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
$=10+(30-1) \times(-3)$
$=10+29(-3)=10-87=-77$
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
$a=-3, d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=\frac{5}{2}$
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 .
$\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
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 an 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 = ?
an = 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, an = 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 = ?
an = 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
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)
⇒ 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)
⇒ 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
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
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.
l = a + (n – 1)d = -5 + (n – 1 )d
21 = -5 + (n – 1)d
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$200=\frac{n}{2}[2 \times(-5)+26]$ [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.
$\mathrm{S}_{8}=\frac{n}{2}[2 a+(n-1) d]$
$192=\frac{8}{2}[2 \times 3+(8-1) d]$
$\Rightarrow 192=4[6+7 d]$
$\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
$\therefore \mathrm{S}_{5}=\frac{n}{2}[2 a+(n-1) d]$
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
⇒ 12 + (n – 1) x 3 = 99
⇒ (n – 1)3 = 99 – 12 = 87
⇒ 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
$\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
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
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.
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
$\therefore$ 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.
Let it be nth term
$\therefore a_{n}=a r^{n-1}$
$\frac{512}{729}=18\left(\frac{-2}{3}\right)^{n-1}$
$ \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
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
$=\frac{1\left[(\sqrt{3})^{8}-1\right]}{\sqrt{3}-1}$
$=\frac{81-1}{\sqrt{3}-1}=\frac{80}{\sqrt{3}-1}$
$=\frac{80(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
(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,
$341=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{1\left(4^{n}-1\right)}{4-1}=\frac{4^{n}-1}{3}$
$341 \times 3=4^{n}-1 \Rightarrow 1023=4^{n}-1$
$\Rightarrow 1023+1=4^{n} \Rightarrow 4^{n}=1024$
$4^{n}=4^{5}$
$\therefore n=5$
Ans (d)
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