Arithmetic Progressions Test

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Arithmetic Progressions Test - 30

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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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Arithmetic Progressions Test - 30

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Arithmetic Progressions Test - 30

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Question 1

−1-1−1

+2+2+2

−2-2−2

+1+1+1

Solution

Question 2

ba\dfrac{b}{a}ab​

000

111

2a2a2a

Solution

Question 3

267026702670

261026102610

−2520-2520−2520

255025502550

Solution

Question 4

a+(n−1)da+(n-1)da+(n−1)d

a+(n−3)da+(n-3)da+(n−3)d

a+(n−2)da+(n-2)da+(n−2)d

None

Solution

Question 5

c3c^{3}c3

c/abc/ abc/ab

abcabcabc

a+b+ca + b + ca+b+c

Solution

Question 6

−775-775−775

−500-500−500

800800800

500500500

Solution

Question 7

4831\dfrac{48}{31}3148​

3148\dfrac{31}{48}4831​

1729\dfrac{17}{29}2917​

2917\dfrac{29}{17}1729​

Solution

Question 8

10,6610,6610,66

12,6612,6612,66

14,6614,6614,66

None of these

Solution

Question 9

500500500

550550550

590590590

620620620

Solution

Question 10

−390-390−390

468468468

−465-465−465

−468-468−468

Solution

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$-1$

$+2$

$-2$

$+1$

$\dfrac{b}{a}$

$0$

$1$

$2a$

$2670$

$2610$

$-2520$

$2550$

$a+(n-1)d$

$a+(n-3)d$

$a+(n-2)d$

$c^{3}$

$c/ ab$

$abc$

$a + b + c$

$-775$

$-500$

$800$

$500$

$\dfrac{48}{31}$

$\dfrac{31}{48}$

$\dfrac{17}{29}$

$\dfrac{29}{17}$

$10,66$

$12,66$

$14,66$

$500$

$550$

$590$

$620$

$-390$

$468$

$-465$

$-468$