Arithmetic Progressions Test

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Arithmetic Progressions Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

How many terms are there in the sequence 3, 6, 9, 12, ..., 111 ?

A
$$32$$

B
$$37$$

C
$$42$$

D
$$49$$

Solution

The given sequence is an A.P. with first term $$a=3$$ and common difference $$d=3$$. Let there be n terms in the given sequence. Then,
$$nth \ term=111$$
$$a+(n-1)d=111$$
$$3+(n-1) \times 3=111$$
$$n=37$$
Thus, the given sequence contains 37 terms.
Question 2
1 / -0

Find the sum of all odd integers between 2 and 100 divisible by 3.

A
452

B
754

C
867

D
765

Solution

$$\textbf{Step-1: Finding the value of n}$$
$$\text{The}$$ $$n^{th}$$ $$\text{term in an A.P. is given by}$$
$$t_n=a+(n-1)d$$
$$\text{We know that the}$$ $$n^{th}$$ $$\text{term is 99}$$
$$99=3+(n-1)6$$
$$96=6(n-1)$$
$$n-1=16$$
$$n=17$$
$$\textbf{Step-2: Finding the required sum}$$
$$\text{Sum of n terms in an A.P. is}$$
$$S_n=\dfrac{n}{2}(a+t_n)$$
$$S_n=\dfrac{17}{2}(3+99)=867$$

$$\textbf{Hence,The correct option is (C)}$$
Question 3
1 / -0

Choose the correct answer from the given four options in the following question:
The first four terms of an A.P., whose first term is $$-2$$ and the common difference is $$-2$$, are

A
$$-2, 0, 2, 4$$

B
$$-2, 4, -8, 16$$

C
$$-2, -4, -6, -8$$

D
$$-2, -4, -8, -16$$

Solution

Hint: use the formula of $$n^{th}$$ term of an A.P.
$$a_n=a+(n-1)d$$

Given:-
first term $$=a=a_1=-2$$
common difference $$=d=-2$$

Step 1 : find the second term by replacing $$n$$ with $$2$$ in the formula of $$n^{th}$$ term of the A.P
$$a_n=a+(n-1)d$$
$$\Rightarrow a_2=(-2)+(2-1)(-2)$$
$$\Rightarrow a_2=-2+(1)(-2)$$
$$\Rightarrow a_2=-2-2$$
$$\therefore a_2=-4$$

Step 2 : find the third term by replacing $$n$$ with $$3$$ in the formula of $$n^{th}$$ term of the A.P
$$a_n=a+(n-1)d$$
$$\Rightarrow a_3=(-2)+(3-1)(-2)$$
$$\Rightarrow a_3=-2+(2)(-2)$$
$$\Rightarrow a_3=-2-4$$
$$\therefore a_3=-6$$

Step 3 : find the fourth term by replacing $$n$$ with $$4$$ in the formula of $$n^{th}$$ term of the A.P
$$a_n=a+(n-1)d$$
$$\Rightarrow a_4=(-2)+(4-1)(-2)$$
$$\Rightarrow a_4=-2+(3)(-2)$$
$$\Rightarrow a_4=-2-6$$
$$\therefore a_4=-8$$

Final step: write down the first four terms of A.P.
first four terms of A.P. are $$a_1,\,a_2,\,a_3,\,a_4$$
i.e, $$-2,\,-4,\,-6,\,-8$$
Question 4
1 / -0

Choose the correct answer from the given four options in the following question:
In an A.P., if $$d = -4, n = 7, a_n = 4,$$ then $$a$$ is

A
6

B
7

C
20

D
28

Solution

Hint : $$n^{th }$$ term of an A.P. is $$a_n=a+(n-1)d$$ where $$a$$ is first term and $$d$$ is common difference.

Given:
common difference $$=d=-4$$
$$n=7$$
$$a_n=4$$ [$$n^{th}$$ term]

As we know, that
general term of an A.P. is given by
$$a_n=a+(n-1)d$$

putting the given values in the above formula
$$\Rightarrow 4=a+(7-1)(-4)$$
$$\Rightarrow 4=a+(6)(-4)$$
$$\Rightarrow 4=a-24\\ \Rightarrow 4+24=a$$
$$\Rightarrow 28=a$$

Hence, value of first term$$,a$$ is $$28$$
Question 5
1 / -0

Choose the correct answer from the followings for the sequence $$-10, -6, -2, 2, ...$$

A
an A.P. with $$d = -16$$

B
an A.P. with $$d = 4$$

C
an A.P. with $$d = -4$$

D
not an A.P.

Solution

Hint : In an A.P. difference between two consecutive terms is always same

Step 1 : find out the difference between two consecutive terms
for the given sequence $$-10,\,-6,\,-2,\,2,...$$
second term - first term $$=-6-(-10)=4$$
third term - second term $$=-2-(-6)=4$$
fourth term - third term $$=2-(-2)=4$$

Step 2 : Comparison of all the difference quantities and conclusion
Thus, difference between two consecutive terms is always $$4$$
$$\therefore \,$$common difference $$=d=4$$

Final step : Hence the given sequence is an A.P. with $$d=4$$.
Question 6
1 / -0

Next three consecutive numbers in the pattern are
11, 8, 5, 2...

A
$$0, – 3, – 6$$

B
$$– 1, – 5, – 8$$

C
$$– 2, – 5, – 8$$

D
$$– 1, – 4, – 7$$

Solution

Option (d) is correct; any two numbers have the common difference of 3,
$$\begin{array}{l}11-3=8,8-3=5, \ldots \\\text { Similarly, }(2-3)=-1,(-1-3)=-4,(-4-3)=-7\end{array}$$
Hence, three consecutive terms will be -1, -4, and -7.
Question 7
1 / -0

Two APs have the same common difference and the 1st term of one AP is -1 and first term of other AP is -8. Then the difference between their 4th terms is

A
-1

B
-8

C
7

D
-9

Solution

Hint: use the formula of $$n^{th}$$ term of an A.P.

Given:-
first term of one A.P. $$=a=-1$$
first term of other A.P. $$=A=-8$$
same common difference of both the A.Ps $$=d$$

Step 1 : find the $$4^{th}$$ term of first A.P.
$$\Rightarrow a_n=a+(n-1)d$$
$$\Rightarrow a_4=-1+(4-1)d$$
$$\therefore a_4=-1+3d$$

Step 2 : find the $$4^{th}$$ term of other A.P.
$$\Rightarrow A_n=A+(n-1)d$$
$$\Rightarrow A_4=-8+(4-1)d$$
$$\therefore A_4=-8+3d$$

Step 3 : find the difference between $$4^{th}$$ terms of both A.P.s
$$|a_4-A_4|=|(-1+3d)-(-8+3d)|=|-1+3d+8-3d|=|-1+8|=|7|=7$$

Final step: Hence, difference between their fourth terms is $$7.$$
Question 8
1 / -0

What is the common difference of an A.P. in which $$a_{18}-a_{14} = 32$$?

A
$$8$$

B
$$-8$$

C
$$-4$$

D
$$4$$

Solution

Hint: The general term formula of an $$A.P$$ is $$\{a_n=a+(n-1)d\}$$ will be used along with the related terms.

Given:
Let $$ \text {a=first term (Let)}$$
$$ \text {d= common difference (Let)}$$
$$a_{18}-a_{14}=32$$

Step 1: Finding the general term expression of $$a_{18}$$
We know,
$$a_n=a+(n-1)d$$

For $$n=18$$
$$ a_{18}=a+(18-1)d=a+17d.....(i)$$

Step 2: Finding the general term expression of $$a_{14}$$
We know,
$$a_n=a+(n-1)d$$

For $$n=14$$
$$ a_{14}=a+(14-1)d=a+13d.....(ii)$$

Step 3: Finding the difference $$a_{18}-a_{14}$$ by subtracting $$eq (i)$$ from $$eq. (ii)$$

$$eq.(ii)-eq. (i)$$

$$a_{18}-a_{14}=a+17d-(a+13d)=a+17d-a-13d=4d.....(iii)$$

Step 4: Comparing and equating the value of $$a_{18}-a_{14}$$ from the question and $$eq(iii)$$

$$a_{18}-a_{14}=4d=32$$
$$\Rightarrow 4d=32 \Rightarrow d=8$$

Hence, the value of 'd' is $$8$$

Final step: The value of $$ \text {common difference, d=8}$$
Question 9
1 / -0

In an A.P $$1^{st}$$ term is $$1$$ and the last term is $$20$$. The sum of all terms is $$=399$$ then $$n=$$....

A
$$42$$

B
$$38$$

C
$$21$$

D
$$19$$

Solution

The is given that,
First term $$(a)=1$$
Last term $$(t_n)=20$$
Sum of terms $$(S_n)=399$$
We know that,
$$t_n=a+(n-1)d$$
$$S_n=\dfrac{n}{2}(2a+(n-1)d)\\$$
$$S_{n}=\dfrac{n}{2} (a+(a+(n-1)d))\\$$
$$\Rightarrow S_n=\dfrac{n}{2}(a+t_n)\\$$
$$\Rightarrow 399=\dfrac{n}{2}(1+20)\\$$
$$\Rightarrow 399=\dfrac{21n}{2}\\$$
$$\Rightarrow 21n=399 \times 2\\$$
$$\Rightarrow n=\dfrac{798}{21}\\$$
$$\Rightarrow n=38$$
Question 10
1 / -0

If $$a$$ is constant then $$a + 2a + 3a + ..... + na$$ is

A
$$\dfrac{an(n + 1)}{4}$$

B
$$\dfrac{an(n + 1)}{2}$$

C
$$\dfrac{n(n + 1)}{2}$$

D
$$\dfrac{an(n + 1)}{6}$$

Solution

For the terms $$a,\ 2a,\ 3a,\ 4a...,\ na$$

$${ a }_{ 2 }-{ a }_{ 1 }=2a-a$$
$$=a$$

$${ a }_{ 3 }-{ a }_{ 2 }=3a-2a$$
$$=a$$

Thus, the difference between consecutive terms of A.P. is the same.

Sum of first n terms of A.P. is given by $${ S }_{ n }=\dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] $$

$$\therefore { S }_{ n }=\dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) a \right] $$

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

$$=\dfrac { n }{ 2 } \left[ a\left( n+2 \right) -a \right] $$

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

$$=\dfrac { n }{ 2 } \left[ a\left( n+1 \right) \right] $$

$$=\dfrac { an\left( n+1 \right) }{ 2 } $$

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Arithmetic Progressions Test - 57

Result

Arithmetic Progressions Test - 57

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SHARING IS CARING

Question 1

$$32$$

$$37$$

$$42$$

$$49$$

Solution

Question 2

452

754

867

765

Solution

Question 3

$$-2, 0, 2, 4$$

$$-2, 4, -8, 16$$

$$-2, -4, -6, -8$$

$$-2, -4, -8, -16$$

Solution

Question 4

6

7

20

28

Solution

Question 5

an A.P. with $$d = -16$$

an A.P. with $$d = 4$$

an A.P. with $$d = -4$$

not an A.P.

Solution

Question 6

$$0, – 3, – 6$$

$$– 1, – 5, – 8$$

$$– 2, – 5, – 8$$

$$– 1, – 4, – 7$$

Solution

Question 7

-1

-8

7

-9

Solution

Question 8

$$8$$

$$-8$$

$$-4$$

$$4$$

Solution

Question 9

$$42$$

$$38$$

$$21$$

$$19$$

Solution

Question 10

$$\dfrac{an(n + 1)}{4}$$

$$\dfrac{an(n + 1)}{2}$$

$$\dfrac{n(n + 1)}{2}$$

$$\dfrac{an(n + 1)}{6}$$

Solution

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