Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.2
Last Updated :
08 Dec, 2020
Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference.
Solution:
Given:
an = 5n – 7
Now putting n = 1, 2, 3, 4,5 we get,
a1 = 5.1 – 7 = 5 – 7 = -2
a2 = 5.2 – 7 = 10 – 7 = 3
a3 = 5.3 – 7 = 15 – 7 = 8
a4 = 5.4 – 7 = 20 – 7 = 13
We can see that,
a2 – a1 = 3 – (-2) = 5
a3 – a2 = 8 – (3) = 5
a4 – a3 = 13 – (8) = 5
Since, the successive difference of list is same i.e 5
∴ The given sequence is in A.P and have common difference of 5
Problem 2: Show that the sequence defined by an = 3n2 – 5 is not an A.P.
Solution:
Given:
an = 3n2 – 5
Now putting n = 1, 2, 3, 4 we get,
a1 = 3.1.1 – 5= 3 – 5 = -2
a2 = 3.2.2 – 5 = 12 – 5 = 7
a3 = 3.3.3 – 5 = 27 – 5 = 22
a4 = 3.4.4 – 5 = 48 – 5 = 43
We can see that,
a2 – a1 = 7 – (-2) = 9
a3 – a2 = 22 – 7 = 15
a4 – a5 = 43 – 22 = 21
Since, the successive difference of list is not the same
∴ The given sequence is not in A.P
Problem 3: The general term of a sequence is given by an = -4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.
Solution:
Given:
an = -4n + 15
Now putting n = 1, 2, 3, 4 we get,
a1 = -4.(1) + 15 = -4 + 15 = 11
a2 = -4.(2) + 15 = -8 + 15 = 7
a3 = -4.(3) + 15 = -12 + 15 = 3
a4 = -4.(4) + 15 = -16 + 15 = -1
We can see that,
a2 – a1 = 7 – (11) = -4
a3 – a2 = 3 – 7 = -4
a4 – a3 = -1 – 3 = -4
Since, the successive difference of list is same i.e -4
∴ The given sequence is in A.P and have common difference of -4
Hence, the 15th term will be
a15 = -4(15) + 15 = -60 + 15 = -45
And, a15 = -45
Problem 4: Write the sequence with nth term :
(i) an = 3 + 4n
Solution:
Given:
an = 3 + 4n
Now putting n = 1, 2, 3, 4 we get,
a1 = 3 + 4.1 = 7
a2 = 3 + 4.2 = 11
a3 = 3 + 4.3 = 15
a4 = 3 + 4.4 = 19
∴ The sequence is 7, 11, 15, 19
We can see that,
a2 – a1 = 11 – (7) = 4
a3 – a2 = 15 – (11) = 4
a4 – a3 = 19 – (15) = 4
Since, the successive difference of list is same i.e 4
∴ The given sequence is in A.P
(ii) an = 5 + 2n
Solution:
Now putting n = 1, 2, 3, 4 we get,
a1 = 5 + 2.1 = 7
a2 = 5 + 2.2 = 9
a3 = 5 + 2.3 = 11
a4 = 5 + 2.4 = 13
∴ The sequence is 7, 9, 11, 13
We can see that,
a2 – a1 = 9 – (7) = 2
a3 – a2 = 11 – (9) = 2
a4 – a3 = 13 – (11) = 2
Since, the successive difference of list is same i.e 2
∴ The given sequence is in A.P
(iii) an = 6 – n
Solution:
Now putting n = 1, 2, 3, 4 we get,
a1 = 6 – 1 = 5
a2 = 6 – 2 = 4
a3 = 6 – 3 = 3
a4 = 6 – 4 = 2
∴ The sequence is 5, 4, 3, 2
We can see that,
a2 – a1 = 4 – (5) = -1
a3 – a2 = 3 – (4) = -1
a4 – a3 = 2 – (3) = -1
Since, the successive difference of list is same i.e -1
∴ The given sequence is in A.P
(iv) an = 9 – 5n
Solution:
Now putting n = 1, 2, 3, 4 we get,
a1 = 9 – 5.1 = 4
a2 = 9 – 5.2 = -1
a3 = 9 – 5.3 = -6
a4 = 9 – 5.4 = -11
∴ The sequence is 4, -1, -6, -11
We can see that,
a2 – a1 = -1 – (4) = -5
a3 – a2 = -6 – (-1) = -5
a4 – a3 = -11 – (-6) = -5
Since, the successive difference of list is same i.e -5
∴ The given sequence is in A.P
Problem 5: The nth term of an A.P. is 6n + 2. Find the common difference.
Solution:
Now putting n = 1, 2, 3, 4 we get,
a1 = 6.1 + 2 = 8
a2 = 6.2 + 2 = 14
a3 = 6.3 + 2 = 20
a4 = 6.4 + 2 = 26
We can see that,
a2 – a1 = 14 – (8) = 6
a3 – a2 = 20 – (14) = 6
a4 – a3 = 26 – (20) = 6
Hence, the common difference is 6
Problem 6: Justify whether it is true to say that the sequence, having following nth term is an A.P.
(i) an = 2n – 1
Solution:
Now putting n = 1, 2, 3, 4 we get,
a1 = 2.1 – 1 = 1
a2 = 2.2 – 1 = 3
a3 = 2.3 – 1 = 5
a4 = 2.4 – 1 = 7
We can see that,
a2 – a1 = 3 – (1) = 2
a3 – a2 = 5 – (3) = 2
a4 – a3 = 7 – (5) = 2
Since, the successive difference of list is same i.e 2
Hence, the given sequence is in A.P
(ii) an = 3n² + 5
Solution:
Now putting n = 1, 2, 3, 4 we get,
a1 = 3.1.1 + 5 = 8
a2 = 3.2.2 + 5 = 17
a3 = 3.3.3 + 5 = 32
a4 = 3.4.4 + 5 = 53
We can see that,
a2 – a1 = 17 – (8) = 9
a3 – a2 = 32 – (17) = 15
a4 – a3 = 53 – (32) = 21
Since, the successive difference of list is not the same
Hence, the given sequence is not in A.P
(iii) an = 1 + n + n²
Solution:
Now putting n = 1, 2, 3 we get,
a1 = 1 + 1 + 1.1 = 3
a2 = 1 + 2 + 2.2 = 7
a3 = 1 + 3 + 3.3 = 13
We can see that,
a2 – a1 = 7 – (3) = 4
a3 – a2 = 13 – (7) = 6
Since, the successive difference of list is not the same
Hence, the given sequence is not in A.P
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