Arithmetic Progressions Test

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Arithmetic Progressions Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

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.

A
$$26$$th

B
$$19$$th

C
$$21$$st

D
$$36$$th

Solution

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
Question 2
1 / -0

The total two-digit numbers which are divisible by $$5$$, are

A
$$17$$

B
$$18$$

C
$$19$$

D
$$20$$

Solution

$$10, 15, 20, ........ 95$$
$$n$$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 $$=18$$
Question 3
1 / -0

If an A.P is given by $$17, 14, ......-40$$, then $$6$$th term from the end is

A
$$-22$$

B
$$-28$$

C
$$-25$$

D
$$26$$

Solution

Given an AP is 17,14...........................-40
Then we get AP is -40,-37,.............14,17
Then d=3,a=-40 n=6
Therefore, $$a_{6}=a+(n-1)d=(-40)+(6-1)3=-40+15=-25$$

,
Question 4
1 / -0

How many terms are there in the sequence $$15\cfrac{1}{2}, 13, ......-47$$?

A
$$27$$

B
$$26$$

C
$$28$$

D
None

Solution

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 $$26$$
Question 5
1 / -0

Find the $$n^{th}$$ term and, hence, the $$25^{th}$$ term of the following AP: $$2, 5, 8, 11$$,.........

A
$$3n-1, 74$$

B
$$2n, 73$$

C
$$3n, 75$$

D
$$3n+1, 72$$

Solution

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
$$\displaystyle { t }_{ n }=a+\left( n-1 \right) d$$
$$\displaystyle =2+\left( n-1 \right) 3$$
$$\displaystyle =2+3n-3$$
$$\displaystyle =3n-1$$
$$\displaystyle { t }_{ 25 }=3\times 25-1$$
$$=75-1=74$$
Question 6
1 / -0

If the first, second and last term of an A.P. are $$x,y$$ and $$2x$$ respectively, its sum is

A
$$\cfrac{xy}{2(y-x)}$$

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

C
$$\cfrac{2xy}{y-x}$$

D
$$\cfrac{xy}{y-x}$$

Solution

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.
Question 7
1 / -0

If in an A.P, $${a}_{9}=0, {a}_{19}=k$$, then $${a}_{29}$$ is

A
$${k}^{2}$$

B
$$10k$$

C
$$9k$$

D
$$2k$$

Solution

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.
Question 8
1 / -0

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

A
10, 13, 16, 19, ...

B
5,7,9, 11, ...

C
8, 12, 16,20, ...

D
All the above

Solution

If $$a_1$$ 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.....
Question 9
1 / -0

Which term of the A.P $$5,15, 25 -----$$ will be $$100$$ more than its $$31^{st}$$ term?

A
$$41^{st}$$

B
$$42^{nd}$$

C
$$40^{th}$$

D
$$43^{rd}$$

Solution

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.
Question 10
1 / -0

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

A
$$22$$

B
$$33$$

C
Both $$22$$ and $$33$$

D
none of these

Solution

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 $$n = 22$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 44

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Arithmetic Progressions Test - 44

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SHARING IS CARING

Question 1

$$26$$th

$$19$$th

$$21$$st

$$36$$th

Solution

Question 2

$$17$$

$$18$$

$$19$$

$$20$$

Solution

Question 3

$$-22$$

$$-28$$

$$-25$$

$$26$$

Solution

Question 4

$$27$$

$$26$$

$$28$$

None

Solution

Question 5

$$3n-1, 74$$

$$2n, 73$$

$$3n, 75$$

$$3n+1, 72$$

Solution

Question 6

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

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

$$\cfrac{2xy}{y-x}$$

$$\cfrac{xy}{y-x}$$

Solution

Question 7

$${k}^{2}$$

$$10k$$

$$9k$$

$$2k$$

Solution

Question 8

10, 13, 16, 19, ...

5,7,9, 11, ...

8, 12, 16,20, ...

All the above

Solution

Question 9

$$41^{st}$$

$$42^{nd}$$

$$40^{th}$$

$$43^{rd}$$

Solution

Question 10

$$22$$

$$33$$

Both $$22$$ and $$33$$

none of these

Solution

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