Arithmetic Progressions Test

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Arithmetic Progressions Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

If the $${2}^{nd}$$ term of an AP is $$13$$ and the $${5}^{th}$$ term is $$25,$$ what is its $${7}^{th}$$ term?

A
$$30$$

B
$$33$$

C
$$37$$

D
$$38$$

Solution

Let the first term of the AP be $$a$$, the common difference be $$d$$.
Then,
$${2}^{nd}$$ term is $$a+d=13.......(1)$$
$${5}^{th}$$ term is $$a+4d=25........(2)$$

Solving for $$d$$,

By $$Eq(2)-Eq(1)$$
$$(a+4d)-(a+d)=25-13$$
$$\Rightarrow 3d=12$$
$$\Rightarrow d=4$$

Hence, From $$(1)$$
$$a=13-d=13-4=9$$

Now, $${7}^{th}$$ term is $$a+6d$$
Substituting values of $$a$$ and $$d$$
$$7^{th} term=9+6\times4=33$$
Question 2
1 / -0

The $$4^{th}$$ term from the end of the AP: $$11, 8, 5, ..., -49$$ is

A
$$-37$$

B
$$-40$$

C
$$-43$$

D
$$-58$$

Solution

In the given A.P. $$a=11,d=-3$$, last term $$(l)=-49$$,
then $$l=a+(n-1)d$$
$$\Rightarrow -49=11+(n-1)(-3)$$
$$\Rightarrow 3n=63\Rightarrow n=21$$
Hence, there are $$21$$ terms in AP.
The $$4^{th}$$ term from the last will be 18th term from beginning.
So, $$a_{18}=a+(18-1)d$$
$$\Rightarrow 11+17(-3)$$
$$\Rightarrow 11-51=-40$$
SO $$4^{th}$$ term from the end of the AP: $$11, 8, 5, ..., -49$$ is $$-40$$.
Question 3
1 / -0

If $$7$$ times the $$7^{th}$$ term of an AP is equal to $$11$$ times its $$11^{th}$$ term, then its $$18^{th}$$ term will be

A
$$7$$

B
$$11$$

C
$$18$$

D
$$0$$

Solution

$${\textbf{Step - 1: Find 7th term of the AP}}$$

  $${a_7} = a + (n - 1)d$$

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

$$ = a + 6d$$

  $${\text{The 11th term of the AP,}}$$

  $${a_{11}} = a + (n - 1)d$$

$$ = a + (11 - 1)d$$

$$ = a + 10d$$

$${\textbf{Step - 2: According to the question 7 times of 7th term is equal to 11 times of 11th term}}$$

  $$7(a + 6d) = 11(a + 10d)$$

  $$\Rightarrow 7a + 42d = 11a + 110d$$

  $$\Rightarrow 4a + 68d = 0$$

$${\textbf{Step - 3: Find value for a}}$$

  $${\text{a = }}\dfrac{{ - 68}}{4}d$$

$${\textbf{Step - 4: Find value of 18th term }}$$

  $$a_{18} = a + (18 - 1)d$$

$$ = a + 17d$$

$${\textbf{Step - 5: Substitute the value of a in equation of }}{{\textbf{a}}_\textbf{18}}$$

  $${a_{18}} = \dfrac{{ - 68}}{4}d + 17d$$

$$ = \dfrac{{ - 68d + 68d}}{4} = 0$$

$${\textbf{Hence , the required value is (D) 0}}$$
Question 4
1 / -0

The $$21^{st}$$ term of the AP whose first term is $$3$$ and second term is $$4$$, is

A
$$17$$

B
$$19$$

C
$$21$$

D
$$23$$

Solution

Given,
$$a_1=3$$, $$a_2=4$$.
$$\therefore$$ Common difference, $$d=1$$.
Using the formula for $$n^{th} $$ term of an $$ A.P. $$
$$a_n=a+(n-1)d$$
For, $$n=21$$,
$$a_{21}=3+(21-1)\times 1$$
$$a_{21}=3+20\times 1$$
$$a_{21}= 23$$
Question 5
1 / -0

The sum of the first $$16$$ terms of the AP: $$10, 6, 2,...$$ is

A
$$-320$$

B
$$320$$

C
$$-400$$

D
$$400$$

Solution

Given, $$10,6,2.........$$
Since, first term $$= 10$$ and common difference $$=6-10=-4$$

$$S_n=\dfrac n2[2a+(n-1)d]$$

$$S_{16}=\dfrac {16}2[2(10)+(16-1)(-4)]$$

$$S_{16}=8[20+(15)(-4)]$$

$$S_{16}=8[20-60]$$

$$S_{16}=8\times (-40) = -320$$
Option A is correct.
Question 6
1 / -0

The sum of first five multiples of $$3$$ is

A
$$45$$

B
$$55$$

C
$$65$$

D
$$75$$

Solution

Sum of an AP is: $${\dfrac{n}{2}}\times({2a+(n-1)d})$$
$$a = 3, d= 3, n=5$$
Then,
Sum = $${\dfrac{5}{2}}\times{(2\times3+(5-1)\times(3)})$$
= $${\dfrac{5}{2}}\times{(6+12)}$$
= $$45$$
Question 7
1 / -0

The sum of $$n$$ terms of an A.P. is $$4n^2-n$$. The common difference $$=$$ ____

A
$$8$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Given, $$ {S}_{n} = 4{n}^{2} - n $$
When $$ n = 1 $$,
$$ {S}_{1} = 4{(1)}^{2} - 1 = 4 - 1 = 3 $$
Now, obviously, the sum of the first term will be the first term itself, as there are no other terms involved.

When $$ n = 2 $$,
$$ {S}_{2} = 4{(2)}^{2} - 1 =16 = 2 = 14 $$

The sum of the first $$2$$ terms is equal to $$a_1+a_2$$

Therefore, $$a_1+a_2=14$$

We have shown, above, that $$a_1=3$$

Substituting it in $$a_1+a_2=14,$$
$$3+a_2=14$$
$$a_2=11.$$

The common difference, $$d,$$ is the difference between any $$2$$ consecutive terms.
$$d=a_2−a_1$$
$$=11−3$$
$$=8$$
Question 8
1 / -0

Given the AP $$10, 7, 4, ....., -62$$. The $$11^{th}$$ term from end of the AP is

A
$$23$$

B
$$53$$

C
$$34$$

D
$$-32$$

Solution

Let $$d$$ be the common difference of the given A.P>

$$d=7-10$$

$$\Rightarrow d=-3$$

Here, $$a=10, d=-3, t_n=-62$$

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

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

Thus,

$$ -62 = 10 +(n-1)(-3)$$

$$ -72 = (n-1)(-3) $$

Thus, $$n=25$$

Hence, in our solution, we are looking for $$11^{th}$$ term from last that is $$15^{th}$$ term from starting.

$$ t_{15}=10+(15-1)(-3)$$
$$=-32$$

Hence option (D) is correct
Question 9
1 / -0

The interior angles of a convex polygon are in Arithmetic

Progression. The smallest angle is $$120^{o}$$ and the common

difference is
$$ 5^{o}$$. What is the number of sides of the polygon?

A
7

B
15

C
17

D
9
Question 10
1 / -0

If the $$n^{th}$$ term of an A.P. is given by $$a_n=5n-3$$, then the sum of first $$10$$ terms is

A
$$225$$

B
$$245$$

C
$$255$$

D
$$270$$

Solution

Given, $$a_n = 5n-3$$
Put $$n=1$$, $$a_1= 5-3= 2$$
Put $$n=10$$, $$a_{10}= 5\times10-3= 47$$
Sum $$=$$ $${\dfrac{n}{2}}{(a+l)}$$
$$=$$ $${\dfrac{10}{2}}{(2+47)}$$
$$=$$ $$49\times5$$
$$= 245$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 34

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Arithmetic Progressions Test - 34

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Rank

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SHARING IS CARING

Question 1

$$30$$

$$33$$

$$37$$

$$38$$

Solution

Question 2

$$-37$$

$$-40$$

$$-43$$

$$-58$$

Solution

Question 3

$$7$$

$$11$$

$$18$$

$$0$$

Solution

Question 4

$$17$$

$$19$$

$$21$$

$$23$$

Solution

Question 5

$$-320$$

$$320$$

$$-400$$

$$400$$

Solution

Question 6

$$45$$

$$55$$

$$65$$

$$75$$

Solution

Question 7

$$8$$

$$10$$

$$11$$

$$12$$

Solution

Question 8

$$23$$

$$53$$

$$34$$

$$-32$$

Solution

Question 9

7

15

17

9

Question 10

$$225$$

$$245$$

$$255$$

$$270$$

Solution

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