Arithmetic Progressions Test

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Arithmetic Progressions Test - 21

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Question 1
1 / -0

For an A.P. $$a = -18.9, d = 2.5 \ ,a_n = 3.6,$$ find $$n$$.

A
$$n=12$$

B
$$n=10$$

C
$$n=8$$

D
$$n=6$$

Solution

Here, $$a=-18.9, d=2.5$$ and $$a_n=3.6$$.
Therefore, $$a_n=a+(n-1)d$$
$$-18.9+(n-1)\times2.5=3.6$$
$$\Rightarrow (n-1)\times2.5=22.5$$
$$\Rightarrow n-1=9$$
$$\therefore n=10$$
Question 2
1 / -0

Check whether $$-150$$ is a term of the AP:
$$11, 8, 5, 2,...$$

A
No

B
Yes

C
Data Insufficient

D
Ambiguous

Solution

Clearly, the given Sequence is an AP with first term $$a=11$$ and common difference $$d=8-11=-3$$.

Let $$-150$$ be the $$nth$$ term of the given AP.

$$\Rightarrow a_n=-150$$

$$\Rightarrow a+(n-1)d=-150$$

$$\Rightarrow 11+(n-1)\times-3=-150$$

$$\Rightarrow (n-1)\times-3=-161$$

$$\Rightarrow -3n+3=-161$$

$$\Rightarrow -3n=-164\Rightarrow n=54\displaystyle \frac23$$

Since, $$n$$ is not a natural number.

Hence, $$-150$$ is not any term of given AP.
Question 3
1 / -0

In AP
Given a =2, d = 8, S$$_n$$ = 90, find n and a$$_n$$

A
$$n = 5$$ and $$a_n = 34$$

B
$$n = 10$$ and $$a_n = 34$$

C
$$n = 5$$ and $$a_n = 54$$

D
$$n = 10$$ and $$a_n = 44$$

Solution

In AP Given a =2, d = 8, S$$_n$$ = 90
Since, $$S_{n}=\frac n2 [2a+(n-1)d] $$
$$\Rightarrow 90=\frac n2[2\times 2+(n-1)8] $$
$$\Rightarrow 90=\frac n2[4+8n-8] $$
$$\Rightarrow 90=\frac n2[8n-4] $$
$$\Rightarrow 180= n[8n-4] $$
$$\Rightarrow 8n^2-4n-180=0 $$
$$\Rightarrow 2n^2-n-45=0 $$
$$\Rightarrow 2n^2-10n+9n-45=0 $$
$$\Rightarrow 2n(n-5)+9(n-5)=0 $$
$$\Rightarrow (2n+9)(n-5)=0 $$
but $$ 2n+9 \neq 0$$
$$\therefore n-5=0$$
$$\Rightarrow n=5 $$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow a_5=2+(5-1)8 $$
$$\Rightarrow a_5=2+32 $$
$$\Rightarrow a_5=34 $$
Question 4
1 / -0

Find the $$20th$$ term from the last term of the AP $$3,8, 13,....253.$$

A
$$156$$

B
$$180$$

C
$$158$$

D
$$188$$

Solution

The AP is $$3, 8, 13, ....,253$$
Its first term $$= 3$$ and the common difference $$= 5$$
The AP in the reverse order will have the first term $$= 253$$ and the common difference $$= -5$$
The $$20th$$ term from the end of the given AP $$=$$ The $$20th$$ term of the AP in the reverse order$$= a + 19d$$
$$= 253 + 19 \times(-5) = 253 - 95 = 158$$
Question 5
1 / -0

Which term of the sequence $$ 3, 8, 13, 18, ........$$ is $$498$$.

A
$$95th$$

B
$$100th$$

C
$$102th$$

D
$$101th$$

Solution

Given series is : $$3,8,13,18,........$$
$$a = 3, l =498, d = 5, n = ? $$
$$l =a + (n - 1)d $$
$$498=3+(n-1)5$$
$$498 - 3 = (n - 1)5$$
$$\dfrac{495}{5}=n-1$$
$$\therefore n =99+1 $$
$$\therefore n = 100 $$
So $$100^{th}$$ term is $$498$$
Question 6
1 / -0

The $$16th$$ term of the AP: $$15,$$ $$\displaystyle \frac{25}{2},10,\frac{15}{2},........$$ is

A
$$\displaystyle \frac{45}{2}$$

B
$$\displaystyle -\frac{45}{2}$$

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

D
$$\displaystyle -\frac{105}{2}$$

Solution

Given AP: $$15,\dfrac{25}{2},10,\dfrac{15}{2},5........$$
First term $$a = 15$$
common difference $$ \displaystyle d=\dfrac{25}{2}-15=\dfrac{25-30}{2}=-\dfrac{5}{2}$$

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

$$\Rightarrow a_{16}=a+15d=15+15\times -\dfrac{5}{2}$$

$$ \Rightarrow\dfrac{30-75}{2}=-\dfrac{45}{2}$$
$$\therefore $$ Option B is correct.
Question 7
1 / -0

If $$\displaystyle \frac{6}{5},$$ $$a,4$$ are in AP, the value of $$a$$ is

A
$$1$$

B
$$13$$

C
$$\displaystyle \frac{13}{5}$$

D
$$\displaystyle \frac{26}{5}$$

Solution

Given, $$\dfrac{6}{5},a,4$$ are in AP

If $$a_1,\ a_2,\ a_3$$ are in AP, the common difference will be the same

$$\Rightarrow d=a_2-a_1=a_3-a_2$$

For the given terms: $$\dfrac{6}{5},\ a,\ 4$$

$$\therefore$$ $$d= a-\dfrac{6}{5} = 4-a$$

$$\Rightarrow$$ $$2a=\dfrac{6}{5}+4$$

$$\Rightarrow$$ $$2a=\dfrac{6+20}{5}$$

$$\Rightarrow$$ $$a=\dfrac{26}{10}$$

$$=\dfrac{13}{5}$$
Question 8
1 / -0

Is $$51$$ a term of the AP, $$5, 8, 11, 14,........?$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

Given series is $$5,8,11,14,....$$
Let $$51$$ be the $$ n^{th} $$ term.
$$\therefore a_n=51$$
$$a = 5, d = 3$$
We know, $$a_n=a+(n-1)d$$
$$\therefore 51 = a + (n - 1)d$$
$$51 = 5 + (n - 1)3$$

$$ \dfrac{46}{3} = n - 1$$

$$ \dfrac{46+3}{3} = n$$

$$ n=\dfrac{49}{3}$$
Since $$n$$ is not a natural no.
$$51$$ is not a term of AP.
Question 9
1 / -0

The $$31st$$ term of the AP whose first two terms are respectively $$-2$$ and $$-7$$ is

A
$$-152$$

B
$$150$$

C
$$148$$

D
$$-148$$

Solution

Given, $$\displaystyle a_{1}=-2 , a_{2}=-7 $$
$$ \Rightarrow d=a_{2}-a_{1}= -7-(-2) = - 7 + 2 = - 5$$
Since, $$ a_n=a+(n-1)d $$
$$\displaystyle \Rightarrow a_{31}=a+30d=-2+30(-5)=-152$$
$$\therefore $$ Option A is correct.
Question 10
1 / -0

The $$11th$$ term of AP whose $$3rd$$ term is $$11$$ and $$8th$$ term is $$26$$ is

A
$$25$$

B
$$-2$$

C
$$-8$$

D
$$35$$

Solution

Given, $$ a_3=11 , a_8=26 $$
if the terms are in $$A.P$$ then $$a_n = a+(n-1)d$$

$$\displaystyle \Rightarrow a_3=a+2d=11\ \ \ \ \ .......(1)$$
$$\displaystyle \Rightarrow a_8=a+7d=26\ \ \ \ \ .......(2)$$

Subtract equation $$(2)$$ from $$(1)$$
we get,
$$\Rightarrow - 5d = -15$$
$$\therefore d = 3$$

By putting $$d=3$$ in equation $$(1)$$ we get,
$$ \Rightarrow a +2(3) = 11 $$
$$\therefore a = 5$$

$$\displaystyle \Rightarrow a_{11}=a+10d$$ (by using formula)
$$=5+10(3)$$
$$=35$$
$$\therefore $$ Option D is correct.

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 21

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Arithmetic Progressions Test - 21

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SHARING IS CARING

Question 1

$$n=12$$

$$n=10$$

$$n=8$$

$$n=6$$

Solution

Question 2

No

Yes

Data Insufficient

Ambiguous

Solution

Question 3

$$n = 5$$ and $$a_n = 34$$

$$n = 10$$ and $$a_n = 34$$

$$n = 5$$ and $$a_n = 54$$

$$n = 10$$ and $$a_n = 44$$

Solution

Question 4

$$156$$

$$180$$

$$158$$

$$188$$

Solution

Question 5

$$95th$$

$$100th$$

$$102th$$

$$101th$$

Solution

Question 6

$$\displaystyle \frac{45}{2}$$

$$\displaystyle -\frac{45}{2}$$

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

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

Solution

Question 7

$$1$$

$$13$$

$$\displaystyle \frac{13}{5}$$

$$\displaystyle \frac{26}{5}$$

Solution

Question 8

Yes

No

Ambiguous

Data insufficient

Solution

Question 9

$$-152$$

$$150$$

$$148$$

$$-148$$

Solution

Question 10

$$25$$

$$-2$$

$$-8$$

$$35$$

Solution

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