Arithmetic Progressions Test

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Arithmetic Progressions Test - 45

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Question 1
1 / -0

The common difference of $$-2, -4, -6, -8,...........$$ is :

A
$$-2$$

B
$$-1$$

C
$$2$$

D
$$3$$

Solution

$$\textbf{Step-1: Write the property of Common difference of an A.P.}$$
$$\text{Given series: -2,-4,-6,-8...}$$
$$\text{We know that Common difference = Second term - Fiirst term = Third term - Second term}$$

$$\therefore$$ $$\text{Common difference=}$$ $$ - 4 - ( - 2) = - 4 + 2 = - 2$$ $$\text{or - 6 - ( - 4) = - 6 + 4 = - 2}$$
$$\textbf{Hence, In the given sequence -2, -4, -6, ......, the common difference is -2, option - A.}$$
Question 2
1 / -0

How many terms of the series $$54, 51, 48,......$$ be taken so that their sum is $$513$$?

A
$$18$$

B
$$19$$

C
$$20$$

D
$$A$$ and $$B$$

Solution

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 $$n=19$$
Question 3
1 / -0

Which term of the A.P $$\ 3, 10, 17, .......$$ will be $$84$$ more than its $$13^{th}$$ term?

A
$$23^{rd}$$

B
$$25^{th}$$

C
$$27^{th}$$

D
$$28^{th}$$

Solution

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.
Question 4
1 / -0

If $$3x, 5x+1$$ and $$8x-1$$ are in$$A.P$$, then the value of $$x$$ is

A
$$2$$

B
$$-3$$

C
$$-4$$

D
$$3$$

Solution

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 $$2b = a+c $$
$$\therefore 2\left(5x+1\right) = 3x+8x-1$$
$$ \therefore 10x+2 = 11x-1$$
$$\therefore 11x-10x = 2+1$$
$$\therefore x = 3$$
Question 5
1 / -0

Find the sum of all three-digit numbers which leave a remainder $$2$$, when divided by $$6$$.

A
$$82656$$

B
$$82658$$

C
$$82650$$

D
$$82654$$

Solution

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.
Question 6
1 / -0

Find the sum of the series up to $$90$$ terms:
$$1+2+3+3+4+6+5+6+9+ $$ ....

A
$$1613$$

B
$$3225$$

C
$$6450$$

D
$$9675$$

Solution

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, $$S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]$$
$$\displaystyle \therefore \quad { S }_{ 30 }=\frac { 30 }{ 2 } \left[ 12+\left( 29 \right) \left( 7 \right) \right] =3225$$
Question 7
1 / -0

If the third and $$11^{th}$$ term of an A.P are $$8$$ and $$20$$ respectively, find the sum of first ten terms.

A
$$\displaystyle 105\frac{1}{2}$$

B
$$108$$

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

D
$$\displaystyle 203\frac{1}{2}$$

Solution

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 $$ a=5,d=\dfrac{3}{2}$$
$$\displaystyle \therefore S_{10}=\frac{10}{2}\left [ 2a+\left ( 10-1 \right )d \right ]$$
$$\displaystyle =5\left [ 2\times 5+9\times \frac{3}{2} \right ]$$
$$\displaystyle =117\frac{1}{2}$$
Question 8
1 / -0

Which of the following list of numbers does form an $$A.P$$?

A
$$2,4,8, 16$$, ....

B
$$2,\dfrac {5}{2},3,\dfrac {7}{2}$$, ...

C
$$0.2, 0.22, 0.222, 0.2222$$, ...

D
$$1,3,9,27$$, ...

Solution

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 $$A.P$$
Question 9
1 / -0

Find the sum of first $$31$$ term term of an A.P whose $$n^{th}$$ term is $$\displaystyle \left ( 3+\frac{2n}{3} \right )$$.

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

B
$$\displaystyle 413\frac{1}{3}$$

C
$$\displaystyle 417\frac{2}{3}$$

D
$$\displaystyle 419\frac{2}{3}$$

Solution

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.
$$\displaystyle a_{1}=3+\frac{2}{3}=\frac{11}{3}$$
$$\displaystyle a_{31=}=3+\frac{2}{3}\times 31=3+\frac{62}{3}=71$$
$$\displaystyle \therefore S_{31}=\frac{31}{2}\times \left ( \frac{11}{3}+\frac{71}{3} \right )=\frac{31}{2}\times \frac{82}{3}=\frac{1271}{3}$$
$$\displaystyle =423\frac{2}{3}$$
Question 10
1 / -0

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

A
$$(99)^2$$

B
$$(49)^2$$

C
$$(50)^2$$

D
$$(100)^2$$

Solution

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

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Arithmetic Progressions Test - 45

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Arithmetic Progressions Test - 45

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SHARING IS CARING

Question 1

$$-2$$

$$-1$$

$$2$$

$$3$$

Solution

Question 2

$$18$$

$$19$$

$$20$$

$$A$$ and $$B$$

Solution

Question 3

$$23^{rd}$$

$$25^{th}$$

$$27^{th}$$

$$28^{th}$$

Solution

Question 4

$$2$$

$$-3$$

$$-4$$

$$3$$

Solution

Question 5

$$82656$$

$$82658$$

$$82650$$

$$82654$$

Solution

Question 6

$$1613$$

$$3225$$

$$6450$$

$$9675$$

Solution

Question 7

$$\displaystyle 105\frac{1}{2}$$

$$108$$

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

$$\displaystyle 203\frac{1}{2}$$

Solution

Question 8

$$2,4,8, 16$$, ....

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

$$0.2, 0.22, 0.222, 0.2222$$, ...

$$1,3,9,27$$, ...

Solution

Question 9

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

$$\displaystyle 413\frac{1}{3}$$

$$\displaystyle 417\frac{2}{3}$$

$$\displaystyle 419\frac{2}{3}$$

Solution

Question 10

$$(99)^2$$

$$(49)^2$$

$$(50)^2$$

$$(100)^2$$

Solution

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