ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions MCQs

 MCQs

Question 1

The list of numbers – 10, – 6, – 2, 2, … is

(a) an A.P. with d = – 16

(b) an A.P with d = 4

(c) an A.P with d = – 4

(d) not an A.P

Sol :

-10, -6, -2, 2, … is

an A.P. with d = – 6 – (-10)

= -6 + 10 = 4

Ans (b)

Question 2

The 10th term of the A.P. 5, 8, 11, 14, … is

(a) 32

(b) 35

(c) 38

(d) 185

Sol :

10th term of A.P. 5, 8, 11, 14, …

{∵ a = 5, d = 3}

a + (n – 1)d = 5 + (10 – 1) x 3

= 5 + 9 x 3

= 5 + 27

= 32 

Ans (a)

Question 3

The 30th term of the A.P. 10, 7, 4, … is

(a) 87

(b) 77

(c) – 77

(d) – 87

Sol :

30th term of A.P. 10, 7, 4, … is

30th term = a + (n – 1)d

$\left\{\begin{aligned} \because a &=10 \\ d &=7-10=-3 \end{aligned}\right\}$

$=10+(30-1) \times(-3)$

$=10+29(-3)=10-87=-77$

Ans (c)

Question 4

The 11th term of the A.P. – 3, $-\frac{1}{2}, 2, \ldots$ is

(a) 28

(b) 22

(c) – 38

(d) – 48

Sol :

Given

$-3,-\frac{1}{2}, 2, \ldots$

$a=-3, d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=\frac{5}{2}$

11 th term $=a+(n-1) d$

$=-3+(11-1) \times \frac{5}{2}$

$=-3+10 \times \frac{5}{2}=-3+25=22$

Ans (b)

Question 5

The 4th term from the end of the A.P. – 11, – 8, – 5, …, 49 is

(a) 37

(b) 40

(c) 43

(d) 58

Sol :

4th term from the end of the A.P. -11, -8, -5, …, 49 is

Here, a = -11, d = -8 – (-11) = -8 + 11 = 3 and l = 49 .

$\therefore l=49=a+(n-1) d$

$\Rightarrow 49=-11+(n-1) \times 3$

$\Rightarrow 49-11=3(n-1)$

$\Rightarrow \frac{60}{3}=n-1$

$\Rightarrow n=20+1=21$

Now, 4 th term from the end $=l-(n-1) d$ 

$=49-(4-1) \times 3=49-9=40$

Ans (b)

Question 6

The 15th term from the last of the A.P. 7, 10, 13, …,130 is

(a) 49

(b) 85

(c) 88

(d) 110

Sol :

15th term from the end of A.P. 7, 10, 13,…, 130

Here, a = 7, d = 10 – 7 = 3, l = 130

15th term from the end = l – (n – 1)d

= 130 – (15 – 1) x 3

= 130 – 42

= 88 

Ans (c)

Question 7

If the common difference of an A.P. is 5, then $a_{18}-a_{13}$ is

(a) 5

(b) 20

(c) 25

(d) 30

Sol :

Common difference of an A.P. (d) = 5

$a_{18}-a_{13}=a+17 d-a-12 d=5 d=5 \times 5=25$

Ans (c)

Question 8

In an A.P., if $a_{18}-a_{14}=32$ then the common difference is

(a) 8

(b) – 8

(c) – 4

(d) 4

Sol :

If $a_{18}-a_{14}=32$, then d = ?

(a + 17d) – a – 13d = 32

⇒ a + 17d – a – 13d = 32

⇒ 4d = 32

$\Rightarrow d \frac{32}{4}=8$

Ans (a)

Question 9

In an A.P., if d = – 4, n = 7, $a_n$ = 4, then a is

(a) 6

(d) 7

(c) 20

(d) 28

Sol :

In an A.P., d = -4, x = 7, $a_n$ = 4 then a = ?

$a_n$ = a(n – 1)d = 4

$a_7$ = a + (7 – 1)d = 4

⇒ a + 6d = 4

⇒ a + 6 x (-4) = 4

a – 24 = 4

⇒ a = 4 + 24 = 28

Ans (d)

Question 10

In an A.P., if a = 3.5, d = 0, n = 101, then a_n will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Sol :

In an A.P.

a = 3.5, d= 0, n = 101, then an = ?

a_n = a₁₀₁ = a + (101 – 1)d

= 3.5 + 100d

= 3.5 + 100×0

= 3.5 + 0

= 3.5 (b)

Question 11

In an A.P., if a = – 7.2, d = 3.6, a_n = 7.2, then n is

(a) 1

(b) 3

(c) 4

(d) 5

Sol :

In an A.P.

a = – 7.2, d = 3.6, an = 7.2, n = ?

a_n = 7.2

a + (n – 1)d = 7.2

– 7.2 + (n – 1) 3.6 = 7.2

(n – 1) x 3.6 = 7.2 + 7.2 = 14.4

$(n-1)=\frac{14.4}{3.6}=4$

n=4+1=5

Ans (d)

Question 12

Which term of the A.P. 21, 42, 63, 84,… is 210?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Sol :

Which term of an A.P. 21, 42, 63, 84, … is 210

Let 210 be nth term, then

Here, a = 21, d = 42 – 21 =21

210 = a + (n – 1)d

210 = 21 + (n – 1) x 21

⇒ 210 – 21 = 21(n – 1)

$\Rightarrow \frac{189}{21}=n-1$

⇒ 9 = n – 1

⇒ n = 9 + 1 = 10

∴ It is 10th term.

Ans (b)

Question 13

If the last term of the A.P. 5, 3, 1, – 1,… is – 41, then the A.P. consists of

(a) 46 terms

(b) 25 terms

(c) 24 terms

(d) 23 terms

Sol :

Last term of an A.P. 5, 3, 1, -1, … is -41

Then A.P. will consist of ……. terms

Here, a = 5, d = 3 – 5 = – 2 and n =?

l = -41

l = -41

l = -41 = a + (n – 1 )d

-41 = 5 + (n – 1) (-2)

-41 – 5 = (n – 1) (-2)

$\Rightarrow \frac{-46}{-2}=n-1$

⇒ n – 1 = 23

⇒ n = 23 + 1 = 24

A.P. consists of 24 terms. 

Ans (c)

Question 14

If k – 1, k + 1 and 2k + 3 are in A.P., then the value of k is

(a) – 2

(b) 0

(c) 2

(d) 4

Sol :

k – 1, k + 1 and 2k + 3 are in A.P.

2(k+ 1) = (k – 1) + (2k + 3)

⇒ 2k + 2 = k – 1 + 2k + 3

⇒ 2k + 2 – 3k + 2

⇒ 3k – 2k = 2 – 2

⇒ k = 0 (b)

Question 15

The 21st term of an A.P. whose first two terms are – 3 and 4 is

(a) 17

(b) 137

(c) 143

(d) – 143

Sol :

First two terms of an A.P. are – 3 and 4

a = -3, d = 4 – (-3) = 4 + 3 = 7

21st term = a + 20d

= -3 + 20(7)

= -3 + 140

= 137

Ans (b)

Question 16

If the 2nd term of an A.P. is 13 and the 5th term is 25, then its 7th term is

(a) 30

(b) 33

(c) 37

(d) 38

Sol :

In an A.P.

2nd term = 13 ⇒ a + d = 13 …(i)

5th term = 25 ⇒ a + 4d = 25 …(ii)

Subtracting (i) and (ii),

3d = 12

$\Rightarrow d=\frac{1}{3}$

Substitute the value of d in eq. (i), we get

a = 13 – 4 = 9

7th term = a + 6d = 9 + 6 x 4 = 9 + 24 = 33 

Ans (b)

Question 17

If the first term of an A.P. is – 5 and the common difference is 2, then the sum of its first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

Sol :

First term (a) of an A.P. = -5

Common difference (d) is 2

Sum of first 6 terms $=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{6}{2}[2 \times(-5)+(6-1) \times 2]$

$=3[-10+5 \times 2]$

$=3 \times[-10-10]$

$=3 \times 0=0$

Ans (a)

Question 18

The sum of 25 terms of the A.P. $-\frac{2}{3},-\frac{2}{3},-\frac{2}{3}$ is 

(a) 0

(b) $-\frac{2}{3}$

(c) $-\frac{50}{3}$

(d) -50

Sol :

Sum of 25 terms of an A.P. $-\frac{2}{3},-\frac{2}{3},-\frac{2}{3}$ is

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{25}{2}\left[2 \times \frac{(-2)}{3}+(25-1) \times 0\right]$

$=\frac{25}{2}\left[\frac{-4}{3}\right]=\frac{-50}{3}$

Ans (c)

Question 19

In an A.P., if $a=1, a_{n}=20$ and $S_{n}=399,$ then n is

(a) 19

(b) 21

(c) 38

(d) 42

Sol :

In an A.P. $\mathrm{a}=1, \mathrm{a}_{n}=20, \mathrm{~S}_{n}=399, \mathrm{n}$ is ?

$a_{n}=a+(n-1) d=20$

$1+(n-1) d=20$

$(n-1) d=20-1=19 \ldots(i)$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$399=\frac{n}{2}[2 \times 1+19]$ [From (i)]

$\Rightarrow 399=\frac{n}{2} \times 21$

$n=\frac{399 \times 2}{21}=38$

$\therefore n=38$

Ans (c)

Question 20

In an A.P., if $a=-5,l=21$. and $S_{n}=200$, then n is equal to

(a) 50

(b) 40

(c) 32

(d) 25

Sol :

In an A.P.

$a=-5,l=21, S_{n}=200, n=?$

l = a + (n – 1)d = -5 + (n – 1 )d

21 = -5 + (n – 1)d

$\Rightarrow(n-1) d=21+5=26$..(i)

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$200=\frac{n}{2}[2 \times(-5)+26]$ [From (i)]

$400=n[-10+26]=n(16)$

$n=\frac{400}{16}=25$

$\therefore n=25$

Ans (d)

Question 21

In an A.P., if a=3 and $S_{8}=192$, then d is

(a) 8

(b) 7

(c) 6

(d) 4

Sol :

In an A.P.

$a=3, S_{8}=192, d=?$

$\mathrm{S}_{8}=\frac{n}{2}[2 a+(n-1) d]$

$192=\frac{8}{2}[2 \times 3+(8-1) d]$

$\Rightarrow 192=4[6+7 d]$

$\frac{192}{4}=6+7 d$

$\Rightarrow 48=6+7 d$

7d=48-6

7d=42

$d=\frac{42}{7}=6$

Ans (c)

Question 22

The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

Sol :

First 5 multiples of 3 :

3, 6, 9, 12, 15

Here, a = 3, d = 6 – 3 = 3

$\therefore \mathrm{S}_{5}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore \mathrm{S}_{5}=\frac{n}{2}[2 a+(n-1) d]$

Ans (a)

Question 23

The number of two digit numbers which are divisible by 3 is

(a) 33

(b) 31

(c) 30

(d) 29

Sol :

Two digit number which are divisible by 3 is 12, 15, 18, 21, … 99

Here, a = 12, d = 3, l = 99

$1=a_{n}=a+(n-1) d$

⇒ 12 + (n – 1) x 3 = 99

⇒ (n – 1)3 = 99 – 12 = 87

$\Rightarrow n-1=\frac{87}{3}=29$

⇒ n = 29 + 1 = 30 

Ans (c)

Question 24

The number of multiples of 4 that lie between 10 and 250 is

(a) 62

(b) 60

(c) 59

(d) 55

Sol :

Multiples of 4 lying between 10 and 250 12, 16, 20, 24, …, 248

Here, a = 12, d = 16 – 12 = 4, l = 248

$l=a_{n}=a+(n-1) d$

$248=12+(n-1) \times 4$

$\Rightarrow 248-12=4(n-1)$

$\Rightarrow \frac{236}{4}=n-1$

$\Rightarrow n-1=59$

$\therefore n=59+1=60$

Ans (b)

Question 25

The sum of first 10 even whole numbers is

(a) 110

(b) 90

(c) 55

(d) 45

Sol :

Sum of first 10 even whole numbers

Even numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18

Here, a = 0, d = 2, n = 10

$\mathrm{S}_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 0+(10-1) \times 2]$

$=5[0+18]=90$

Ans (b)

Question 26

The list of number $\frac{1}{9}, \frac{1}{3}, 1,-3, \ldots$ is a

(a) GP. with r=-3

(b) G.P. with $r=-\frac{1}{3}$

(c) GP. with r = 3

(d) not a G.P.

Sol:

The given list of numbers

$\frac{1}{9}, \frac{1}{3}, 1,-3, \ldots$

Here, $a=\frac{1}{9}, r=-\frac{1}{3} \div \frac{1}{9}=-\frac{1}{3} \times \frac{9}{1}=-3$

It is a G.P. with r=-3

Ans (a)

Question 27

The 11 th of the G.P. $\frac{1}{8},-\frac{1}{4}, 2,-1, \ldots$ is

(a) 64

(b) – 64

(c) 128

(d) – 128

Sol :

11th of the G.P.

$\frac{1}{8},-\frac{1}{4}, 2,-1, \ldots$ is

Here $a=\frac{1}{8}, r=-\frac{1}{4} \div \frac{1}{8}=-\frac{1}{4} \times \frac{8}{1}=-2$

$\therefore a_{11}=a r^{n-1}=\frac{1}{8}(-2)^{10}$

$=\frac{1}{2^{3}} \times(-1)^{10} \times 2^{10}=1 \times 2^{10-3}=2^{7}$

=128

Ans (c)

Question 28

The 5th term from the end of the G.P. 2, 6, 18, …, 13122 is

(a) 162

(b) 486

(c) 54

(d) 1458

Sol :

5th term from the end of the G.P. 2, 6, 18, …, 13122 is

Here, $a=2, r=\frac{6}{2}=3,1=13122$

$\therefore$ 5th term from the end $=l\left(\frac{1}{r}\right)^{m-n}$

Here, $l=a_{n}=a r^{n-1}$

$13122=2(3)^{n-1}$

$\Rightarrow \frac{13122}{2}=3^{(n-1)}$

$\Rightarrow 6561=3^{(n-1)}$

$\begin{array}{l|l}3 & 6561 \\\hline 3 & 2187 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline 3 & 3 \\\hline & 1\end{array}$

$\Rightarrow(3)^{8}=3^{n-1}$

Comparing, we get

$n-1=8 \Rightarrow n=9$

5th term from the end 

$=l\left(\frac{1}{r}\right)^{m-n}$

$=13122\left(\frac{1}{3}\right)^{9-5}$

$=13122 \times\left(\frac{1}{3}\right)^{4}=\frac{13122}{3 \times 3 \times 3 \times 3}$

$\begin{array}{l}81\overline{)13122(}162\\\phantom{81)}81\\\phantom{81)}\overline{502}\\\phantom{81)}486\\\phantom{81)}\overline{\phantom{4}162}\end{array}$

=162

Ans (a)

Question 29

If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is

(a) – 1

(b) – 4

(c) 1

(d) 4

Sol :

k, 2(k + 1), 3(k + 1) are in G.P.

[2(k + 1)]² = k x 3(k + 1)

⇒ 4(k + 1)² = 3k(k + 1)

⇒ 4 (k + 1) = 3 k

(Dividing by k + 1 if k + 1 ≠ 0) 4

⇒ 4k + 4 = 3k

⇒ 4k – 3k = -4

⇒ k = -4 

Ans (b)

Question 30

Which term of the G.P. 18, – 12, 8, … is $\frac{512}{729}$ ?

(a) 12th

(b) 11th

(c) 10th

(d) 9th

Sol :

Which term of the G.P.

$18,-12,8, \ldots \frac{512}{729}$

Let it be nth term

Here, $a=18, d=\frac{-12}{18}=\frac{-2}{3}$

$\therefore a_{n}=a r^{n-1}$

$\frac{512}{729}=18\left(\frac{-2}{3}\right)^{n-1}$

$\Rightarrow \frac{512}{729} \times \frac{1}{18}=\left(\frac{-2}{3}\right)^{n-1}$

$\frac{256}{729 \times 9}=\left(\frac{-2}{3}\right)^{n-1}$

$\Rightarrow\left(\frac{-2}{3}\right)^{8}=\left(\frac{-2}{3}\right)^{n-1}$

Comparing, n-1=8

$\Rightarrow n=8+1=9$

$\therefore$ It is 9 th term.

Ans (d)

Question 31

The sum of the first 8 terms of the series 1 + √3 + 3 + … is

Sol :

Sum of first 8 terms of 1 + √3 + 3 + … is

Here $a=1, r=\frac{\sqrt{3}}{1}=\sqrt{3}, n=8$

$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

$=\frac{1\left[(\sqrt{3})^{8}-1\right]}{\sqrt{3}-1}$

$=\frac{81-1}{\sqrt{3}-1}=\frac{80}{\sqrt{3}-1}$

$=\frac{80(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

(Rationalising the denominator)

$\frac{80(\sqrt{3}+1)}{3-1}=\frac{80(\sqrt{3}+1)}{2}$

$=40(\sqrt{3}+1)$

Ans (a)

Question 32

The sum of first 6 terms of the G.P. 1, $-\frac{2}{3}, \frac{4}{9}, \ldots$ is

(a) $-\frac{133}{243}$ 3

(b) $\frac{133}{243}$

(c) $\frac{793}{1215}$

(d) none of these

Sol :

Sum of first 6 terms of G.P.

$1,-\frac{2}{3}, \frac{4}{9}, \ldots$

Here, $a=1, r=-\frac{2}{3}, n=6$

$\mathrm{S}_{6}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{1\left[1-\left(\frac{2}{3}\right)^{6}\right]}{1+\frac{2}{3}}$

$=\left[1-\frac{64}{729}\right] \frac{3}{5}$

$=\frac{729-64}{729} \times \frac{3}{5}$

$=\frac{665}{243 \times 5}=\frac{133}{243}$

Ans (b)

Question 33

If the sum of the GP., 1,4, 16, … is 341, then the number of terms in the GP. is

(a) 10

(b) 8

(c) 6

(d) 5

Sol :

The sum of G.P. 1, 4, 16, … is 341

Let n be the number of terms,

Here, $a=1, r=\frac{4}{1}=4, \mathrm{~S}_{n}=341$

$341=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{1\left(4^{n}-1\right)}{4-1}=\frac{4^{n}-1}{3}$

$341 \times 3=4^{n}-1 \Rightarrow 1023=4^{n}-1$

$\Rightarrow 1023+1=4^{n} \Rightarrow 4^{n}=1024$

$4^{n}=4^{5}$

$\therefore n=5$

Ans (d)