In Exercise 19.2, we shall discuss problems based on the general terms of an A.P., which includes finding the nth term in the given expression. The solutions to the exercise-wise problems under this chapter are prepared by the experienced faculty team at BYJU’S after conducting vast research on each concept. The main aim of designing solutions is to help students with their board exam preparation and, thereby, secure good marks. Through regular practice, students can achieve their goals. RD Sharma Class 11 Maths Solutions PDFs are available here, students can make use of it by downloading them from the links provided below.
RD Sharma Solutions for Class 11 Maths Exercise 19.2 Chapter 19 – Arithmetic Progressions
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1. Find:
(i) 10th term of the A.P. 1, 4, 7, 10, …..
(ii) 18th term of the A.P. √2, 3√2, 5√2, …
(iii) nth term of the A.P 13, 8, 3, -2, ….
Solution:
(i) 10th term of the A.P. 1, 4, 7, 10, …..
Arithmetic Progression (AP) whose common difference is = an – an-1 where n > 0
Let us consider, a = a1 = 1, a2 = 4 …
So, Common difference, d = a2 – a1 = 4 – 1 = 3
To find the 10th term of A.P, firstly, find an
By using the formula,
an = a + (n-1) d
= 1 + (n-1) 3
= 1 + 3n – 3
= 3n – 2
When n = 10:
a10 = 3(10) – 2
= 30 – 2
= 28
Hence, the 10th term is 28.
(ii) 18th term of the A.P. √2, 3√2, 5√2, …
Arithmetic Progression (AP) whose common difference is = an – an-1 where n > 0
Let us consider, a = a1 = √2, a2 = 3√2 …
So, Common difference, d = a2 – a1 = 3√2 – √2 = 2√2
To find the 18th term of A.P, firstly, find an
By using the formula,
an = a + (n-1) d
= √2 + (n – 1) 2√2
= √2 + 2√2n – 2√2
= 2√2n – √2
When n = 18:
a18 = 2√2(18) – √2
= 36√2 – √2
= 35√2
Hence, the 10th term is 35√2
(iii) nth term of the A.P 13, 8, 3, -2, ….
Arithmetic Progression (AP) whose common difference is = an – an-1 where n > 0
Let us consider, a = a1 = 13, a2 = 8 …
So, Common difference, d = a2 – a1 = 8 – 13 = -5
To find the nth term of A.P, firstly, find an
By using the formula,
an = a + (n-1) d
= 13 + (n-1) (-5)
= 13 – 5n + 5
= 18 – 5n
Hence, the nth term is 18 – 5n
2. In an A.P., show that am+n + am–n = 2am.
Solution:
We know the first term is ‘a’, and the common difference of an A.P. is d.
Given:
am+n + am–n = 2am
By using the formula,
an = a + (n – 1)d
Now, let us take LHS: am+n + am-n
am+n + am-n = a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a + md – nd – d
= 2a + 2md – 2d
= 2(a + md – d)
= 2[a + d(m – 1)] {∵ an = a + (n – 1)d}
am+n + am-n = 2am
Hence proved.
3. (i) Which term of the A.P. 3, 8, 13,… is 248?
(ii) Which term of the A.P. 84, 80, 76,… is 0?
(iii) Which term of the A.P. 4, 9, 14,… is 254?
Solution:
(i) Which term of the A.P. 3, 8, 13,… is 248?
Given A.P is 3, 8, 13,…
Here, a1 = a = 3, a2 = 8
Common difference, d = a2 – a1 = 8 – 3 = 5
We know, an = a + (n – 1)d
an = 3 + (n – 1)5
= 3 + 5n – 5
= 5n – 2
Now, to find which term of A.P is 248.
Put an = 248
∴ 5n – 2 = 248
= 248 + 2
= 250
= 250/5
= 50
Hence, the 50th term of the given A.P. is 248.
(ii) Which term of the A.P. 84, 80, 76,… is 0?
Given A.P is 84, 80, 76,…
Here, a1 = a = 84, a2 = 88
Common difference, d = a2 – a1 = 80 – 84 = -4
We know, an = a + (n – 1)d
an = 84 + (n – 1)-4
= 84 – 4n + 4
= 88 – 4n
Now, to find which term of A.P is 0.
Put an = 0
88 – 4n = 0
-4n = -88
n = 88/4
= 22
Hence, the 22nd term of the given A.P. is 0.
(iii) Which term of the A.P. 4, 9, 14,… is 254?
Given A.P. is 4, 9, 14,…
Here, a1 = a = 4, a2 = 9
Common difference, d = a2 – a1 = 9 – 4 = 5
We know, an = a + (n – 1)d
an = 4 + (n – 1)5
= 4 + 5n – 5
= 5n – 1
Now, to find which term of A.P is 254,
Put an = 254
5n – 1 = 254
5n = 254 + 1
5n = 255
n = 255/5
= 51
Hence, the 51st term of the given A.P. is 254.
4. (i) Is 68 a term of the A.P. 7, 10, 13,…?
(ii) Is 302 a term of the A.P. 3, 8, 13,…?
Solution:
(i) Is 68 a term of the A.P. 7, 10, 13,…?
Given A.P is 7, 10, 13,…
Here, a1 = a = 7, a2 = 10
Common difference, d = a2 – a1 = 10 – 7 = 3
We know, an = a + (n – 1)d [where a is the first term or a1 and d is the common difference and n is any natural number]
an = 7 + (n – 1)3
= 7 + 3n – 3
= 3n + 4
Now, to find whether 68 is a term of this A.P. or not,
Put an = 68
3n + 4 = 68
3n = 68 – 4
3n = 64
n = 64/3
64/3 is not a natural number.
Hence, 68 is not a term of the given A.P.
(ii) Is 302 a term of the A.P. 3, 8, 13,…?
Given A.P is 3, 8, 13,…
Here, a1 = a = 3, a2 = 8
Common difference, d = a2 – a1 = 8 – 3 = 5
We know, an = a + (n – 1)d
an = 3 + (n – 1)5
= 3 + 5n – 5
= 5n – 2
To find whether 302 is a term of this A.P. or not,
Put an = 302
5n – 2 = 302
5n = 302 + 2
5n = 304
n = 304/5
304/5 is not a natural number.
Hence, 304 is not a term of the given A.P.
5. (i) Which term of the sequence 24, 23 ¼, 22 ½, 21 ¾ is the first negative term?
Solution:
Given:
AP: 24, 23 ¼, 22 ½, 21 ¾, … = 24, 93/4, 45/2, 87/4, …
Here, a1 = a = 24, a2 = 93/4
Common difference, d = a2 – a1 = 93/4 – 24
= (93 – 96)/4
= – 3/4
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
We know, an = a + (n – 1) d
an = 24 + (n – 1) (-3/4)
= 24 – 3/4n + ¾
= (96+3)/4 – 3/4n
= 99/4 – 3/4n
Now, we need to find the first negative term.
Put an < 0
an = 99/4 – 3/4n < 0
99/4 < 3/4n
3n > 99
n > 99/3
n > 33
Hence, the 34th term is the first negative term of the given AP.
(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, … is (a) purely real (b) purely imaginary?
Solution:
Given:
AP: 12 + 8i, 11 + 6i, 10 + 4i, …
Here, a1 = a = 12 + 8i, a2 = 11 + 6i
Common difference, d = a2 – a1
= 11 + 6i – (12 + 8i)
= 11 – 12 + 6i – 8i
= -1 – 2i
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
an = 12 + 8i + (n – 1) -1 – 2i
= 12 + 8i – n – 2ni + 1 + 2i
= 13 + 10i – n – 2ni
= (13 – n) + (10 – 2n) i
To find the purely real term of this A.P., the imaginary part has to be zero.
10 – 2n = 0
2n = 10
n = 10/2
= 5
Hence, the 5th term is purely real.
To find the purely imaginary term of this A.P., the real part has to be zero.
∴ 13 – n = 0
n = 13
Hence, the 13th term is purely imaginary.
6. (i) How many terms are in A.P. 7, 10, 13,…43?
Solution:
Given:
AP: 7, 10, 13,…
Here, a1 = a = 7, a2 = 10
Common difference, d = a2 – a1 = 10 – 7 = 3
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
an = 7 + (n – 1)3
= 7 + 3n – 3
= 3n + 4
To find the total terms of the A.P., put an = 43 as 43 is the last term of A.P.
3n + 4 = 43
3n = 43 – 4
3n = 39
n = 39/3
= 13
Hence, total 13 terms exist in the given A.P.
(ii) How many terms are there in the A.P. -1, -5/6, -2/3, -1/2, …, 10/3?
Solution:
Given:
AP: -1, -5/6, -2/3, -1/2, …
Here, a1 = a = -1, a2 = -5/6
Common difference, d = a2 – a1
= -5/6 – (-1)
= -5/6 + 1
= (-5+6)/6
= 1/6
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
an = -1 + (n – 1) 1/6
= -1 + 1/6n – 1/6
= (-6-1)/6 + 1/6n
= -7/6 + 1/6n
To find the total terms of the AP,
Put an = 10/3 [Since, 10/3 is the last term of AP]
an = -7/6 + 1/6n = 10/3
1/6n = 10/3 + 7/6
1/6n = (20+7)/6
1/6n = 27/6
n = 27
Hence, total 27 terms exist in the given A.P.
7. The first term of an A.P. is 5, the common difference is 3, and the last term is 80; find the number of terms.
Solution:
Given:
First term, a = 5; last term, l = an = 80
Common difference, d = 3
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
an = 5 + (n – 1)3
= 5 + 3n – 3
= 3n + 2
To find the total terms of the A.P., put an = 80 as 80 is the last term of A.P.
3n + 2 = 80
3n = 80 – 2
3n = 78
n = 78/3
= 26
Hence, total 26 terms exist in the given A.P.
8. The 6th and 17th terms of an A.P. are 19 and 41, respectively. Find the 40th term.
Solution:
Given:
The 6th term of an A.P. is 19, and the 17th term of an A.P. is 41.
So, a6 = 19 and a17 = 41
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
When n = 6:
a6 = a + (6 – 1) d
= a + 5d
Similarly, When n = 17:
a17 = a + (17 – 1)d
= a + 16d
According to question,
a6 = 19 and a17 = 41
a + 5d = 19 ……………… (i)
And a + 16d = 41………….. (ii)
Let us subtract equation (i) from (ii), and we get,
a + 16d – (a + 5d) = 41 – 19
a + 16d – a – 5d = 22
11d = 22
d = 22/11
= 2
Put the value of d in equation (i):
a + 5(2) = 19
a + 10 = 19
a = 19 – 10
= 9
As, an = a + (n – 1)d
a40 = a + (40 – 1)d
= a + 39d
Now, put the value of a = 9 and d = 2 in a40, and we get,
a40 = 9 + 39(2)
= 9 + 78
= 87
Hence, the 40th term of the given A.P. is 87.
9. If the 9th term of an A.P. is Zero, prove that its 29th term is double the 19th term.
Solution:
Given:
The 9th term of an A.P. is 0.
So, a9 = 0
We need to prove: a29 = 2a19
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
When n = 9,
a9 = a + (9 – 1)d
= a + 8d
According to the question,
a9 = 0
a + 8d = 0
a = -8d
When n = 19,
a19 = a + (19 – 1)d
= a + 18d
= -8d + 18d
= 10d
When n = 29,
a29 = a + (29 – 1)d
= a + 28d
= -8d + 28d [Since, a = -8d]
= 20d
= 2×10d
a29 = 2a19 [Since, a19 = 10d]
Hence proved.
10. If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that the 25th term of the A.P. is Zero.
Solution:
Given:
10 times the 10th term of an A.P. is equal to 15 times the 15th term.
So, 10a10 = 15a15
We need to prove: a25 = 0
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
When n = 10,
a10 = a + (10 – 1)d
= a + 9d
When n = 15,
a15 = a + (15 – 1)d
= a + 14d
When n = 25,
a25 = a + (25 – 1)d
= a + 24d ………(i)
According to question,
10a10 = 15a15
10(a + 9d) = 15(a + 14d)
10a + 90d = 15a + 210d
10a – 15a + 90d – 210d = 0
-5a – 120d = 0
-5(a + 24d) = 0
a + 24d = 0
a25 = 0 [From (i)]
Hence oroved.
11. The 10th and 18th terms of an A.P. are 41 and 73, respectively. Find the 26th term.
Solution:
Given:
The 10th term of an A.P. is 41, and the 18th term of an A.P. is 73.
So, a10 = 41 and a18 = 73
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
When n = 10:
a10 = a + (10 – 1)d
= a + 9d
When n = 18:
a18 = a + (18 – 1)d
= a + 17d
According to question,
a10 = 41 and a18 = 73
a + 9d = 41 ………………(i)
And a + 17d = 73…………..(ii)
Let us subtract equation (i) from (ii) , andwe get,
a + 17d – (a + 9d) = 73 – 41
a + 17d – a – 9d = 32
8d = 32
d = 32/8
d = 4
Put the value of d in equation (I), and we get,
a + 9(4) = 41
a + 36 = 41
a = 41 – 36
a = 5
we know, an = a + (n – 1)d
a26 = a + (26 – 1)d
= a + 25d
Now, put the value of a = 5 and d = 4 in a26
a26 = 5 + 25(4)
= 5 + 100
= 105
Hence, the 26th term of the given A.P. is 105.
12. In a certain A.P., the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Solution:
Given:
The 24th term is twice the 10th term.
So, a24 = 2a10
We need to prove: a72 = 2a34
We know, an = a + (n – 1) d [where a is the first term or a1 and d is the common difference and n is any natural number]
When n = 10:
a10 = a + (10 – 1)d
= a + 9d
When n = 24:
a24 = a + (24 – 1)d
= a + 23d
When n = 34:
a34 = a + (34 – 1)d
= a + 33d ………(i)
When n = 72:
a72 = a + (72 – 1)d
= a + 71d
According to question,
a24 = 2a10
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
a – 2a + 23d – 18d = 0
-a + 5d = 0
a = 5d
Now, a72 = a + 71d
a72 = 5d + 71d
= 76d
= 10d + 66d
= 2(5d + 33d)
= 2(a + 33d) [since, a = 5d]
a72 = 2a34 (From (i))
Hence proved.
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