Selina Concise Class 10 Math Chapter 10 Arithmetic Progression Solutions
Exercise 10A
(Q1) Which of the following sequence are in arithmetic progression?
(i) 2, 6, 10, 14
Solution:
Given sequence is
2, 6, 10, 14, —–
Let, a = t1 = 2, t2 = 6, t3 = 10, t4 = 14,
t2 – t1 = 6 – 2 = 4 = d1
t3 – t2 = 10 – 6 = 4 = d2
t4 – t3 = 14 – 10 = 4 = d3
When the difference is common then it’s a arithmetic progression.
∴ The given sequence is a common difference
d1 = 4
d2 = 4
and d3 = 4, —-
So, the given sequence is in arithmetic progression.
(ii) 15, 12, 9, 6, —-
Solution:
The given sequence is 15, 12, 9, 6, —-
∴ t1, = 15, t2 = 12, t3 = 9, t4 – 6, ——
∴ The difference between the two terms is,
t2 – t1 = 12 – 15 = -3 = d1
t3 – t2 = 9 – 12 = -3 = d1
t4 – t3 = 6 – 9= – 3 = d3
∴ When the difference is common then it an Arithmetic progression.
∴ The given sequence of common difference.
d1 = d2 = d3 = -3
∴ So, the given sequence is in arithmetic progression.
(iii) 5, 9, 12, 18, —–
Solution:
The given sequence is 5, 9, 12, 18, —–
∴ t1 = 5, t2 = 9, t3 = 12, t4 = 18, ——
∴ The difference between two terms is,
t2 – t1 = 9 – 5 = 4 = d1
t3 – t2 = 12 – 9 = 3 = d2
t4 – t3 = 18 – 12 = 6 = d3
∴ When the difference is common then it is an Arithmetic progression.
But we can see that there is no common difference.
So, d1 ≠ d2 ≠ d3 —–
∴ The given sequence is not in Arithmetic progression.
(iv) ½, 1/3, ¼, 1/5, ——
= Solution:
Given sequence is 1/2, 1/3, ¼, 1/5, —–
∴ t1 = 1/2, t2 = 1/3, t3 = ¼, t4 = 1/5
The difference between the two terms is,
t2 – t1 = 1/3 – ½ = 2-3/6 = -1/6
t3 – t2 = ¼ – 1/3 = 3-4/12 = -1/12 = -1/12
t4 – t3 = 1/5 – ¼ = 4-5/20 = -1/20
When the difference is common then it is an arithmetic progression.
But we can see that there is no common difference.
So, d1 ≠ d2 ≠ d3, ——-
So, the given sequence is an arithmetic progression.
(Q2) The nth term of sequence is (2n – 3), find it’s fifteenth term.
= Solution:
The given nth term is (2n – 3).
To find: 15th term.
t1, t2, t3 —— t15,
So, we have to find 15th term,
tn = (2n – 3)
t15 = (2 (15) – 3) ∵ n = 15
t15 = 30 – 3
t15 = 27
∴ The fifteenth term is 27.
(Q3) If the Pth of an Arithmetic progression is (2P + 3); find the Arithmetic progression.
Solution:
Given Pth term is (2P + 3)
We have to find the arithmetic progression.
We know that arithmetic progression means, first we have to find terms and after we have to find common difference,
t1 = 2P + 3
= 2 (1) + 3
= 2 + 3 (∵ P = 1)
t2 = 2(2) + 3
= 4 + 3
t2 =7
t3 = 2 (3) + 3
= 6 + 3
t3 = 9
t4 = 2 (4) + 3
= 8 + 3
t4 = 11
The difference between these arithmetic progression is
t2 – t1 = 7 – 5 = 2 = d1
t3 – t2 = 9 – 7 = 2 = d2
t4 – t3 = 11 – 9 = 2 = d3
∴ d1 = d2 = d3
∴ The required sequence is,
t1 = 5, t2 = 7, t3 = 9, t4 = 11, ——
∴ 5, 7, 9, 11, —-
(Q4) Find the 24th term of the sequence 12, 10, 8, 6, —–
Solution:
Given sequence is 12, 10, 8, 6, —–
We have to find 24th term.
First we have to find common difference.
a = t1 = 12, t2 = 10, t3 = 8, t4 = 6, ——
t2 – t1 = 10 – 12 = – 2 = d1
t3 – t2 = 8 – 10 = -2 = d2
t4 – t3 = 6 – 8 = -2 = d3
∴ d1 = d2 = d3
∴ a = t1 = 12
We know that, the general nth term of an arithmetic progression is.
tn = a + (n – 1) d
Given that n = 24 and d = -2
t24 = 12 + (24 – 1 (-2)
= 12 + (23) (-2)
= 12 – 46
t24 = -34
∴ The 24th term is – 34
(Q5) Find the 30th term of the sequence ½, 1, 3/2, —-
Solution:
Given sequence is ½, 1, 3/2, ——
Firs we have to find difference of given sequence.
t1= ½, t2 = 1, t3 = 3/2, —–
t2 – t1 = 1 – ½ = 2-1/2 = 1/2 = d1
∴ d1 = d2 = d
Now, we have to find 30th term.
a = t1 = 1/2, and d = ½
∴ The general term of arithmetic progression is, —-
tn = a + (n – 1) d
T30 = ½ + (30 – 1) ½ (∵ n = 30)
= ½ + (29) 1/2
= ½ + 29/2 = 1 + 29/2
= 30/2
t30 = 15
∴ The 30th term is 15.
(Q6) Find the 100th term of the sequence.
√3, 2√3, 3√3, —–
Solution:
The given sequence is √3, 2√3, 3√3, —–
First we have to find difference of these sequence.
t1 = √3, t2 = 2√3, t3 = 3√3, ——
d1 = t2 – t1 = 2√3 – √3 = √3(2 – 1) = √3
d2 = t3 – t2 = 3√3 – 2√3 = √3 (3 – 2 = √3
d1 = d2 = d
∴ We have to find 100th term, we know that the genera term of arithmetic progression.
tn = a + (n – 1) d
∴ a = t1 = √3 and d = √3
t100 = √3 + (100 – 1) √3 (∵ n = 100)
t100 = √3 + (99) √3
= √3 + 99√3
= √3 (1 + 99)
t100 = 100√3
∴ The 100th term is 100√3
(Q7) Find the 50th term of the sequence 1/n, (n+1)/n, (2n+1)/n —–
Solution:
The given sequence is 1/n, (n+1)/n, (2n+1)/n, —–
First we have to find terms and common difference of given sequence.
t1 = 1/n, t2 = (n +1)/n, t3 = (2n + 1)/n
t2 – t1 = (n + 1)/n – 1/n
= (n + 1) – 1/n
= n + 1 – 1/n
= n/n
t2 – t1 = 1 = d1
t3 – t2 (2n+1)/n – (n+1)/n
= (2n + 1) – (n + 1)/n
= 2n + 1 – n – 1/n
= 2n – n/n
= n/n
T3 – t2 = 1 = d2
∴ d1 = d2 = d
∴ We have to find 50th term.
We know that the general term of arithmetic progression.
tn = a + (n – 1) d
∴ a = t1 = 1/n and d = 1
t50 = 1/n + (50 – 1) 1 (∵ n = 50)
= 1/n + (49)
t50 = 1/n + 49
t50 = 1 + 49n/n
∴ The 50th term is 1+49n/n
(Q9) Find the common difference and 99th term of the arithmetic progression 7 ¾, 9 ½, 11 ¼, —–
Solution:
Given sequence is
7 3/2, 9 ½, 11 ¼, —–
This is the mix number, this mix number convert into rational number.
7 3/4 = 4×7+3/4 = 28+3/4 = 31/4
9 ½ = 2×9+1/2 = 18+1/2 = 19/2
11 ½ = 4×11+1/4 = 44+1/4 = 45/4
So, The new given sequence is 31/4, 19/2, 45/4, —–
T1 = 31/4, t2 = 19/2, t3 = 45/4, —–
T2 – t1 = 19/2 – 31/4 = 2×19-31/4
= 38-31/4
= 7/4 = d1
T3 – t2 = 45/4 – 19/2 = 45-19×2/4
= 45-38/4
= 7/4 = d2
∴ d1 = d2 = d
∴ Common difference (d) = 7/4
And the first term a = t1 =31/4
We know that the general term is,
tn = a + (n – 1) d
To find 99th term.
t99 = 31/4 + (99 – 1) 7/4 (∵ n = 99)
= 31/4 + (98) 7/4
= 31/4 + 686/4
= 717/4
T99 = 179.25
T99 = 179 1/4 (∵ 0.25 = ¼)
∴ The 99th term is 179 ¼