CHAPTER-5 ARITHMETIC PROGESSIONS
QUESTION -4
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
ANSWER:
(i) 2, 4, 8, 16 …
a2 − a1 = 4 − 2 = 2
a3 − a2 = 8 − 4 = 4
a4 − a3 = 16 − 8 = 8
i.e., ak+1− ak is not the same every time. Therefore, the given numbers are not forming an A.P.
It can be observed that
i.e., ak+1− ak is same every time.
Therefore,
and the given numbers are in A.P.
Three more terms are
(iii) −1.2, −3.2, −5.2, −7.2 …
It can be observed that
a2 − a1 = (−3.2) − (−1.2) = −2
a3 − a2 = (−5.2) − (−3.2) = −2
a4 − a3 = (−7.2) − (−5.2) = −2
i.e., ak+1− Ak is same every time. Therefore, d = −2
The given numbers are in A.P.
Three more terms are
a5 = − 7.2 − 2 = −9.2
a6 = − 9.2 − 2 = −11.2
a7 = − 11.2 − 2 = −13.2
(iv) −10, −6, −2, 2 …
It can be observed that
a2 − a1 = (−6) − (−10) = 4
a3 − a2 = (−2) − (−6) = 4
a4 − a3 = (2) − (−2) = 4
i.e., ak+1 − ak is same every time. Therefore, d = 4
The given numbers are in A.P.
Three more terms are
a5 = 2 + 4 = 6
a6 = 6 + 4 = 10
a7 = 10 + 4 = 14
It can be observed that
i.e., ak+1 − ak is same every time. Therefore,
The given numbers are in A.P.
Three more terms are
(vi) 0.2, 0.22, 0.222, 0.2222 ….
It can be observed that
a2 − a1 = 0.22 − 0.2 = 0.02
a3 − a2 = 0.222 − 0.22 = 0.002
a4 − a3 = 0.2222 − 0.222 = 0.0002
i.e., ak+1 − ak is not the same every time.
Therefore, the given numbers are not in A.P.
(vii) 0, −4, −8, −12 …
It can be observed that
a2 − a1 = (−4) − 0 = −4
a3 − a2 = (−8) − (−4) = −4
a4 − a3 = (−12) − (−8) = −4
i.e., ak+1 − ak is same every time. Therefore, d = −4
The given numbers are in A.P.
Three more terms are
a5 = − 12 − 4 = −16
a6 = − 16 − 4 = −20
a7 = − 20 − 4 = −24
It can be observed that
i.e., ak+1 − ak is same every time. Therefore, d = 0
The given numbers are in A.P.
Three more terms are
(ix) 1, 3, 9, 27 …
It can be observed that
a2 − a1 = 3 − 1 = 2
a3 − a2 = 9 − 3 = 6
a4 − a3 = 27 − 9 = 18
i.e., ak+1 − ak is not the same every time.
Therefore, the given numbers are not in A.P.
(x) a, 2a, 3a, 4a …
It can be observed that
a2 − a1 = 2a − a = a
a3 − a2 = 3a − 2a = a
a4 − a3 = 4a − 3a = a
i.e., ak+1 − ak is same every time. Therefore, d = a
The given numbers are in A.P.
Three more terms are
a5 = 4a + a = 5a
a6 = 5a + a = 6a
a7 = 6a + a = 7a
(xi) a, a2, a3, a4 …
It can be observed that
a2 − a1 = a2 − a = a (a − 1)
a3 − a2 = a3 − a2 = a2 (a − 1)
a4 − a3 = a4 − a3 = a3 (a − 1)
i.e., ak+1 − ak is not the same every time.
Therefore, the given numbers are not in A.P.
It can be observed that
i.e., ak+1 − ak is same every time.
Therefore, the given numbers are in A.P.
And,
Three more terms are
It can be observed that
i.e., ak+1 − ak is not the same every time.
Therefore, the given numbers are not in A.P.
(xiv) 12, 32, 52, 72 …
Or, 1, 9, 25, 49 ……
It can be observed that
a2 − a1 = 9 − 1 = 8
a3 − a2 = 25 − 9 = 16
a4 − a3 = 49 − 25 = 24
i.e., ak+1 − ak is not the same every time.
Therefore, the given numbers are not in A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
It can be observed that
a2 − a1 = 25 − 1 = 24
a3 − a2 = 49 − 25 = 24
a4 − a3 = 73 − 49 = 24
i.e., ak+1 − ak is same every time.
Therefore, the given numbers are in A.P.
And, d = 24
Three more terms are
