MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 14

The common difference of an Arithmatic Progression, whose

$3^{rd}$ term is

$5$ and

$7^{th}$ term is

$9$ , is ...............

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

Explanation

We know that

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

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

where

$a=$ First term,

$d=$ Common difference and

$n=$ Position of term.

Given

$3^{rd}$ term is

$5$ ,

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

$\Rightarrow 5 = a+2d$

$\Rightarrow a = 5-2d$ ........

$(1)$

Given

$7^{th}$ term is

$9$ ,

${a}_{7} = a+(7-1)d$

$\Rightarrow 9 = a+6d$

$\Rightarrow a =9 -6d$ .......

$(2)$

From

$(1)$ and

$(2)$

$5-2d = 9 -6d$

$\Rightarrow 6d-2d = 9-5$

$\Rightarrow 4d = 4$

$\Rightarrow d = 1$
So , common difference is

$1$

The sum of first

$10$ terms and

$20$ terms of an AP are

$120$ and

$440$ respectively. What is the common difference?

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

Explanation

Let the first term be

$a$ and common difference be

$d$ .
Sum of first n terms in an A.P.

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

So, sum of first

$10$ terms

$=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$\implies 120 =5(2a+9d)$

$\implies 24=2a+9d$ ..............

$(i)$
Sum of first 20 terms

$=\frac { 20 }{ 2 } (2a+(20-1)d)$

$\implies 440 =10(2a+19d)$

$\implies 44=2a+19d$ .........

$(ii)$
Subtracting equation (i) from (ii) gives

$20=10d$

$\implies d=2$
Common difference =2

Hence, option B is correct.

In a given A.P.,

${ T }_{ 25 }-{ T }_{ 20 }=15$ .

$\therefore$

$d=$ ............. for the A.P.

0%
$5$
0%
$3$
0%
$25$
0%
$120$

Explanation

Given:

$T_{25}-T_{20}=15$ ...........

$(i)$

We know that,

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

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

$\therefore { T }_{ 25 }={ a }_{ 1 }+(25-1)d={ a }_{ 1 }+24d$ and

${ T }_{ 20 }={ a }_{ 1 }+(20-1)d={ a }_{ 1 }+19d$
Now,

${ T }_{ 25 }-{ T }_{ 20 }={ a }_{ 1 }+24d-{ a }_{ 1 }-19d=5d$

$\implies 15=5d$ ...... From

$(i)$

$\implies d=3$

Hence, the answer is

$3$ .

................ can be one of the term in Arithmetic progression 4, 7, 10, ...............

0%
103
0%
123
0%
171
0%
99

Explanation

The nth term of an AP is given by

$t_n = a + (n - 1) d$
In the given AP,

$a = 4, d = 3$
So,

$a + (n - 1) d = 4 + (n - 1) 3$

$= 4 + 3n - 3$

$= 3n + 1$
Now, if we equate the first option

$103 = 3n - 1$
we get

$n = 34$ which is a possible value for

$n$
However, if we equate any of the other values we do not get a positive integer value of

$n$ .
This cannot be possible since

$n$ , which is the number of terms has to be a positive integer.

$1 + 6 + 11 + 16 + ... + x = 148$ , then the value of

$x$ is

0%
$36$
0%
$35$
0%
$-36$
0%
None of these

Explanation

Given series is

$1 + 6 + 11 + 16 + ..... + x = 148$
Here,

$a = 1, d = 6 - 1 = 11 - 6 = 16 - 11 = 5$ and

$S_{n} = 148$

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

$\therefore 148 = \dfrac {n}{2}[2 \times 1 + (n - 1)5]$

$\Rightarrow 296 = 2n + 5n^{2} - 5n$

$\Rightarrow 296 = 5n^{2} - 3n$

$\Rightarrow 5n^{2} - 3n - 296 = 0$

$\Rightarrow (n - 8)(5n + 37) = 0$

$\Rightarrow n = 8, n = \dfrac {-37}{5}$

$\therefore n = 8$

$(\because n$ cannot be negative)
Now,

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

$x = 1 + (8 - 1)\times 5$

$\Rightarrow x = 1 + 35 \Rightarrow x = 36$ .

Find the sum of first 25 terms of an A.P. whose

$n^{th}$ term is given by

$T_n = (7 - 3n)$

0%
800
0%
-803
0%
735
0%
None of these

Explanation

Given,

$T_{n}=7-3n$ .

Let us try to reaarange it into the form,

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

where

$a$ is the first term,

$n$ is the number of terms and

$d$ is the common difference.

$T_{n}=7-3n$

$\Rightarrow$

$T_{n}=4+3-3n$

$\Rightarrow$

$T_{n}=4-3(n-1)$

So,here

$a=4,d=-3$

Sum of

$n$ terms in an AP can be calculated using the formula

$S_{n}=\frac{n}{2}(a+T_{n})$

$\Rightarrow$

$S_{n}=\frac{n}{2}(4+7-3n)$

$\Rightarrow$

$S_{n}=\frac{n}{2}(11-3n)$

$\Rightarrow$

$S_{n}=\frac{25}{2}(-64)$ (

$\because n=25$ )

$\Rightarrow$

$S_{n}=-800$

$\therefore$ No Option

The

$54^{th}$ and

$4^{th}$ terms of an A.P. are

$-61$ and

$64$ respectively. Find the

$23^{rd}$ term.

0%
$\dfrac {35}2$
0%
$\dfrac {33}2$
0%
$\dfrac {37}2$
0%
$\dfrac {39}2$

Explanation

Here we use the formula for nth term of an AP:

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

So we have

$a + 53d = -61 ....(1)$
&

$a + 3d = 64 ....(2)$

$50d = -125$ [Subtracting

$(2)$ from

$(1)$ ]

$d = \dfrac {-5}{2},$ using this value in

$(2)$

$\implies a = 64 - 3d = 64 + \dfrac {15}{2} = \dfrac {143}{2}$

$a_{23} = \dfrac {143}{2} - 22 \times \dfrac {5}{2} = \dfrac {33}{2}$ .

The sum of all numbers of the form

$2k + 1$ , where

$k$ takes on integral values from

$1$ to

$n$ is:

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

Explanation

The sum of numbers from

$2K+1$ , where

$K$ started with

$1,2,3,4,5......$

then the sum will be

$3+5+7+9+11+.......$

Now using sum of series

$S_n$ =

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

Now put the value for

$a=3\ and\ d=2$

$S_n$ =

$\frac{n}{2}[2\times3+(n-1)\times 2]$

$S_n$ =

$\frac{n}{2}$

$[6+2n-2]$

$S_n$ =

$\frac{n}{2}[4+2n]$

$S_n$ =

$n(n+2)$ .

so option

$C$ is correct.

The tenth term of an A.P. :

$\sqrt {2}, 3\sqrt {2}, 5\sqrt {2}, 7\sqrt {2}$ , ..... is:

0%
$10\sqrt {2}$
0%
$12$
0%
$19\sqrt {2}$
0%
$11\sqrt {2}$

Explanation

Given A.P. is

$\sqrt {2}, 3\sqrt {2}, 5\sqrt {2}, 7\sqrt {2}....$

Its first term,

$a = \sqrt {2}$

Common difference,

$d = 3\sqrt {2} - \sqrt {2} = 2\sqrt {2}$

To find out: The tenth term of the A.P.

We know that, the

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

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

$\therefore \ T_{10} = a + (10 - 1)d$

$\Rightarrow a + 9d$

$\Rightarrow \sqrt {2} + 9(2\sqrt {2})$

$\Rightarrow \sqrt {2} + 18\sqrt {2}$

$\therefore \ t_{10}= 19\sqrt {2}$

Hence, the tenth term of the given A.P. is

$19\sqrt {2}$ .

$0$ (zero) is a term of the A.P

$40,37,34,31,....$ . True or false?

0%
True
0%
False

Explanation

We have,

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

Let's check

$0=40+(n-1)(-3)$

solving the above equation, we get,

$n=\dfrac{43}{3}$

as,

$n$ is not an integer value,

$0$ cannot be a term of series .

Match the APs given in column A with suitable common differences given in column B.

Column A Column B

$-2, 2, 6, 10,...$ 1.

$\dfrac{2}{3}$

B.

$a = 18, n = 10, a_n = 0$

$-5$

$a = 0, a_{10} = 6$ 3.

$4$

$a_2 = 13, a_4 =3$

$-4$

$-2$

$\dfrac{1}{2}$

$5$

0%
A-3, B-5, C-1, D-2
0%
A-4, B-1, C-3, D-2
0%
A-2, B-7, C- 5, D-6
0%
A-7, B-4, C-2, D- 2

Explanation

1. Given sequence

$A:-2,\ 2,\ 6,\ 10,\ ...$

Here, first term

$a=-2$ and common difference

$d=6-2=4$

$\therefore\ A\rightarrow 3$

2. Given,

$B:a=18,n=10,a_n=0$
Now,

$a_n=a+(n-1)d=0$

$\Rightarrow 9d=-18$

$\Rightarrow d=-2$

$\therefore\ B\rightarrow 5$

3. Given,

$C:a=0,a_{10}=6$

$\Rightarrow a+9d=6$

$\Rightarrow d=\dfrac69$

$\Rightarrow d=\dfrac23$

$\therefore C\rightarrow1$

4. Given,

$D:a_2=13,a_{4}=3$

$\Rightarrow a+d=13$ ...(i)

$\Rightarrow a+3d=3$ ...(ii)
On subtracting (i) from (ii), we get

$2d=-10$

$\Rightarrow d=-5$

$\therefore D\rightarrow 2$

Find the sum of the first

$15$ terms of the sequences having

$n$ th term as

${ x }_{ n }=6-n$

0%
$30$
0%
$-30$
0%
$20$
0%
$-20$

Explanation

Formula,

$S_n$ =

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

Given,

$a_n=6-n$

$a_1=6-1=5$

$a_{2}=6-2=4$

$d= 4-5 = -1$

$S_{15}$ =

$\frac{15}{2}[2(5)+(15-1) \times (-1)]$

$\therefore S_{15}=-30$

The sum of an A.P. whose first term is a, second term is b and the last term is c is equal to

$\dfrac{(a+c)(b+c-2a)}{2(b-a)}$ .

0%
True
0%
False

Explanation

$T_1=a, T_2=a+d=b$

$\therefore d=b-a$

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

$n-1=\dfrac{c-a}{d}=\dfrac{c-a}{b-a}$

$\therefore n=\dfrac{c-a}{b-a}+1=\dfrac{c+b-2a}{b-a}$ ..

$(1)$

$S_n=\dfrac{n}{2}[a+T_n]=\dfrac{(c+b-2a)}{2(b-a)}(a+c)$

Identify which of the following list of numbers is an arithmetic progression?

0%
$1,1,2,3,5,...$
0%
$2,3,5,7,11,...$
0%
$10,100,1000,...$
0%
$12,18,24,30,...$

Explanation

The sequence given in option '

$a$ ' to '

$b$ ' do not have a fixed difference between its consecutive numbers. Only the sequence

$12,18,24, 30,...$ has a difference of

$6$ between its two consecutive numbers. Thus the correct answer is option '

$(D)$ '.

$7th$ and

$13th$ terms of an

$A.P$ . Be

$34$ and

$64$ , respectively, then its

$18th$ terms is:

0%
$87$
0%
$88$
0%
$89$
0%
$90$

Explanation

The general term of an

$AP$ is given by

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

$a$ &

$d$ are the first and common difference of an

$AP$ respectively.

${a}_{7}=34$

${a}_{13}=64$

$a+6d=34$

$a+12d=64$

$-6d=-30$

$d=\cfrac{30}{6}=5$

$\therefore a= 4.$

${a}_{18}=a+17d=4+17(5)=89$

If two terms of an arithmetic progression are known, then the two terms can be represented using which of the formula below?

0%
$t_{n1} = a - (n_1-1)d\$ and $\ t_{n2} = a - (n_2-1)d$ .
0%
$t_{n1} = 2a + (n_1-1)d$ and $t_{n1} = 2a + (n_2-1)d$ .
0%
$t_{n1} = a + (n_1-1)d$ and $t_{n1} = a + (n_2-1)d$ .
0%
$t_{n1} = a + (n_1-2)d$ and $t_{n1} = a + (n_2-2)d$ .

Explanation

If two terms of an arithmetic progression are known, then the two terms can be represented using

$t_n = a + (n-1)d$ as the values for tn will be known.

The third term of an arithmetic progression is

$18$ , and the seventh term is

$30$ , then the sum of

$17$ terms is

0%
$612$
0%
$600$
0%
$656$
0%
None of these

Explanation

Let

$T_n$ be the

$n^{th}$ term of the AP.

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

Where

$a$ and

$d$ are first term and common difference respectively.

According to the given condition

$T_{3} = a+(3-1)d$

$T_{3} = a+2d$

$18 = a+2d$ ........

$(1)$

$T_{7} = a+(7-1)d$

$T_{7} = a+6d$

$30 = a+6d$ ........

$(2)$

solving

$(1)$ and

$(2)$ we get,

$a = 12$

$d = 3$

Sum of

$n$ terms of an AP is given as,

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

$S_{17}= \dfrac {17}{2} [2(12)+(17-1)3]$

$S_{17}=612$

Hence option (A) is correct option

Sum of the first

$20$ odd natural numbers is

0%
$400$
0%
$40$
0%
$200$
0%
$0$

Explanation

An odd number is one that is not divisible by

$2$ and the first odd natural number is

$1.$ Since two consecutive odd numbers differ by two, the first

$20$ odd natural numbers are

$1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39$

which is an A.P. with the first term

$a=1,$ the common difference

$d=2$ and total number of terms

$n = 20.$ Applying the general formula for the sum of an A.P.

$S_{n}= \dfrac{n}{2} [{2a + (n–1)d}]$

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

$\Rightarrow S_{20}= 10(2+38)$

$\Rightarrow S_{20}= 400$

The sum of first

$p$ terms of an arithmetic progression is

$q$ , and the sum of first

$q$ term is

$p$ , then the sum of first

$p + q$ terms is

0%
$(p + q)$
0%
$-(p + q)$
0%
$(q - p)$
0%
$(p - q)$

If fourth therm of an

$A.P$ is thrice its first term and seventh term is one less than the twice of third term, then its common difference is ?

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

The first term of an

$A.P.$ is

$10$ and its

$41^{st}$ term is

$81$ . Find the sum of the series, if

$81$ is the last term.

0%
$1025$
0%
$1865.5$
0%
$858$
0%
$962$

Explanation

Given, first term, $a=10$

last term, $l=81$

total number of terms, $n=41$

Sum, $S_n=\dfrac{n}{2}[a+l]$

$=\dfrac{41}{2}[10+81]$

$=\dfrac{41}{2}\times91$

$=1865.5$

Hence, the sum of the series is $1865.5$

Sum of the first

$n$ terms of an A.P. having first term

$a$ and last term

$\ell$ is ____

0%
$\dfrac{n}{2} (2a - \ell)$
0%
$\dfrac{n}{2} (2a + \ell)$
0%
$\dfrac{n}{2} (a + \ell)$
0%
$\dfrac{n}{2} (a - \ell)$

Explanation

Given: First term of an A.P

$= a$

Last term

$=2 l$

$\therefore$

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

$...(1)$

Sum of

$n$ terms

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

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

$=\dfrac{n}{2}(a+l)$ [From

$(1)$ ]

Sum of first

$n$ natural numbers is

$S_{1}$ , sum of first

$n$ odd natural numbers is

$S_{2}$ , and sum of first

$n$ even numbers is

$S_{3}$ . Then

$S_{1}:S_{2}:S_{3}$ =

0%
$n-1\ :\ n\ :\ n+1$
0%
$n\ :\ 2n-1-1\ :\ 2n$
0%
$n+1\ :\ 2n\ :\ 2(n+1)$
0%
$n+1\ :\ n\ :\ 2n-1\ :\ 2n$

Explanation

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

$S_{1}^{{}}=1+2+3+4+........n$

$a=1,d=1$

$S_{1}^{{}}=\frac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)1 \right]$

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

$S_{2}^{{}}=1+3+5+..........+n$

$a=1,d=2$

$S_{2}^{{}}=\frac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)2 \right]$

$\,\,\,\,\,\,\,=\frac{n}{2}\left[ 2n \right]$

$\,\,\,\,\,={{n}^{2}}$

$S_{3}^{{}}=2+4+6+..........n$

$a=2,d=2$

$S_{3}^{{}}=\frac{n}{2}\left[ 2\left( 2 \right)+\left( n-1 \right)2 \right]$

$\,\,\,\,\,=n\left[ 2+n-1 \right]$

$\,\,\,\,\,=n\left( n+1 \right)$

$S_{1}^{{}}:S_{2}^{{}}:S_{3}^{{}}$

$\frac{n\left( n+1 \right)}{2}:{{n}^{2}}:n\left( n+1 \right)$

$\Rightarrow \frac{n+1}{2}:n:n+1$

$\Rightarrow \left( n+1 \right):2n:2\left( n+1 \right)$

If the first, second and last term of an

$A.P.$ are

$a,\ b$ and

$2a$ respectively. then its sum is

0%
$\dfrac { ab }{ 2\left( b-a \right) }$
0%
$\dfrac { ab }{ \left( b-a \right) }$
0%
$\dfrac { 3ab }{ 2\left( b-a \right) }$
0%
$none\ of\ these$

Explanation

Common difference

$=b-a$

As we know the nth term,

$a_n = a+(n-1)d$ , where a and d are the first term and common difference respectively.

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

$\Rightarrow \dfrac{a}{b-a}=n-1$

$\Rightarrow \dfrac{a}{b-a}+1=n$

$\Rightarrow \dfrac{a+b-a}{b-a}=n$

$\Rightarrow \dfrac{b}{b-a}=n$

$\therefore$ Sum of

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

$=\dfrac{3ab}{2(b-a)}$

If the ratio of

${n^{th}}$ terms of two A.P.'s is

$\left( {2n + 8} \right):\left( {5n - 3} \right)$ , then the ratio of the sums of their

$n$ terms is

0%
$\,\,\left( {2n + 18} \right):\left( {5n + 3} \right)$
0%
$\,\left( {5n - 1} \right):\left( {2n + 18} \right)$
0%
$\,\,\left( {2n + 18} \right):\left( {5n - 1} \right)$
0%
$\,\left( {3n + 18} \right):\left( {4n + 1} \right)$

Explanation

Given ratio of

$n^{th}$ terms of A.P.:

$(2n+8):(5n-3)=\dfrac{2n-2+2+8}{5n-5+5-3}=\dfrac{10+(n-1)\times2}{2+(n-1)\times5}$

Since

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

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

For the A.P. in numerator,

$a_1=10$ and

$d_1=2$ and denominator

$a_2=2$ and

$d_2=5$

Also, Sum of n terms of an A.P. is

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

Sum of n terms of the A.P. in numerator

$=\dfrac{n}{2}[2\times10+(n-1)\times2]$

Sum of n terms of the A.P. in denominator

$=\dfrac{n}{2}[2\times2+(n-1)\times5]$

Then, the ratio of the sums of their

n nterms is

$\dfrac{\dfrac{n}{2}[20+2(n-1)]}{\dfrac{n}{2}[4+5(n-1)]}=\dfrac{2n+18}{5n-1}=(2n+18):(5n-1)$

The formula to find

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

0%
$a + (n - 1)d$
0%
$ar^{n - 1}$
0%
$\dfrac{1}{a + (n - 1)d}$
0%
$a - (n + 1)d$

Explanation

nth term of the A.P. is calculated by the formula,

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

where,

$a$ = first term of A.P.

$n$ = total number of terms in A.P.

$d$ = common difference between two consecutive terms of A.P.

A man saved Rs.

$200$ in each of the first three months of his service. In each of the subsequent months his saving increases by Rs.

$40$ more than the saving of immediately previous month. His total saving from the start of service will be Rs.

$11040$ after.

0%
$18$ months
0%
$19$ months
0%
$20$ months
0%
$21$ months

Explanation

$\dfrac{\overset { 1st }{ Month } }{200}\ \dfrac{\overset { 2nd }{ Month } }{200}\ \dfrac{\overset { 3rd }{ Month } }{200} \ \dfrac{\overset { 4th }{ Month } }{240}\ \dfrac{\overset { 5th }{ Month } }{280}$

So let

$n$ be the number of monthe.

Final savings

$=11040$

$\underbrace { 200+200 } +200+240+280+.... =11040$ .

$\Rightarrow 200+240+280+...+200+ (n-1)40=11040-400=10640$

Here

$a=200, d=40$ .

We know

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

$\Rightarrow \dfrac{n}{2} (2 \times 200+ (n-1) \times 40)=10640$

$\Rightarrow n(n+q) =532$

$\Rightarrow n^{2}+qn-532=0$ .

$\Rightarrow (n+28)(n-19)=0$ .

$\Rightarrow n=19$ (Since

$n=-28$ not possible)

For the A.P. if

$a = 7$ and

$d = 2.5 ,$ then

$t _ { 12 } = ?$

0%
37.5
0%
34.5
0%
28.2
0%
44.5

Explanation

${\textbf{Step - 1: Find 12th term}}$

${\text{Given, a = 7 and d = 2}}{\text{.5}}$

${\text{a = first term and }}$

${\text{d = difference between two terms}}$

${\text{It is known that in an AP}}\;{{\text{n}}^{th}}{\text{ term will be}}$

${{\text{t}}_n}{\text{ = a + (n - 1)d}}$

${\text{Put the value of n as 12}}{\text{.}}$

$\text{t}_{12} = 7 + (12 - 1)(2.5)$

${ = 7 + 11 \times 2.5}$

${ = 34.5}$

${\textbf{Hence, the correct answer is option B}}{\text{.}}$

Three numbers are in

$AP$ such that their sum is

$18$ and sum of their squares is

$158$ . The greatest number among them is

0%
$10$
0%
$11$
0%
$12$
0%
$none\ of\ these$

Explanation

Let the nos be ,

$a-d, a, a+d$ then

$a-d+a+a+d = 18$

$3a = 18 \Rightarrow a = 6$

$\therefore 6-d, 6, 6+d$

Sum of square = 158

$(6-d)^{2}+6^{2}+(6+d)^{2} = 158$

$6^{2}+d^{2}-12d+6^{2}+6^{2}+d^{2}+12d = 158$

$108+2d^{2} = 158$

$2d^{2} = 50$

$d^{2} = 25 \Rightarrow d = \pm 5$

when

$d=+5$ , Numbers are : 1, 6, 11
when

$d=-5$ , Numbers are : 11, 6, 1

Greatest : 11

The first term of an AP is unity and common difference is

$5 .$ Which of the following will be a term of this AP.

0%
$4880$
0%
$7881$
0%
$5890$
0%
$7891$

Explanation

$\\a=1,d=5\\\therefore\>a_n=a+(n-1)d$

$\\=1+(n-1)5\\=5n-4$

Consider option (A),

$\\4880=5n-4\\5n=4884\\n=(\dfrac{4884}{5})=not\>an\>integer$

Consider option (B),

$\\7881=5n-4\\5n=7885\\n=(\dfrac{7885}{5})=1577=integer$

Consider option (C),

$\\5890=5n-4\\5n=5894\\n=(\dfrac{5894}{5})=not\>an\>integer$

Consider option (D),

$\\7891=5n-4\\5n=7895\\n=(\dfrac{7895}{5})=1579=integer$

$\\\ast\>two\>options\>possible.$

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 14 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 14 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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