Arithmetic Progressions Test

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Arithmetic Progressions Test - 54

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Question 1
1 / -0

The mean of five numbers in AP is $$89$$. The product of first and last terms is $$7885$$. The AM of first, third and fifth term is

A
$$83$$

B
$$86$$

C
$$89$$

D
$$90$$

Solution

Let the numbers in A.P. be $$ (a-2d), (a-d), a, (a+d), (a+2d)$$
$$\therefore \dfrac{(a-2d)+(a-d)+a+(a+d)+(a+2d)}{5}=89$$
$$\Rightarrow \dfrac{5a}{5}=89$$
$$\therefore a=89$$
$$\therefore$$ The AM of 1st, 3rd & 5th term is:
$$=\dfrac{(a-2d)+a+(a+2d)}{3}$$
$$=a=89$$
Question 2
1 / -0

The common difference of an Arithmatic Progression, whose $$3^{rd}$$ term is $$5$$ and $$7^{th}$$ term is $$9$$, is ...............

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

We know that
$$n^{th}$$ term of an AP ,$${a}_{n} = a+(n-1)d$$
where $$a=$$First term, $$d=$$Common difference and $$n=$$Position of term.

Given $$3^{rd}$$ term is $$5$$ ,
$${a}_{3} = a+(3-1)d$$
$$\Rightarrow 5 = a+2d$$
$$\Rightarrow a = 5-2d$$........$$(1)$$

Given $$7^{th}$$ term is $$9$$,
$${a}_{7} = a+(7-1)d$$
$$\Rightarrow 9 = a+6d$$
$$\Rightarrow a =9 -6d$$ .......$$(2)$$

From $$(1)$$ and $$(2)$$
$$5-2d = 9 -6d$$
$$\Rightarrow 6d-2d = 9-5$$
$$\Rightarrow 4d = 4$$
$$\Rightarrow d = 1$$
So , common difference is $$1$$
Question 3
1 / -0

The Sum of three numbers in AP is $$75$$, and product of extremities is $$609$$. The AM of first and second number is

A
$$\{21,25,29\}$$, AM $$= 23$$

B
$$\{13,17,21\}$$, AM $$= 22$$

C
$$\{21,25,29\}$$, AM $$= 25$$

D
$$\{21,22,29\}$$, AM $$= 23$$

Solution

Let the numbers in A.P. be (a-d), a, (a+d)
$$\therefore (a-d)+a+(a+d)=75$$
$$\Rightarrow a=25$$
Also, $$(a-d)(a+d)=609$$
$$\Rightarrow a^{2}-d^{2}=609$$
$$\Rightarrow 25^{2}-d^{2}=609$$
$$\therefore d=\pm4$$
The numbers are {21,25,29} or {29,25,21}
According to option, we take {21,25,29}
A.M. of first and second number $$=\dfrac{21+25}{2}=23$$
Question 4
1 / -0

The sum of first $$10$$ terms and $$20$$ terms of an AP are $$120$$ and $$440$$ respectively. What is the common difference?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let the first term be $$a$$ and common difference be $$d$$.
Sum of first n terms in an A.P. $$=\dfrac{n}{2}(2a+(n-1)d)$$
So, sum of first $$10$$ terms $$=\dfrac { 10 }{ 2 } (2a+(10-1)d)$$
$$\implies 120 =5(2a+9d)$$
$$\implies 24=2a+9d$$ .............. $$(i)$$
Sum of first 20 terms $$=\frac { 20 }{ 2 } (2a+(20-1)d)$$
$$\implies 440 =10(2a+19d)$$
$$\implies 44=2a+19d$$ ......... $$(ii)$$
Subtracting equation (i) from (ii) gives
$$20=10d$$
$$\implies d=2$$
Common difference =2
Hence, option B is correct.
Question 5
1 / -0

In a given A.P., $${ T }_{ 25 }-{ T }_{ 20 }=15$$. $$\therefore$$ $$d=$$............. for the A.P.

A
$$5$$

B
$$3$$

C
$$25$$

D
$$120$$

Solution

Given: $$T_{25}-T_{20}=15$$ ........... $$(i)$$
We know that, $$n^{th}$$ term of an A.P. is given by $$T_n=a+(n-1)d$$
$$\therefore { T }_{ 25 }={ a }_{ 1 }+(25-1)d={ a }_{ 1 }+24d$$ and
$${ T }_{ 20 }={ a }_{ 1 }+(20-1)d={ a }_{ 1 }+19d$$
Now, $${ T }_{ 25 }-{ T }_{ 20 }={ a }_{ 1 }+24d-{ a }_{ 1 }-19d=5d$$
$$\implies 15=5d$$ ...... From $$(i)$$
$$\implies d=3$$
Hence, the answer is $$3$$.
Question 6
1 / -0

................ can be one of the term in Arithmetic progression 4, 7, 10, ...............

A
103

B
123

C
171

D
99

Solution

The nth term of an AP is given by $$t_n = a + (n - 1) d$$
In the given AP,
$$a = 4, d = 3$$
So,
$$a + (n - 1) d = 4 + (n - 1) 3$$
$$= 4 + 3n - 3$$
$$= 3n + 1$$
Now, if we equate the first option
$$103 = 3n - 1$$
we get $$n = 34$$ which is a possible value for $$n$$
However, if we equate any of the other values we do not get a positive integer value of $$n$$.
This cannot be possible since $$n$$, which is the number of terms has to be a positive integer.
Question 7
1 / -0

If $$1 + 6 + 11 + 16 + ... + x = 148$$, then the value of $$x$$ is

A
$$36$$

B
$$35$$

C
$$-36$$

D
None of these

Solution

Given series is
$$1 + 6 + 11 + 16 + ..... + x = 148$$
Here, $$a = 1, d = 6 - 1 = 11 - 6 = 16 - 11 = 5$$ and $$S_{n} = 148$$
$$\because S_{n} = \dfrac {n}{2} [2a + (n - 1) d]$$
$$\therefore 148 = \dfrac {n}{2}[2 \times 1 + (n - 1)5]$$
$$\Rightarrow 296 = 2n + 5n^{2} - 5n$$
$$\Rightarrow 296 = 5n^{2} - 3n$$
$$\Rightarrow 5n^{2} - 3n - 296 = 0$$
$$\Rightarrow (n - 8)(5n + 37) = 0$$
$$\Rightarrow n = 8, n = \dfrac {-37}{5}$$
$$\therefore n = 8$$
$$(\because n$$ cannot be negative)
Now, $$T_{n} = a + (n - 1)d$$
$$x = 1 + (8 - 1)\times 5$$
$$\Rightarrow x = 1 + 35 \Rightarrow x = 36$$.
Question 8
1 / -0

Find the sum of first 25 terms of an A.P. whose $$n^{th}$$ term is given by $$T_n = (7 - 3n)$$

A
800

B
-803

C
735

D
None of these

Solution

Given, $$T_{n}=7-3n$$.
Let us try to reaarange it into the form, $$T_{n}=a+(n-1)d$$
where $$a$$ is the first term,$$n$$ is the number of terms and $$d$$ is the common difference.
$$T_{n}=7-3n$$
$$\Rightarrow$$ $$T_{n}=4+3-3n$$
$$\Rightarrow$$ $$T_{n}=4-3(n-1)$$
So,here $$a=4,d=-3$$
Sum of $$n $$ terms in an AP can be calculated using the formula
$$S_{n}=\frac{n}{2}(a+T_{n})$$
$$\Rightarrow$$ $$S_{n}=\frac{n}{2}(4+7-3n)$$
$$\Rightarrow$$ $$S_{n}=\frac{n}{2}(11-3n)$$
$$\Rightarrow$$ $$S_{n}=\frac{25}{2}(-64)$$ ($$\because n=25$$)
$$\Rightarrow$$ $$S_{n}=-800$$
$$\therefore$$ No Option
Question 9
1 / -0

The $$54^{th}$$ and $$4^{th}$$ terms of an A.P. are $$-61$$ and $$64$$ respectively. Find the $$23^{rd}$$ term.

A
$$\dfrac {35}2$$

B
$$\dfrac {33}2$$

C
$$\dfrac {37}2$$

D
$$\dfrac {39}2$$

Solution

Here we use the formula for nth term of an AP: $$a_n = a+(n-1)d$$
So we have $$a + 53d = -61 ....(1)$$
& $$a + 3d = 64 ....(2)$$
$$50d = -125$$ [Subtracting $$(2)$$ from $$(1)$$]
$$d = \dfrac {-5}{2}, $$ using this value in $$(2)$$
$$\implies a = 64 - 3d = 64 + \dfrac {15}{2} = \dfrac {143}{2}$$
$$a_{23} = \dfrac {143}{2} - 22 \times \dfrac {5}{2} = \dfrac {33}{2}$$.
Question 10
1 / -0

The sum of all numbers of the form $$2k + 1$$, where $$k$$ takes on integral values from $$1$$ to $$n$$ is:

A
$$n^2$$

B
$$n(n + 1)$$

C
$$n(n + 2)$$

D
$$(n + 1)^2$$

E
$$(n + 1)(n + 2)$$

Solution

The sum of numbers from $$2K+1$$, where $$K$$ started with $$1,2,3,4,5......$$
then the sum will be
$$3+5+7+9+11+.......$$
Now using sum of series $$S_n $$=$$\frac{n}{2}[2a+(n-1) \times d]$$
Now put the value for $$a=3\ and\ d=2$$
$$S_n$$=$$\frac{n}{2}[2\times3+(n-1)\times 2]$$
$$S_n$$=$$\frac{n}{2}$$$$[6+2n-2]$$
$$S_n$$=$$\frac{n}{2}[4+2n]$$
$$S_n$$=$$n(n+2)$$.
so option $$C$$ is correct.

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 54

Result

Arithmetic Progressions Test - 54

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SHARING IS CARING

Question 1

$$83$$

$$86$$

$$89$$

$$90$$

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

$$\{21,25,29\}$$, AM $$= 23$$

$$\{13,17,21\}$$, AM $$= 22$$

$$\{21,25,29\}$$, AM $$= 25$$

$$\{21,22,29\}$$, AM $$= 23$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

$$5$$

$$3$$

$$25$$

$$120$$

Solution

Question 6

103

123

171

99

Solution

Question 7

$$36$$

$$35$$

$$-36$$

None of these

Solution

Question 8

800

-803

735

None of these

Solution

Question 9

$$\dfrac {35}2$$

$$\dfrac {33}2$$

$$\dfrac {37}2$$

$$\dfrac {39}2$$

Solution

Question 10

$$n^2$$

$$n(n + 1)$$

$$n(n + 2)$$

$$(n + 1)^2$$

$$(n + 1)(n + 2)$$

Solution

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