Arithmetic Progressions Test

Result

Arithmetic Progressions Test - 56

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If fourth therm of an $$A.P$$ is thrice its first term and seventh term is one less than the twice of third term, then its common difference is ?

A
$$1$$

B
$$2$$

C
$$-2$$

D
$$3$$

Solution

Let a be the first term and d be the common difference.

The 4th term can be written as ‘a+3d’ and 7th term can be written as ‘a+6d’.

As per the question,

1. a+3d = 3a → 2a-3d=0

2. a+6d=2(a+2d)+1 →a+6d=2a+4d+1 →a-2d+1=0

Now solve the 2 quations, you will get:

d=2 & a=3

Hence, the first term is 3 and the common difference is 2.

The AP would be,

3,5,7,9,11,…..
Question 2
1 / -0

Sum of first $$n$$ natural numbers is $$S_{1}$$, sum of first $$n$$ odd natural numbers is $$S_{2}$$, and sum of first $$n$$ even numbers is $$S_{3}$$. Then $$S_{1}:S_{2}:S_{3}$$=

A
$$n-1\ :\ n\ :\ n+1$$

B
$$n\ :\ 2n-1-1\ :\ 2n$$

C
$$n+1\ :\ 2n\ :\ 2(n+1)$$

D
$$n+1\ :\ n\ :\ 2n-1\ :\ 2n$$

Solution

$$ S_{n}^{{}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] $$
$$ S_{1}^{{}}=1+2+3+4+........n $$
$$ a=1,d=1 $$
$$ S_{1}^{{}}=\frac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)1 \right] $$
$$ \,\,\,\,\,\,=\frac{n\left( n+1 \right)}{2} $$
$$ S_{2}^{{}}=1+3+5+..........+n $$
$$ a=1,d=2 $$
$$ S_{2}^{{}}=\frac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)2 \right] $$
$$ \,\,\,\,\,\,\,=\frac{n}{2}\left[ 2n \right] $$
$$ \,\,\,\,\,={{n}^{2}} $$
$$ S_{3}^{{}}=2+4+6+..........n $$
$$ a=2,d=2 $$
$$ S_{3}^{{}}=\frac{n}{2}\left[ 2\left( 2 \right)+\left( n-1 \right)2 \right] $$
$$ \,\,\,\,\,=n\left[ 2+n-1 \right] $$
$$ \,\,\,\,\,=n\left( n+1 \right) $$
$$ S_{1}^{{}}:S_{2}^{{}}:S_{3}^{{}} $$
$$ \frac{n\left( n+1 \right)}{2}:{{n}^{2}}:n\left( n+1 \right) $$
$$ \Rightarrow \frac{n+1}{2}:n:n+1 $$
$$ \Rightarrow \left( n+1 \right):2n:2\left( n+1 \right) $$
Question 3
1 / -0

Sum of the first $$n$$ terms of an A.P. having first term $$a$$ and last term $$\ell$$ is ____

A
$$\dfrac{n}{2} (2a - \ell)$$

B
$$\dfrac{n}{2} (2a + \ell)$$

C
$$\dfrac{n}{2} (a + \ell)$$

D
$$\dfrac{n}{2} (a - \ell)$$

Solution

Given: First term of an A.P $$= a$$
Last term $$=2 l $$

$$\therefore$$ $$l = a+(n-1)d$$ $$...(1)$$

Sum of $$n$$ terms $$=\dfrac{n}{2}(2a+(n-1)d)$$

$$=\dfrac{n}{2}(a+a+(n-1)d) $$

$$=\dfrac{n}{2}(a+l)$$ [From $$(1)$$]
Question 4
1 / -0

If the first, second and last term of an $$A.P.$$ are $$a,\ b$$ and $$2a$$ respectively. then its sum is

A
$$\dfrac { ab }{ 2\left( b-a \right) } $$

B
$$\dfrac { ab }{ \left( b-a \right) } $$

C
$$\dfrac { 3ab }{ 2\left( b-a \right) } $$

D
$$none\ of\ these$$

Solution

Common difference $$=b-a$$
As we know the nth term, $$a_n = a+(n-1)d$$, where a and d are the first term and common difference respectively.
$$\therefore 2a=a+(n-1)(b-a)$$
$$\Rightarrow \dfrac{a}{b-a}=n-1$$
$$\Rightarrow \dfrac{a}{b-a}+1=n$$
$$\Rightarrow \dfrac{a+b-a}{b-a}=n$$
$$\Rightarrow \dfrac{b}{b-a}=n$$
$$\therefore$$ Sum of $$AP=\dfrac{n}{2}\left[ a+2a\right]$$
$$=\dfrac{3ab}{2(b-a)}$$
Question 5
1 / -0

If the ratio of $${n^{th}}$$ terms of two A.P.'s is $$\left( {2n + 8} \right):\left( {5n - 3} \right)$$, then the ratio of the sums of their $$n$$ terms is

A
$$\,\,\left( {2n + 18} \right):\left( {5n + 3} \right)$$

B
$$\,\left( {5n - 1} \right):\left( {2n + 18} \right)$$

C
$$\,\,\left( {2n + 18} \right):\left( {5n - 1} \right)$$

D
$$\,\left( {3n + 18} \right):\left( {4n + 1} \right)$$

Solution

Given ratio of $$n^{th}$$ terms of A.P.:
$$(2n+8):(5n-3)=\dfrac{2n-2+2+8}{5n-5+5-3}=\dfrac{10+(n-1)\times2}{2+(n-1)\times5}$$
Since $$n^{th}$$ terms of an A.P. is $$t_n=a+(n-1)d$$
For the A.P. in numerator, $$a_1=10$$ and $$d_1=2$$ and denominator $$a_2=2$$ and $$d_2=5$$
Also, Sum of n terms of an A.P. is $$S_n=\dfrac{n}{2}[2a+(n-1)d]$$
Sum of n terms of the A.P. in numerator $$=\dfrac{n}{2}[2\times10+(n-1)\times2]$$
Sum of n terms of the A.P. in denominator $$=\dfrac{n}{2}[2\times2+(n-1)\times5]$$

Then, the ratio of the sums of their n terms is $$\dfrac{\dfrac{n}{2}[20+2(n-1)]}{\dfrac{n}{2}[4+5(n-1)]}=\dfrac{2n+18}{5n-1}=(2n+18):(5n-1)$$
Question 6
1 / -0

A man saved Rs. $$200$$ in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. $$40$$ more than the saving of immediately previous month. His total saving from the start of service will be Rs. $$11040$$ after.

A
$$18$$ months

B
$$19$$ months

C
$$20$$ months

D
$$21$$ months

Solution

$$\dfrac{\overset { 1st }{ Month } }{200}\ \dfrac{\overset { 2nd }{ Month } }{200}\ \dfrac{\overset { 3rd }{ Month } }{200} \ \dfrac{\overset { 4th }{ Month } }{240}\ \dfrac{\overset { 5th }{ Month } }{280}$$
So let $$n$$ be the number of monthe.
Final savings $$=11040$$
So $$\underbrace { 200+200 } +200+240+280+.... =11040$$.
$$\Rightarrow 200+240+280+...+200+ (n-1)40=11040-400=10640$$
Here $$a=200, d=40$$.
We know $$a+ (a+d)+....+af(n-1)d= \dfrac{n}{2} (2a+(n-1)d)$$
$$\Rightarrow \dfrac{n}{2} (2 \times 200+ (n-1) \times 40)=10640$$
$$\Rightarrow n(n+q) =532$$
$$\Rightarrow n^{2}+qn-532=0$$.
$$\Rightarrow (n+28)(n-19)=0$$.
$$\Rightarrow n=19$$ (Since $$n=-28$$ not possible)
Question 7
1 / -0

For the A.P. if $$a = 7$$ and $$d = 2.5 ,$$ then $$t _ { 12 } = ?$$

A
37.5

B
34.5

C
28.2

D
44.5

Solution

$$ {\textbf{Step - 1: Find 12th term}} $$

$$ {\text{Given, a = 7 and d = 2}}{\text{.5}} $$

$$ {\text{a = first term and }} $$

$$ {\text{d = difference between two terms}} $$

$$ {\text{It is known that in an AP}}\;{{\text{n}}^{th}}{\text{ term will be}} $$

$$ {{\text{t}}_n}{\text{ = a + (n - 1)d}} $$

$$ {\text{Put the value of n as 12}}{\text{.}} $$

$$ \text{t}_{12} = 7 + (12 - 1)(2.5) $$

$$ { = 7 + 11 \times 2.5} $$

$$ { = 34.5} $$

$$ {\textbf{Hence, the correct answer is option B}}{\text{.}} $$
Question 8
1 / -0

Three numbers are in $$AP$$ such that their sum is $$18$$ and sum of their squares is $$158$$. The greatest number among them is

A
$$10$$

B
$$11$$

C
$$12$$

D
$$none\ of\ these$$

Solution

Let the nos be , $$ a-d, a, a+d$$ then $$ a-d+a+a+d = 18 $$
$$ 3a = 18 \Rightarrow a = 6 $$
$$ \therefore 6-d, 6, 6+d $$
Sum of square = 158
$$ (6-d)^{2}+6^{2}+(6+d)^{2} = 158 $$
$$ 6^{2}+d^{2}-12d+6^{2}+6^{2}+d^{2}+12d = 158 $$
$$ 108+2d^{2} = 158 $$
$$ 2d^{2} = 50 $$
$$ d^{2} = 25 \Rightarrow d = \pm 5 $$
when $$d=+5$$, Numbers are : 1, 6, 11
when $$d=-5$$, Numbers are : 11, 6, 1
Greatest : 11
Question 9
1 / -0

Find the sum of $$2,4,6,8,,,........2n$$

A
$$n(n+1)$$

B
$$n^2$$

C
$$n(n-1)$$

D
$$n^2+n$$

Solution

The given sequence is $$2,4,6,8,..... 2n$$
$$\therefore a= 2, d = 2$$ & $$a_k= 2n$$
$$\therefore 2n = a+(k-1)d $$
$$\Rightarrow 2n = 2+(k-1)2$$
$$\Rightarrow k = n$$
Sum of $$n$$ terms $$=\dfrac{k}{2}(2a+(k-1)d)$$
$$= \dfrac {n}{2} \times [2(2)+(n-1)(2)] $$
$$=n(n+1)$$
Question 10
1 / -0

The $$27^{th}$$ term of AP $$7,9,11,13,15,17,19,\ldots$$ is

A
$$59$$

B
$$61$$

C
$$63$$

D
None of these

Solution

Given series is $$7,9,11,13,15,17,19,\dots$$
$$a=7$$
$$d=9-7$$
$$=2$$

Formula for $$n^{th}$$ term of an AP whose first term is $$a$$ and common difference is $$d$$ is given by,
$$a_n=a+(n-1)d$$

So, $$27^{th} $$ term is given as,
$$a_{27}=7 + (27-1)2$$
$$=7+26\times2$$
$$=7+52$$
$$=59$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 56

Result

Arithmetic Progressions Test - 56

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1$$

$$2$$

$$-2$$

$$3$$

Solution

Question 2

$$n-1\ :\ n\ :\ n+1$$

$$n\ :\ 2n-1-1\ :\ 2n$$

$$n+1\ :\ 2n\ :\ 2(n+1)$$

$$n+1\ :\ n\ :\ 2n-1\ :\ 2n$$

Solution

Question 3

$$\dfrac{n}{2} (2a - \ell)$$

$$\dfrac{n}{2} (2a + \ell)$$

$$\dfrac{n}{2} (a + \ell)$$

$$\dfrac{n}{2} (a - \ell)$$

Solution

Question 4

$$\dfrac { ab }{ 2\left( b-a \right) } $$

$$\dfrac { ab }{ \left( b-a \right) } $$

$$\dfrac { 3ab }{ 2\left( b-a \right) } $$

$$none\ of\ these$$

Solution

Question 5

$$\,\,\left( {2n + 18} \right):\left( {5n + 3} \right)$$

$$\,\left( {5n - 1} \right):\left( {2n + 18} \right)$$

$$\,\,\left( {2n + 18} \right):\left( {5n - 1} \right)$$

$$\,\left( {3n + 18} \right):\left( {4n + 1} \right)$$

Solution

Question 6

$$18$$ months

$$19$$ months

$$20$$ months

$$21$$ months

Solution

Question 7

37.5

34.5

28.2

44.5

Solution

Question 8

$$10$$

$$11$$

$$12$$

$$none\ of\ these$$

Solution

Question 9

$$n(n+1)$$

$$n^2$$

$$n(n-1)$$

$$n^2+n$$

Solution

Question 10

$$59$$

$$61$$

$$63$$

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.