Arithmetic Progressions Test

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Arithmetic Progressions Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The common difference of A.P.: $$1, -1, -3, ....$$ is _______

A
$$-1$$

B
$$+2$$

C
$$-2$$

D
$$+1$$

Solution

The common difference d of A.P. is given by the difference between second term and first term.
$$\therefore d=-1-1=-2$$
Option $$C$$ is correct.
Question 2
1 / -0

If $$a, b$$ and $$c$$ arc in arithmetic progression then $$\dfrac{b- a}{c - b}$$ is equal to

A
$$\dfrac{b}{a}$$

B
$$0$$

C
$$1$$

D
$$2a$$

Solution

Solution:
If $$a,b$$ and $$c$$ are in A.P. then,
$$b-a=c-b=$$ common difference
or, $$\cfrac{b-a}{c-b}=1$$
Therefore, $$C$$ is the correct option.
Question 3
1 / -0

Firs term of an arithmetic progression is $$2$$. If the sum of its first five terms is equal to one-fourth of the sum of the next five terms, then the sum of its first $$30$$ terms is

A
$$2670$$

B
$$2610$$

C
$$-2520$$

D
$$2550$$

Solution

Given,
First term $$a=2$$
According to the problem, first 10 are
$$a,a+d, a+2d, a+3d,a+4d, a+5d, a+6d, a+7d, a+8d, a+9d$$
$$5a+10d=\dfrac{1}{4}(5a+35d)$$
$$20a+40d = 5a+35d$$
$$15a = - 5d$$
$$3a = -d$$
As $$a=2$$
$$d =-6$$ and $$n=30$$
$$S_{30} = \dfrac{n}{2}[2(a)+(n-1)d]$$
$$S_{30} = \dfrac{30}{2}[2(2)+(29)(-6)]$$
$$=-2550$$
Question 4
1 / -0

$$(n-2)^{th}$$ term of an arithmetic progression will be __________.

A
$$a+(n-1)d$$

B
$$a+(n-3)d$$

C
$$a+(n-2)d$$

D
None

Solution

$$n^{th}$$ term of an A.P. is
$$T_n=a+(n-1)d$$
$$\therefore (n-2)^{th}$$ term of an A.P. is
$$T_{n-2}=a+(n-2-1)d$$
$$=a+(n-3)d$$
Question 5
1 / -0

If $$S_{r}$$ denotes the sum of $$r$$ terms of an AP and $$\dfrac {S_{a}}{a^{2}} = \dfrac {S_{b}}{b^{2}} = c$$ then $$S_{c}$$ is

A
$$c^{3}$$

B
$$c/ ab$$

C
$$abc$$

D
$$a + b + c$$

Solution

$${ S }_{ r }$$ denotes sum of $$r$$ terms.
$$\cfrac { { S }_{ a } }{ { a }^{ 2 } } =\cfrac { { S }_{ b } }{ { b }^{ 2 } } \\ \cfrac { \cfrac { a }{ 2 } (2{ a }_{ 1 }+(a-1)d) }{ { a }^{ 2 } } =\cfrac { \cfrac { b }{ 2 } (2{ a }_{ 1 }+(b-1)d) }{ { b }^{ 2 } } \\ \cfrac { 2{ a }_{ 1 } }{ a } +\cfrac { (a-1)d }{ a } =\cfrac { 2{ a }_{ 1 } }{ b } +\cfrac { (b-1)d }{ b } \\ or,\cfrac { 2{ a }_{ 1 } }{ a } +d-\cfrac { d }{ a } =\cfrac { 2{ a }_{ 1 } }{ b } +d-\cfrac { d }{ b } \\ or,2{ a }_{ 1 }(\cfrac { 1 }{ a } -\cfrac { 1 }{ b } )=d(\cfrac { 1 }{ a } -\cfrac { 1 }{ b } )\\ So,\cfrac { a }{ 2 } (\cfrac { 2{ a }_{ 1 }+(a-1)d }{ { a }^{ 2 } } )=c\\ or,2{ a }_{ 1 }\cfrac { (1+a-a) }{ 2a } =c\\ \cfrac { { a }_{ 1 }a }{ a } =c\\ c={ a }_{ 1 }\\ { S }_{ c }=\cfrac { c }{ 2 } (2{ a }_{ 1 }+(c-1)d)\\ =\cfrac { c }{ 2 } (2{ a }_{ 1 }+(c-1)2{ a }_{ 1 })\\ =\cfrac { c }{ 2 } (2c+(c-1)2c)\\ =\cfrac { c }{ 2 } \times 2c(1+c-1)\\ ={ c }^{ 3 }$$
Answer $$(A)$$
Question 6
1 / -0

Find the sum of the first $$25$$ terms of an A.P whose $$n$$th term is given by $${a}_{n}=8-3n$$

A
$$-775$$

B
$$-500$$

C
$$800$$

D
$$500$$

Solution

$$a_{n}=8-3n$$

$$a_{1}=5$$

$$a_{2}=2$$

$$d=2-5=-3$$

Thus, $$S_{25}=\dfrac{25}{2}[10+24(-3)]$$

$$=\dfrac{25}{2} \times -62=-775$$

Hence, option A is correct.
Question 7
1 / -0

If $$\dfrac{2}{3},k, \dfrac{5}{8}$$ are in AP, find the value of k.

A
$$\dfrac{48}{31}$$

B
$$\dfrac{31}{48}$$

C
$$\dfrac{17}{29}$$

D
$$\dfrac{29}{17}$$

Solution

$$\dfrac{2}{3},k, \dfrac{5}{8}$$ are in AP.

Thus, $$k-\dfrac {2}{3} = \dfrac{5}{8}-k$$

$$2k=\dfrac{2}{3}+\dfrac{5}{8}$$

$$2k=\dfrac{31}{24}$$

$$k=\dfrac{31}{48}$$.
Question 8
1 / -0

Find the number of terms of the A.P $$-12,-9,-6,.....,21$$. If $$1$$ is added to each term of this A.P., then find the sum of all terms of the A.P thus obtained.

A
$$10,66$$

B
$$12,66$$

C
$$14,66$$

D
None of these

Solution

From given series, we have,

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

$$21=-12+(n-1)3$$

$$n=\dfrac{21+12}{3}+1$$

$$\therefore n=12$$

upon adding $$1$$ to the series, we get,

$$-11,-8,-5,.............,22$$

$$S_n=\dfrac{n}{2}(a+l)$$

$$S_n=\dfrac{12}{2}(-11+22)$$

$$\therefore S_n=66$$
Question 9
1 / -0

There are $$20$$ rows of seats in a conference hall with $$20$$ seats in the first row, $$21$$ seats in the second row, $$22$$ seats in the third row and so on. In total, how many seats are there in the conference hall?

A
$$500$$

B
$$550$$

C
$$590$$

D
$$620$$

Solution

Given:-
Total number of rows $$(n) = 20$$
Seats in first row $$(a) = 20$$
Seats in second row $$({a}_{2}) = 21$$
Seats in third row $$({a}_{3})= 22$$
Common difference $$(d) = {a}_{3} - {a}_{2}= 22-21 = 1$$
$$\Rightarrow \text{Total number of seats in conference hall } ({S}_{n}) = \cfrac{n}{2} \left(2a + \left(n-1\right)d \right)$$
$$\Rightarrow {S}_{20} = \cfrac{20}{2} \left(2 \times 20 + \left(20 - 1 \right) \times 1 \right)$$
$$\Rightarrow {S}_{20} = 10 \left(40 +19 \right) = 10 \times 59 = 590$$
Hence correct answer is 590.
Question 10
1 / -0

Find the sum of the first $$15$$ terms of the sequence having $$n$$th term as
$${ y }_{ n }=9-5n\quad $$

A
$$-390$$

B
$$468$$

C
$$-465$$

D
$$-468$$

Solution

$$y_{n}=9-5n$$

$$y_{1}=9-5=4$$

$$y_{2}=9-10=-1$$

$$d=y_{2}-y_{1}=-1-4=-5$$

Thus, $$S_{15}=\dfrac{15}{2}[2(4)+14(-5)]$$

$$=\dfrac{15}{2}[8-70]=-465$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 30

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Arithmetic Progressions Test - 30

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SHARING IS CARING

Question 1

$$-1$$

$$+2$$

$$-2$$

$$+1$$

Solution

Question 2

$$\dfrac{b}{a}$$

$$0$$

$$1$$

$$2a$$

Solution

Question 3

$$2670$$

$$2610$$

$$-2520$$

$$2550$$

Solution

Question 4

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

$$a+(n-3)d$$

$$a+(n-2)d$$

None

Solution

Question 5

$$c^{3}$$

$$c/ ab$$

$$abc$$

$$a + b + c$$

Solution

Question 6

$$-775$$

$$-500$$

$$800$$

$$500$$

Solution

Question 7

$$\dfrac{48}{31}$$

$$\dfrac{31}{48}$$

$$\dfrac{17}{29}$$

$$\dfrac{29}{17}$$

Solution

Question 8

$$10,66$$

$$12,66$$

$$14,66$$

None of these

Solution

Question 9

$$500$$

$$550$$

$$590$$

$$620$$

Solution

Question 10

$$-390$$

$$468$$

$$-465$$

$$-468$$

Solution

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