MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 11

$S=18+17\cfrac{1}{3}+16\cfrac{2}{3}+.......=242$ , then the number of terms are:

0%
$22$
0%
$33$
0%
Both $22$ and $33$
0%
none of these

Explanation

Given Series in A.P.
Here ,

$a = 18, d = \dfrac{52}{3}-18 =\dfrac{ 52-54}{3} = -\dfrac{2}{3}$ and

$S_{n} = 242$
We know,

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

$\Rightarrow 242= \dfrac{n}{2} \left[2\times 18+\left(n-1\right)\dfrac{-2}{3}\right]$

$\Rightarrow 484 = n\left[36+\left(\dfrac{-2}{3}n+\dfrac{2}{3}\right)\right]$

$\Rightarrow 484 = n\left[\dfrac{108-2n+2}{3}\right]$

$\Rightarrow 1452 = n\left[110-2n\right]$

$\Rightarrow 1452= 110n-2n^2$

$\Rightarrow 2n^2-110n+1452 =0$

$\Rightarrow 2 \left(n^2-55n+726\right) =0$

$\Rightarrow n^2-55n+726 =0$

$\Rightarrow n^2-33n-22n+726=0$

$\Rightarrow n\left(n-33\right)-22\left(n-33\right)=0$

$\Rightarrow \left(n-33\right)\left(n-22\right) =0$

$n = 33$ and

$n = 22$

How many terms of the series

$54, 51, 48,......$ be taken so that their sum is

$513$ ?

0%
$18$
0%
$19$
0%
$20$
0%
$A$ and $B$

Explanation

Given series is

$54,51, 48,....$

Sum of

$n$ terms of an

$A.P.$ is given by:

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

$513=\dfrac{n}{2}[2\times 54+(n-1)3]$

$1026=n[108-3n+3]$

$1026=111n-3n^2$

$3n^2-111n+1026=0$

$n^2-37n+342=0$

$n^2-18n-19n+342=0$

$n(n-18)-19(n-18)=0$

$(n-18)(n-19)=0$
Then,

$n=18$ and

$n=19$

Which term of the A.P

$\ 3, 10, 17, .......$ will be

$84$ more than its

$13^{th}$ term?

0%
$23^{rd}$
0%
$25^{th}$
0%
$27^{th}$
0%
$28^{th}$

Explanation

Given A.P is

$1,10,17.....$

$13^{ th}$ term

$a_{13}=a+12d$

$\Rightarrow 3+12\times 7$

$\Rightarrow 87$
According to the question

$a_n=a_{13}+84$

$\Rightarrow 87+84$

$\Rightarrow 171$

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

$171=3+(n-1)7$

$171=3+7n-7$

$7n=175$

$\therefore n=25$
Hence

$25^{th}$ term is

$84$ more than

$13^{ th}$ term.

$3x, 5x+1$ and

$8x-1$ are in

$A.P$ , then the value of

$x$ is

0%
$2$
0%
$-3$
0%
$-4$
0%
$3$

Explanation

Given:

$3x, 5x+1, 8x-1$ are in AP.

We know if

$a, b, c$ are in A.P. then

$b-a = b-c$ or

$2b = a+c$

$\therefore 2\left(5x+1\right) = 3x+8x-1$

$\therefore 10x+2 = 11x-1$

$\therefore 11x-10x = 2+1$

$\therefore x = 3$

Find the sum of all three-digit numbers which leave a remainder

$2$ , when divided by

$6$ .

0%
$82656$
0%
$82658$
0%
$82650$
0%
$82654$

Explanation

All three-digit numbers and lying between

$100$ and

$999$ which leave a remainder of

$2$ when divided by

$6$ .

$104, 110, 116, .........., 998$

As we know nth term,

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

& Sum of first

$n$ terms,

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

$a$ &

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

$\therefore 998=104+(n-1)\times 6$

$n=150$

Now,

$S_{150} =\dfrac{150}{2}[2\times 104 +(150-1)\times 6 ]$

$=75(208+894)$

$=82650$

Hence, this is the answer.

Find the sum of the series up to

$90$ terms:

$1+2+3+3+4+6+5+6+9+$ ....

0%
$1613$
0%
$3225$
0%
$6450$
0%
$9675$

Explanation

Series

$\displaystyle =(1+2+3)+(3+4+6)+(5+6+9)+.....$

$\displaystyle =6+13+20+....(30 terms)$
Here,

$\displaystyle d=7,a=6,n=30$

As we know,

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

$\displaystyle \therefore \quad { S }_{ 30 }=\frac { 30 }{ 2 } \left[ 12+\left( 29 \right) \left( 7 \right) \right] =3225$

If the third and

$11^{th}$ term of an A.P are

$8$ and

$20$ respectively, find the sum of first ten terms.

0%
$\displaystyle 105\frac{1}{2}$
0%
$108$
0%
$\displaystyle 117\frac{1}{2}$
0%
$\displaystyle 203\frac{1}{2}$

Explanation

As we know nth term,

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

& Sum of first

$n$ terms,

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

$a$ &

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

Since,

$a_3=a+2d=8$ ------(1)

$a_{11}=a+10d=20$ -----(2)

By solving above two equations, we get

$a=5,d=\dfrac{3}{2}$

$\displaystyle \therefore S_{10}=\frac{10}{2}\left [ 2a+\left ( 10-1 \right )d \right ]$

$\displaystyle =5\left [ 2\times 5+9\times \frac{3}{2} \right ]$

$\displaystyle =117\frac{1}{2}$

Which of the following list of numbers does form an

$A.P$ ?

0%
$2,4,8, 16$ , ....
0%
$2,\dfrac {5}{2},3,\dfrac {7}{2}$ , ...
0%
$0.2, 0.22, 0.222, 0.2222$ , ...
0%
$1,3,9,27$ , ...

Explanation

Option

$B$ is the answer

$\displaystyle 2,\frac { 5 }{ 2 } ,3,\frac { 7 }{ 2 } ,...$ is in the form of

$x_1,x_2,x_3,x_4,.....$

Any series is said to be in AP if and only if all the terms in series have same common difference

$c.d$ .

$\Rightarrow c.d = x_2-x_1=x_3-x_2=x_4-x_3$

$\Rightarrow c.d=\dfrac{5}{2}-2= 3-\dfrac{5}{2}=\dfrac{7}{2}-3 = \dfrac{1}{2}$

$\Rightarrow c.d=\dfrac { 1 }{ 2 }$ (same common difference)

Hence this series of numbers form

$A.P$

Find the sum of first

$31$ term term of an A.P whose

$n^{th}$ term is

$\displaystyle \left ( 3+\frac{2n}{3} \right )$ .

0%
$\displaystyle 423\frac{2}{3}$
0%
$\displaystyle 413\frac{1}{3}$
0%
$\displaystyle 417\frac{2}{3}$
0%
$\displaystyle 419\frac{2}{3}$

Explanation

As we know that the sum of first

$n$ terms,

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

$a$ &

$l$ are the first term & last term of an AP.

$\displaystyle a_{1}=3+\frac{2}{3}=\frac{11}{3}$

$\displaystyle a_{31=}=3+\frac{2}{3}\times 31=3+\frac{62}{3}=71$

$\displaystyle \therefore S_{31}=\frac{31}{2}\times \left ( \frac{11}{3}+\frac{71}{3} \right )=\frac{31}{2}\times \frac{82}{3}=\frac{1271}{3}$

$\displaystyle =423\frac{2}{3}$

$1+3+5+\cdots\cdots+99=$

0%
$(99)^2$
0%
$(49)^2$
0%
$(50)^2$
0%
$(100)^2$

Explanation

The given series is an AP
Let

$a$ be the first term and

$d$ be their common difference of the AP.
Then,

$n th$ term of the is AP

$a_n = a + (n-1)d$
Here, in the given AP,.

$a = 1; d = 3-1 = 2$
Given,

$a_n = 99$

$=> 1 + (n-1)2 = 99$

$=> n = 50$
Also, Sum to

$n$ terms of an AP,

$S_n = \frac { n }{ 2 } (2a+(n-1)d)$
So, sum to

$50$ terms,

$S_{50}= \frac { 50 }{ 2 } (2(1)+(50-1)2) = 50^2$

A man saved Rs.

$320$ during the first month Rs.

$360$ in the second month Rs.

$400$ in the third month, if he continues his saving in this sequence in how many months will be save Rs.

$20,000$ ?

0%
$28$
0%
$25$
0%
$22$
0%
$20$

Explanation

a=320, d=40

$\displaystyle S_{n}=20000$
It is required to find n

$\displaystyle S_{n}=20000=\frac{n}{2}\left [ 2\times 320+\left ( n-1 \right )\times 40 \right ]$

$\displaystyle \Rightarrow 500=\frac{n}{2}\left [ 15+\left ( n-1 \right ) \right ]=\frac{n}{2}\left ( n+15 \right )$

$\displaystyle n^{2}+15n-1000=0$

$\displaystyle \Rightarrow n=25$

Find the number of terms in an arithmetic progression for which the first term is

$4$ , last term is

$22$ and the common difference is

$\displaystyle\dfrac{1}{4}$

0%
$70$
0%
$71$
0%
$72$
0%
$73$

Explanation

Let

$a$ be the first term and

$d$ be their common difference of the AP.
Then,

$n th$ term of the AP

$, T_n = a + (n-1)d$
Here, in the given AP,.

$a = 4; d = \dfrac {1}{4}$

Given

$T_n= 22$

$\Rightarrow 4+ (n-1)\times \dfrac {1}{4} = 22$

$\Rightarrow (n-1)\times \dfrac {1}{4} = 18$

$\Rightarrow n-1 =72$

$n = 73$

Hence there are

$73$ terms in the series .

If the seventh term of an AP is 25 and the common difference is 4, then find the 15th term of AP.

0%
55
0%
50
0%
57
0%
52

Explanation

Let

$a$ be the first term and

$d$ be their common difference of the AP.
Then,

$n th$ term of the AP

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

Given,

$T_7 = 25$

$=> a + (7-1)d = 25$

$=> a + 6d = 25$

$=> a + 6 \times 4 = 25$

$=> a = 1$

So,

$T_{15} = a + (15-1)d = 1 + 14(4) = 57$

$8$ times the

$8th$ term of an arithmetic progression is equal to

$12$ times the

$12th$ term, then the

$20th$ term is

0%
$12$ times to $8th$ term
0%
$8$ times to $12th$ term
0%
$0$
0%
Cannot be determined

Explanation

Let

$a$ be the first term and

$d$ be their common difference of the AP.
Then,

$n th$ term of the AP

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

Given,

$8 \times T_8 = 12 \times T_{12}$

$\Rightarrow 8(a + (8-1)d) = 12(a + (12-1)d)$

$\Rightarrow 8(a + 7d) = 12(a + 11d)$

$\Rightarrow 2(a + 7d) = 3(a + 11d)$

$\Rightarrow 2a + 14d = 3a+33d$

$\Rightarrow a = - 19d$

So,

$T_{20} = a + (20-1)d = -19d + 19d = 0$

What is the common difference of an AP whose n

$^{th}$ term is

$xn+y$ ?

0%
$x+y$
0%
$y$
0%
$x$
0%
$-x$

Explanation

Given that, the

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

$xn+y$ .

To find out: The common difference of the A.P.

We know that, the common difference of an A.P. is the difference between any two consecutive terms.

That is,

$d=a_{n+1}-a_n$

We have

$a_n=xn+y$

$\therefore \ a_{n+1}=x(n+1)+y=nx+x+y$

Hence,

$d=(nx+x+y)-(nx+y)$

$\Rightarrow nx+x+y-nx-y$

$\therefore \ d=x$

Hence, the common difference of the given A.P. is

$x$ .

Find the common difference of an AP

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

Explanation

Let

$a$ be the first term and

$d$ be their common difference of the AP.
Then,

$n th$ term of the AP

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

Given,

$a_3 = 6$

$=> a + (3-1)d = 6$

$=> a + 2d = 6$ -- (1)

Also,

$a_17 = 34$

$=> a + (17-1)d = 34$

$=> a + 16d = 34$ -- (2)

Subtracting eqn 2 from eqn 1, we get

$14d = 28$

$=> d = 2$

Which of the following does not belong to the series 8, 11, 14, 17,20

$\dots\dots$ ?

0%
120
0%
131
0%
144
0%
150

The first term and last term number is

$12$ and

$30$ . The total number of term is

$50$ . Find its sum.

0%
$1,100$
0%
$1,250$
0%
$1,150$
0%
$1,050$

Explanation

Given,

$n=50$ , first term

$=12$ and last term

$=30$

Sum of first

$n$ numbers of an A.P. is

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

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

$+$ Last term]

$= \dfrac{50}{2} [12 + 30]$

$= 25 [42]$

$S_{50} = 1,050$

Find the sum of first 20 terms of the sequence whose nth term is

$2n+3n$

0%
1050
0%
645
0%
1100
0%
1000

Explanation

Given:

$a_n=2n+3n$

Put

$n=1$ in

$a_n$ , we get

$a_1=2(1)+3(1)=2+3=5$

Put

$n=2$ in

$a_n$ , we get

$a_2=2(2)+3(2)=4+6=10$

Put

$n=3$ in

$a_n$ , we get

$a_3=2(3)+3(3)=6+9=15$

So, Given A.P is

$5, 10, 15,....$

Hence,

$a=5$ and

$d=a_2-a_1=10-5=5$

Now, Sum of first

$20$ terms is given by

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

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

$\Rightarrow S_{20}=10[10+19(5)]$

$\Rightarrow S_{20}=1050$

So,

$\text{A}$ is the correct option.

Find the sums given below:

$-5+(-8)+(-11)+......+(-230)$

0%
$-8930$
0%
$8769$
0%
$6778$
0%
$6667$

Explanation

Given series can also be written as,

$-5-8-11-14-....-230$

It can be observed that difference between any two consecutive terms of the given series is constant and equal to

$-3$ .

Thus given series is an AP whose common difference and first term as

$-5$ and

$-3$ respectively.

As we know

$n^{th}$ term is given as

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

& Sum of first

$n$ terms,

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

$a$ &

$l$ are the first term & last term of an AP.

So,

$a=-5, d=-3$

$\Rightarrow -230=-5-3(n-1)$

$\Rightarrow 3(n-1)=225$

$\Rightarrow n-1=75$

$\Rightarrow n=76$

Sum of first

$n$ terms is given as,

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

$a$ &

$l$ are the first term & last term of an AP.
Thus,

$S=\dfrac {76}{2}(-5+-230)\\=38\times (-235)\\=-8930$

The sum of the three numbers in A.P is

$21$ and the product of the first and third number of the sequence is

$45$ . What are the three numbers?

0%
$3, 7,$ and $11$
0%
$9, 7,$ and $5$
0%
$3, 9,$ and $15$
0%
Both $A$ and $B$
0%
None of these

Explanation

Let the three numbers in A.P be

$a - d, a, a + d.$
Then,

$\begin{aligned}{}a - d + a + a + d &= 21\\3a &= 21\\a &= 7\end{aligned}$

And,

$\begin{aligned}{}(a - d)(a + d) &= 45\\{a^2} - {d^2}& = 45\\(7)^2-d^2&=45\\49-d^2&=45\\{d^2} &= 4\\d &= \pm 2\end{aligned}$

Hence, the numbers are

$5, 7$ and

$9$ when

$d = 2$ and

$9, 7$ and

$5$ when

$d = -2$ .

In both cases, numbers are the same.

Find the sum of the AP:

$2,5,8,11,\dots$ to first

$11$ terms.

0%
$187$
0%
$178$
0%
$192$
0%
$171$

Explanation

The sum of first

$n$ terms of an AP whose first term is

$a$ and common difference is

$d$ is given by,

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

Here

$d = 5 - 2$

$= 3$

$a = 2$

So,

$S_{11}= \dfrac{11}{2} [4 + (11 - 1) \times 3]$

$= \dfrac{11}{2} [4 + 30]$

$= \dfrac{11}{2} \times 34$

$= 11 \times 17$

$= 187$

Hence, option

$A$ is correct.

The sum of the third and ninth terms of an A.P is

$8$ . Find the sum of the first

$11$ terms of the progression.

0%
$44$
0%
$22$
0%
$19$
0%
$18$
0%
None of these

Explanation

As we know that,

$n^{th}$ term,

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

Sum of first

$n$ terms,

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

$a$ &

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

It is given that sum of third and ninth terms is

$8$ .

Therefore,

$(a+2d)+(a+8d)=8$

$\Rightarrow 2a+10d=8$ ...(1)

Therefore, sum of first

$11$ terms:

$S_{11}=\dfrac{11}2[2a+(11-1)d]$

$\Rightarrow S_{11}=\dfrac{11}2[2a+10d]$

$\Rightarrow S_{11}=\dfrac{11}2[8]$ {From (1)}

$\Rightarrow S_{11}=44$

Hence, option A is correct.

What is the sum of all positive integers up to

$1000$ , which are divisible by

$5$ and are not divisible by

$2$ ?

0%
$10,050$
0%
$5050$
0%
$5000$
0%
$50,000$

Explanation

The positive integers, which are divisible by

$5$ , are

$5, 10, 15, \dots, 1000.$ Out of these

$10,20,30,\dots, 1000$ are divisible by

$2.$

Thus, we have to find the sum of the positive integers

$5, 15, 25, ...., 995$

Clearly above sequence is an AP with the first term

$5$ and the common difference of

$10.$

$n$ is the number of terms in this sequence then,

$\begin{aligned}{}5 + 10\left( {n - 1} \right) &= 995\\5 + 10n - 10 &= 995\\10n& = 1000\\n &= 100\end{aligned}$

Now by using the formula for the sum of the first

$n$ terms of an AP,

$\begin{aligned}{}{S_{100}} &= \frac{{100}}{2}\left[ {2 \times 5 + \left( {100 - 1} \right)10} \right]\\ &= 50\left( {10 + 990} \right)\\& = 50 \times 1000\\ &= 50000\end{aligned}$

If the

$5th$ and

$8th$ term of an AP are

$6$ and

$15$ respectively, find the

$19th$ term

0%
$48$
0%
$54$
0%
$19$
0%
$57$

Explanation

Given:

$a_5=6$ and

$a_8=15$

$\Rightarrow$

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

$\Rightarrow$

$6=a+(5-1)d$

$\Rightarrow$

$6=a+4d$

$\Rightarrow$

$a+4d=6$ ----- ( 1 )

Now,

$\Rightarrow$

$a_8=a+(8-1)d$

$\Rightarrow$

$15=a+7d$

$\Rightarrow$

$a+7d=15$ ----- ( 2 )

Subtracting equation ( 1 ) from ( 2 ) we get,

$\Rightarrow$

$3d=9$

$\Rightarrow$

$d=3$

Substituting value of

$d$ in equation ( 1 ) we get,

$\Rightarrow$

$a+(4)(3)=6$

$\Rightarrow$

$a+12=6$

$\Rightarrow$

$a=-6$

Now,

$\Rightarrow$

$a_{19}=a+(19-1)d$

$\Rightarrow$

$a_{19}=-6+18(3)$

$\Rightarrow$

$a_{19}=-6+54$

$\therefore$

$a_{19}=48$

The

$17$ th term of an AP exceeds its

$10$ th term by

$7$ . Find the common difference

0%
$\dfrac{3}{7}$
0%
$\dfrac{-3}{7}$
0%
$\dfrac{7}{3}$
0%
$\dfrac{-7}{3}$

Explanation

$a_7=a+6d$
And,

$a_{10}=a+9d$
Or,

$a+9d-a-6d=7$
Or,

$3d=7$
Or,

$d=7/3$

Subba Rao started work in

$1995$ at an annual salary of Rs.

$5000$ and received an increment of Rs.

$200$ each year. In which year did his income reached Rs.

$7000$ ?

0%
$2005$
0%
$2002$
0%
$2004$
0%
$2007$

Explanation

From the given problem we can form the AP as given below

$5000, 5200, 5400, 5800, ........, 7000.$

Here

$a_n=7000$

$a=5000$ and

$d=200$

$7000=5000+200(n-1)$

Or,

$200(n-1)=2000$

Or,

$n-1=10$

Or,

$n=11$

So in

$2005$ his income reaches

$Rs7000$

Find the sum of first 10 terms of the arithmetic sequence

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

0%
90
0%
95
0%
96
0%
100

Explanation

The sum of first n terms of arithmetic series formula is given by the formula,

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

$= 10$

$a =$ first term

$=1$

$d =$ common difference of A.P.

$=3-1=2$
On substituting the given values in the formula, we get

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

$= 5 [2 + 18]$

$= 5[20]$

$= 100$
So, the sum of first 10 terms of the A.P. is

$100$

Find the sum of first 10 terms of the arithmetic series if

$a_1 = 2$ and

$a_{10} = 22$ .

0%
$155$
0%
$120$
0%
$165$
0%
$130$

Explanation

The sum of $n$ terms of arithmetic sequence can be calculated by the formula

$S_n =\dfrac{n}{2}(a_1+a_n)$

$n=10$

Given,

First term $a_1=2$

Tenth term $a_{10}=22$

$\Rightarrow S_n=\dfrac{10}{2}(22+2)$

$\Rightarrow S_n=10(12)=120$

Hence, option $B$ is correct

What is the sum of first 20 multiples of 2?

0%
$225$
0%
$420$
0%
$210$
0%
$217$

Explanation

The sum of first n terms of arithmetic series formula can be written as,

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

Where n = number of terms $\Rightarrow$ 20

a = first odd number $\Rightarrow$ 2

d = common difference of A.P. $\Rightarrow$ 2

Apply the given data in the formula,

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

$= 10[4 + (19)2]$

$= 10 [4 + 38]$

$= 10 [42]$

$= 420$

So, the sum of first 20 multiples of 2 is 420.

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 11 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 11 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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