Arithmetic Progressions Test

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Arithmetic Progressions Test - 46

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Question 1
1 / -0

A man saved Rs. $$320$$ during the first month Rs. $$360$$ in the second month Rs. $$400$$ in the third month, if he continues his saving in this sequence in how many months will be save Rs. $$20,000$$?

A
$$28$$

B
$$25$$

C
$$22$$

D
$$20$$

Solution

a=320, d=40
$$\displaystyle S_{n}=20000$$
It is required to find n
$$\displaystyle S_{n}=20000=\frac{n}{2}\left [ 2\times 320+\left ( n-1 \right )\times 40 \right ]$$
$$\displaystyle \Rightarrow 500=\frac{n}{2}\left [ 15+\left ( n-1 \right ) \right ]=\frac{n}{2}\left ( n+15 \right )$$
$$\displaystyle n^{2}+15n-1000=0$$
$$\displaystyle \Rightarrow n=25$$
Question 2
1 / -0

Find the number of terms in an arithmetic progression for which the first term is $$4$$, last term is $$22$$ and the common difference is $$\displaystyle\dfrac{1}{4}$$

A
$$70$$

B
$$71$$

C
$$72$$

D
$$73$$

Solution

Let $$ a $$ be the first term and $$ d $$ be their common difference of the AP.
Then, $$ n th $$ term of the AP$$, T_n = a + (n-1)d $$
Here, in the given AP,. $$ a = 4; d = \dfrac {1}{4} $$

Given
$$ T_n= 22 $$
$$\Rightarrow 4+ (n-1)\times \dfrac {1}{4} = 22 $$
$$\Rightarrow (n-1)\times \dfrac {1}{4} = 18 $$
$$\Rightarrow n-1 =72 $$
$$ n = 73 $$

Hence there are $$ 73 $$ terms in the series .
Question 3
1 / -0

If the seventh term of an AP is 25 and the common difference is 4, then find the 15th term of AP.

A
55

B
50

C
57

D
52

Solution

Let $$ a $$ be the first term and $$ d $$ be their common difference of the AP.
Then, $$ n th $$ term of the AP $$ T_n = a + (n-1)d $$

Given, $$ T_7 = 25 $$
$$ => a + (7-1)d = 25 $$
$$ => a + 6d = 25 $$
$$ => a + 6 \times 4 = 25 $$
$$ => a = 1 $$

So, $$ T_{15} = a + (15-1)d = 1 + 14(4) = 57 $$
Question 4
1 / -0

If $$8$$ times the $$8th$$ term of an arithmetic progression is equal to $$12$$ times the $$12th$$ term, then the $$20th$$ term is

A
$$12$$ times to $$8th$$ term

B
$$8$$ times to $$12th$$ term

C
$$0$$

D
Cannot be determined

Solution

Let $$ a $$ be the first term and $$ d $$ be their common difference of the AP.
Then, $$ n th $$ term of the AP $$ T_n = a + (n-1)d $$

Given, $$ 8 \times T_8 = 12 \times T_{12} $$
$$\Rightarrow 8(a + (8-1)d) = 12(a + (12-1)d) $$
$$ \Rightarrow 8(a + 7d) = 12(a + 11d) $$
$$\Rightarrow 2(a + 7d) = 3(a + 11d) $$
$$\Rightarrow 2a + 14d = 3a+33d $$
$$\Rightarrow a = - 19d $$

So, $$ T_{20} = a + (20-1)d = -19d + 19d = 0 $$
Question 5
1 / -0

What is the common difference of an AP whose n$$^{th}$$ term is $$xn+y$$?

A
$$x+y$$

B
$$y$$

C
$$x$$

D
$$-x$$

Solution

Given that, the $$n^{th}$$ term of an A.P. is $$xn+y$$.

To find out: The common difference of the A.P.

We know that, the common difference of an A.P. is the difference between any two consecutive terms.
That is, $$d=a_{n+1}-a_n$$
We have $$a_n=xn+y$$
$$\therefore \ a_{n+1}=x(n+1)+y=nx+x+y$$

Hence, $$d=(nx+x+y)-(nx+y)$$
$$\Rightarrow nx+x+y-nx-y$$
$$\therefore \ d=x$$

Hence, the common difference of the given A.P. is $$x$$.
Question 6
1 / -0

Directions For Questions

The 3rd term is 6 and the 17th term is 34.

...view full instructions

Find the common difference of an AP

A
2

B
3

C
1

D
4

Solution

Let $$ a $$ be the first term and $$ d $$ be their common difference of the AP.
Then, $$ n th $$ term of the AP $$ a_n = a + (n-1)d $$

Given, $$ a_3 = 6 $$
$$ => a + (3-1)d = 6 $$
$$ => a + 2d = 6 $$ -- (1)

Also, $$ a_17 = 34 $$
$$ => a + (17-1)d = 34 $$
$$ => a + 16d = 34 $$ -- (2)

Subtracting eqn 2 from eqn 1, we get
$$ 14d = 28 $$
$$ => d = 2 $$
Question 7
1 / -0

Find the sum of first 20 terms of the sequence whose nth term is $$2n+3n$$

A
1050

B
645

C
1100

D
1000

Solution

Given:$$a_n=2n+3n$$
Put $$n=1$$ in $$a_n$$, we get
$$a_1=2(1)+3(1)=2+3=5$$
Put $$n=2$$ in $$a_n$$, we get
$$a_2=2(2)+3(2)=4+6=10$$
Put $$n=3$$ in $$a_n$$, we get
$$a_3=2(3)+3(3)=6+9=15$$
So, Given A.P is $$5, 10, 15,....$$
Hence, $$a=5$$ and $$d=a_2-a_1=10-5=5$$
Now, Sum of first $$20$$ terms is given by
$$S_{20}=\dfrac{n}{2}[2a+(n-1)d]$$
$$\Rightarrow S_{20}=\dfrac{20}{2}[2(5)+(20-1)(5)]$$
$$\Rightarrow S_{20}=10[10+19(5)]$$
$$\Rightarrow S_{20}=1050$$
So, $$\text{A}$$ is the correct option.
Question 8
1 / -0

Find the sums given below:
$$-5+(-8)+(-11)+......+(-230)$$

A
$$-8930$$

B
$$8769$$

C
$$6778$$

D
$$6667$$

Solution

Given series can also be written as,
$$-5-8-11-14-....-230$$
It can be observed that difference between any two consecutive terms of the given series is constant and equal to $$-3$$.
Thus given series is an AP whose common difference and first term as $$-5$$ and $$-3$$ respectively.
As we know $$n^{th}$$ term is given as
, $$a_n = a+(n-1)d$$
& Sum of first $$n$$ terms, $$S_n = \dfrac{n}{2} (a+l)$$, where $$a$$ & $$l$$ are the first term & last term of an AP.
So, $$a=-5, d=-3$$
$$\Rightarrow -230=-5-3(n-1)$$
$$\Rightarrow 3(n-1)=225$$
$$\Rightarrow n-1=75$$
$$\Rightarrow n=76$$
Sum of first $$n$$ terms is given as,
$$S_n = \dfrac{n}{2} (a+l)$$, where $$a$$ & $$l$$ are the first term & last term of an AP.
Thus,
$$S=\dfrac {76}{2}(-5+-230)\\=38\times (-235)\\=-8930$$
Question 9
1 / -0

The sum of the three numbers in A.P is $$21$$ and the product of the first and third number of the sequence is $$45$$. What are the three numbers?

A
$$3, 7,$$ and $$11$$

B
$$9, 7,$$ and $$5$$

C
$$3, 9,$$ and $$15$$

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

E
None of these

Solution

Let the three numbers in A.P be $$a - d, a, a + d.$$
Then,
$$\begin{aligned}{}a - d + a + a + d &= 21\\3a &= 21\\a &= 7\end{aligned}$$

And,
$$\begin{aligned}{}(a - d)(a + d) &= 45\\{a^2} - {d^2}& = 45\\(7)^2-d^2&=45\\49-d^2&=45\\{d^2} &= 4\\d &= \pm 2\end{aligned}$$

Hence, the numbers are $$5, 7$$ and $$9$$ when $$d = 2$$ and $$9, 7$$ and $$5$$ when $$d = -2$$.
In both cases, numbers are the same.
Question 10
1 / -0

Find the sum of the AP: $$2,5,8,11,\dots$$ to first $$11$$ terms.

A
$$187$$

B
$$178$$

C
$$192$$

D
$$171$$

Solution

The sum of first $$n$$ terms of an AP whose first term is $$a$$ and common difference is $$d$$ is given by,
$$S_n = \dfrac{n}{2} [2a + (n - 1) d]$$

Here
$$d = 5 - 2 $$
$$= 3$$
$$a = 2$$

So,
$$S_{11}= \dfrac{11}{2} [4 + (11 - 1) \times 3]$$
$$= \dfrac{11}{2} [4 + 30]$$
$$= \dfrac{11}{2} \times 34$$
$$= 11 \times 17$$
$$= 187$$

Hence, option $$A$$ is correct.

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Arithmetic Progressions Test - 46

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Arithmetic Progressions Test - 46

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SHARING IS CARING

Question 1

$$28$$

$$25$$

$$22$$

$$20$$

Solution

Question 2

$$70$$

$$71$$

$$72$$

$$73$$

Solution

Question 3

55

50

57

52

Solution

Question 4

$$12$$ times to $$8th$$ term

$$8$$ times to $$12th$$ term

$$0$$

Cannot be determined

Solution

Question 5

$$x+y$$

$$y$$

$$x$$

$$-x$$

Solution

Question 6

2

3

1

4

Solution

Question 7

1050

645

1100

1000

Solution

Question 8

$$-8930$$

$$8769$$

$$6778$$

$$6667$$

Solution

Question 9

$$3, 7,$$ and $$11$$

$$9, 7,$$ and $$5$$

$$3, 9,$$ and $$15$$

Both $$A$$ and $$B$$

None of these

Solution

Question 10

$$187$$

$$178$$

$$192$$

$$171$$

Solution

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