Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression
Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression Exercise 10(A)
Question 1. Which of the following sequences are in arithmetic progression?
(i) 2, 6, 10, 14, ….
(ii) 15, 12, 9, 6, ….
(iii) 5, 9, 12, 18, ….
(iv) 1/2, 1/3, 1/4, 1/5, ….
Solution:
(i) 2, 6, 10, 14, ….
Finding the difference between the terms,
d1 = 6 – 2 = 4
d2 = 10 – 6 = 4
d3 = 14 – 10 = 4
As d1 = d2 = d3, the given sequence is in arithmetic progression.
(ii) 15, 12, 9, 6, ….
Finding the difference between the terms,
d1 = 12 – 15 = -3
d2 = 9 – 12 = -3
d3 = 6 – 9 = -3
As d1 = d2 = d3, the given sequence is in arithmetic progression.
(iii) 5, 9, 12, 18, ….
Finding the difference between the terms,
d1 = 9 – 5 = 4
d2 = 12 – 9 = 3
d3 = 18 – 12 = 6
As d1 ≠ d2 ≠ d3, the given sequence is not in arithmetic progression.
(iv) 1/2, 1/3, 1/4, 1/5, ….
Finding the difference between the terms,
d1 = 1/3 – 1/2 = -1/6
d2 = 1/4 – 1/3 = -1/12
d3 = 1/5 – 1/4 = -1/20
As d1 ≠ d2 ≠ d3, the given sequence is not in arithmetic progression.
Question 2. The nth term of sequence is (2n – 3), find its fifteenth term.
Solution:
Given, nth term of sequence is (2n – 3)
So, the 15th term is when n = 15
t15 = 2(15) – 3 = 30 – 3 = 27
Thus, the 15th term of the sequence is 27.
Question 3. If the pth term of an A.P. is (2p + 3); find the A.P.
Solution:
Given, pth term of an A.P. = (2p + 3)
So, on putting p = 1, 2, 3, …, we have
t1 = 2(1) + 3 = 5
t2 = 2(2) + 3 = 7
t3 = 2(3) + 3 = 9
…..
Hence, the sequence A.P. is 5, 7, 9, …
Question 4. Find the 24th term of the sequence:
12, 10, 8, 6,……
Solution:
Given sequence,
12, 10, 8, 6,……
The common difference:
10 – 12 = -2
8 – 10 = -2
6 – 8 = -2 ….
So, the common difference(d) of the sequence is -2 and a = 12.
Now, the general term of this A.P. is given by
tn = a + (n – 1)d = 12 + (n – 1)(-2) = 12 – 2n + 2 = 14 – 2n
For 24th term, n = 24
tn = 14 – 2(24) = 14 – 48 = -34
Therefore, the 24th term is -34
Question 5. Find the 30th term of the sequence:
1/2, 1, 3/2, …….
Solution:
Given sequence,
1/2, 1, 3/2, …….
So,
a = ½
d = 1 – ½ = ½
We know that,
tn = a + (n – 1)d
Hence, the 30th term will be
t30 = ½ + (30 – 1)(1/2) = ½ + 29/2 = 30/2 = 15
Therefore, the 30th term is 15.
Question 6. Find the 100th term of the sequence:
√3, 2√3, 3√3, ….
Solution:
Given A.P. is √3, 2√3, 3√3, ….
So,
a = √3
d = 2√3 – √3 = √3
The general term is given by,
tn = a + (n – 1)d
For 100th term
t100 = √3 + (100 – 1) √3 = √3 + 99√3 = 100√3
Therefore, the 100th term of the given A.P. is 100√3.
Question 7. Find the 50th term of the sequence:
1/n, (n+1)/n, (2n+1)/n, ……
Solution:
Given sequence,
1/n, (n+1)/n, (2n+1)/n, ……
So,
a = 1/n
d = (n+1)/n – 1/n = (n+1-1)/n = 1
Then, the general term is given by
tn = a + (n – 1)d
For 50th term, n = 50
t50 = 1/n + (50 – 1)1 = 1/n + 49 = (49n + 1)/n
Hence, the 50th term of the given sequence is (49n + 1)/n.
Question 8. Is 402 a term of the sequence: 8, 13, 18, 23,………….?
Solution:
Give sequence, 8, 13, 18, 23,………….
d = 13 – 8 = 5 and a = 8
So, the general term is given by
tn = a + (n – 1)d
tn = 8 + (n – 1)5 = 8 + 5n – 5 = 3 + 5n
Now,
If 402 is a term of the sequence whose nth is given by (3 + 5n) then n must be a non-negative integer.
3 + 5n = 402
5n = 399
n = 399/5
So, clearly n is a fraction.
Thus, we can conclude that 402 is not a term of the given sequence.
Question 9. Find the common difference and 99th term of the arithmetic progression:
Solution:
Given, A.P,
i.e., 31/4, 19/2, 45/4, …….
So,
a = 31/4
Common difference, d = 19/2 – 31/4 = (38 – 31)/ 4 = 7/4
Then the general term of the A.P
tn = a + (n – 1)d
t99 = (31/4) + (99 – 1) x (7/4)
= (31/4) + 93 x (7/4)
= (31/4) + (686 / 4)
= (31 + 686)/ 4
= (717)/ 4
= 179 ¼
Hence, the 99th term of A.P. is
Question 10. How many terms are there in the series :
(i) 4, 7, 10, 13, …………, 148?
(ii) 0.5, 0.53, 0.56, ……………, 1.1?
(iii) 3/4, 1, 1 ¼, ……., 3?
Solution:
(i) Given series, 4, 7, 10, 13, …………, 148
Here,
a = 4 and d = 7 – 4 = 3
So, the given term is given by tn = 4 + (n – 1)3 = 4 + 3n – 3
tn = 1 + 3n
Now,
148 = 1 + 3n
147 = 3n
n = 147/3 = 49
Thus, there are 49 terms in the series.
(ii) Given series, 0.5, 0.53, 0.56, ……………, 1.1
Here,
a = 0.5 and d = 0.53 – 0.5 = 0.03
So, the given term is given by tn = 0.5 + (n – 1)0.03 = 0.5 + 0.03n – 0.03
tn = 0.47 + 0.03n
Now,
1.1 = 0.47 + 0.03n
1.1 – 0.47 = 0.03n
n = 0.63/0.03 = 21
Thus, there are 21 terms in the series.
(iii) Given series, 3/4, 1, 1 ¼, ……., 3
Here,
a = 3/4 and d = 1 – 3/4 = 1/4
So, the given term is given by tn = 3/4 + (n – 1)1/4 = (3 + n – 1)/4
tn = (2 + n)/ 4
Now,
3 = (2 + n)/ 4
12 = 2 + n
n = 10
Thus, there are 10 terms in the series.
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