given condition,

Thus,

${ t }_{ 5 } = a + 4d$

${ t }_{ 1 } = a$

${ t }_{ 8 } = a + 7d$

${ t }_{ 4 } = a + 3d$

Given,

${ t }_{ 5 } = 8{ t }_{ 1 }$

$\Longrightarrow a+ 4d= 8a\\ \Longrightarrow \quad 4d= 7a\\ \Longrightarrow d= \dfrac { 7a }{ 4 }$

Also given,

${ t }_{ 8 }$

$-2{ t }_{ 4 } = 3$

$\Longrightarrow a+ 7d- 2(a+ 3d)= 3\\ \Longrightarrow d - a= 3\\$

Substituting

$d= \dfrac { 7a }{ 4 }$ in the above equation,

$\dfrac{7a}{4}-a=3$

$\dfrac{3a}{4}=3$

we get

$a = 4$ and hence,

$d = 7$ .

$9^{th}$ term of an A.P. is

$47$ and

$16^{th}$ term is

$82$ , then the general term is

0%
$9n+2$
0%
$6n-7$
0%
$5n+2$
0%
none

Explanation

$a_9=a+8d=47$ ........(1)

$a_{16}=a+15d=82$ .....(2)
Subtract eq (1) by (2)

$7d=35$

$\therefore d=5$

$a+8d=47$

$a+8\times 5=47$

$\therefore a=7$
Hence general term is

$a_n = a+(n-1)d = 5n+2$

Find

${a}_{25}-{a}_{15}$ for the A.P

$-6, -10, -14, -18,..........$

0%
$-40$
0%
$-48$
0%
$40$
0%
$-44$

Explanation

Given series is:

$-6,-10,-14,-18.....$

$a=-6, d=-4$

As we know

$nth$ term,

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

$\therefore$

$a_{25}=a+24d$
&

$a_{15}=a+14d$

$a_{25}-a_{15}=(a+24d)-(a+14d)=10d$
In the following series

$d=-4$

$\therefore a_{25}-a_{15}=10\times (-4)=-40$

If an A.P is given by

$17, 14, ......-40$ , then

$6$ th term from the end is

0%
$-22$
0%
$-28$
0%
$-25$
0%
$26$

Explanation

Given an AP is 17,14...........................-40

Then we get AP is -40,-37,.............14,17

Then d=3,a=-40 n=6

Therefore,

$a_{6}=a+(n-1)d=(-40)+(6-1)3=-40+15=-25$

If three times the

$9$ th term of an A.P is equal to five times its

$13$ th term, then which of the following term will be zero.

0%
$26$ th
0%
$19$ th
0%
$21$ st
0%
$36$ th

Explanation

According to the question

$3(a+8d)=5(a+12d)$

$\Rightarrow 3a+24d=5a+60d$

$\Rightarrow 2a+36d=0$

$\Rightarrow a+18d=0$

$\Rightarrow a+(19-1)d=0$
Hence

$19^{th}$ term is zero

The total two-digit numbers which are divisible by

$5$ , are

0%
$19$
0%
$20$
0%
$17$
0%
$18$

Explanation

$10, 15, 20, ........ 95$

$n$ th term of AP

$=a+(n-1)d$

$\Rightarrow 95=10+(n-1)5$

$\Rightarrow 95=10+5n-5$

$\Rightarrow 95=5n+5$

$\Rightarrow n=18$
Hence, required numbers

$=18$

How many terms are there in the sequence

$15\cfrac{1}{2}, 13, ......-47$ ?

0%
$27$
0%
$26$
0%
$28$
0%
None

Explanation

Given series is

$15 \dfrac {1}{2}, 13, ....., -47$

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

$\therefore -47=\dfrac{31}{2}+(n-1)\left(\dfrac{-5}{2}\right)$

$-47=\dfrac{31}{2}-\dfrac{5n}{2}+\dfrac{5}{2}$

$-47=18-\dfrac{5n}{2}$

$\dfrac{5n}{2}=65$

$\therefore n=\dfrac{65\times 2}{5}=26$
Hence number of terms are

$26$

Find the

$n^{th}$ term and, hence, the

$25^{th}$ term of the following AP:

$2, 5, 8, 11$ ,.........

0%
$3n-1, 74$
0%
$2n, 73$
0%
$3n, 75$
0%
$3n+1, 72$

Explanation

Given series, 2, 5, 8, 11, ... is in AP

Here
First term

$a = 2$

Common difference

$d = 5 - 2$

$= 8 - 5$

$= 3$

We know that, general term

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

$\displaystyle =2+\left( n-1 \right) 3$

$\displaystyle =2+3n-3$

$\displaystyle =3n-1$

$\displaystyle { t }_{ 25 }=3\times 25-1$

$=75-1=74$

If the first, second and last term of an A.P. are

$x,y$ and

$2x$ respectively, its sum is

0%
$\cfrac{xy}{2(y-x)}$
0%
$\cfrac{3xy}{2(y-x)}$
0%
$\cfrac{2xy}{y-x}$
0%
$\cfrac{xy}{y-x}$

Explanation

Here

$a_1=x,$

$a_2=y$ and

$a_n=2x$ .

$\therefore \$ First term,

$a=x$ and common difference,

$d=y-x$
Since,

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

$\Rightarrow 2x=x+(n-1)(y-x)$

$\Rightarrow x=(n-1)(y-x)$

$\displaystyle \Rightarrow \frac {x}{y-x}=n-1$

$\displaystyle \Rightarrow \frac {x}{y-x}+1=n$

$\displaystyle \Rightarrow n=\frac {y}{y-x}$

Now, we know that,

$S_n= \dfrac n2(a+a_n)$

$\therefore \ S_n= \displaystyle \dfrac {\frac {y}{y-x}}2(x+2x)$

$\Rightarrow S_n= \displaystyle \dfrac {\frac {y}{y-x}}2(3x)$

$\Rightarrow S_n=\displaystyle \dfrac {3xy}{2(y-x)}$

Hence, option B is correct.

If in an A.P,

${a}_{9}=0, {a}_{19}=k$ , then

${a}_{29}$ is

0%
${k}^{2}$
0%
$10k$
0%
$9k$
0%
$2k$

Explanation

Let

$a$ and

$d$ be the first term and common difference of given AP.
Since,

$a_9 = 0 \Rightarrow a+8d=0$ ....(1) [using

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

$a_{19} = k \Rightarrow a+18d=k$ ....(2)
On subtracting (1) from (2). we get

$a+18d-a-8d = k$

$\Rightarrow 10d=k$

$\Rightarrow d=\dfrac k{10}$
By substituting

$d=\dfrac k{10}$ in (1), we get

$a+\dfrac {8k}{10}=0$

$\Rightarrow a= -\dfrac {8k}{10}$
Now,

$a_{29}=a+28d = -\dfrac {8k}{10}+\dfrac {28k}{10} = 2k$
Hence, option D is correct.

The common difference of

$-2, -4, -6, -8,...........$ is :

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

Explanation

$\textbf{Step-1: Write the property of Common difference of an A.P.}$

$\text{Given series: -2,-4,-6,-8...}$

$\text{We know that Common difference = Second term - Fiirst term = Third term - Second term}$

$\therefore$

$\text{Common difference=}$

$- 4 - ( - 2) = - 4 + 2 = - 2$

$\text{or - 6 - ( - 4) = - 6 + 4 = - 2}$

$\textbf{Hence, In the given sequence -2, -4, -6, ......, the common difference is -2, option - A.}$

The AP whose first term is 10 and common difference is 3 is

0%
10, 13, 16, 19, ...
0%
5,7,9, 11, ...
0%
8, 12, 16,20, ...
0%
All the above

Explanation

$a_1$ is the first term, then the nth term is

$a_1 + (n-1)d$ , where

$d$ is the common difference
Here

$a_1 = 10, d = 3,$
Hence,

$a_2 = a_1 + (2-1)d = 10 + 3 = 13$

$a_3 = a_1 + (3-1)d = 10 + 6 = 16$ and so on.....

Which term of the A.P

$5,15, 25 -----$ will be

$100$ more than its

$31^{st}$ term?

0%
$41^{st}$
0%
$42^{nd}$
0%
$40^{th}$
0%
$43^{rd}$

Explanation

The given A.P is

$5,15,25.......$ where the first term is

$a=5$ and the common difference

$d$ will be calculated by subtracting the first term from the second term as shown below:

$d=15-5=10$

Now,

$31$ st term

$=a+30d$ , substituting the values of

$a$ and

$d$ , we get:

$31$ st term

$=5+(30\times10)=5+300=305$

It is given that the A.P of the given terms is

$100$ more than its

$31$ st term, thus, we have:

$a_{ n }=305+100=405\\ \Rightarrow a+(n-1)d=405\\ \Rightarrow 5+(n-1)10=405\\ \Rightarrow (n-1)10=405-5\\ \Rightarrow (n-1)10=400\\ \Rightarrow n-1=\dfrac { 400 }{ 10 } \\ \Rightarrow n-1=40\\ \Rightarrow n=40+1\\ \Rightarrow n=41$

Hence, the

$41$ st term of the given A.P will be

$100$ more than its

$31$ st term.

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

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Practice Class 10 Maths Quiz Questions and Answers

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

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Practice Class 10 Maths Quiz Questions and Answers

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