Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16, ……. (ii) 2, 52 , 3, 72 , ……. (iii) -1.2, -3.2, -5.2, -7.2, …… (iv) -10, -6, -2,2, ….. (v) 3, 3 + 2–√, 3 + 22–√, 3 + 32–√, ….. (vi) 0.2, 0.22, 0.222, 0.2222, …… (vii) 0, -4, -8, -12, ….. (viii) −12 , −12 , −12 , −12 , ……. (ix) 1, 3, 9, 27, ……. (x) a, 2a, 3a, 4a, ……. (xi) a, a2, a3, a4, ……. (xii) 2–√, 8–√, 18−−√, 32−−√, ….. (xiii) 3–√, 6–√, 9–√, 12−−√, ….. (xiv) 12, 32, 52, 72, …… (xv) 12, 52, 72, 73, ……
Answers
(i) 2,4,8,16 …
Here,
a2 - a1 = 4 - 2 = 2
a3 - a2 = 8 - 4 = 4
a4 - a3 = 16 - 8 = 8
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(ii) 2, 5/2, 3, 7/2 ....
Here,
a2 - a1 = 5/2 - 2 = 1/2
a3 - a2 = 3 - 5/2 = 1/2
a4 - a3 = 7/2 - 3 = 1/2
⇒ an+1 - an is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a5 = 7/2 + 1/2 = 4
a6 = 4 + 1/2 = 9/2
a7 = 9/2 + 1/2 = 5
(iii) -1.2, - 3.2, -5.2, -7.2 …
Here,
a2 - a1 = ( -3.2) - ( -1.2) = -2
a3 - a2 = ( -5.2) - ( -3.2) = -2
a4 - a3 = ( -7.2) - ( -5.2) = -2
⇒ an+1 - an is same every time.
Therefore, d = -2 and 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 …
Here,
a2 - a1 = (-6) - (-10) = 4
a3 - a2 = (-2) - (-6) = 4
a4 - a3 = (2) - (-2) = 4
⇒ an+1 - an is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a5 = 2 + 4 = 6
a6 = 6 + 4 = 10
a7 = 10 + 4 = 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a2 - a1 = 3 + √2 - 3 = √2
a3 - a2 = (3 + 2√2) - (3 + √2) = √2
a4 - a3 = (3 + 3√2) - (3 + 2√2) = √2
⇒ an+1 - an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a5 = (3 + √2) + √2 = 3 + 4√2
a6 = (3 + 4√2) + √2 = 3 + 5√2
a7 = (3 + 5√2) + √2 = 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2 - a1 = 0.22 - 0.2 = 0.02
a3 - a2 = 0.222 - 0.22 = 0.002
a4 - a3 = 0.2222 - 0.222 = 0.0002
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(vii) 0, -4, -8, -12 …
Here,
a2 - a1 = (-4) - 0 = -4
a3 - a2 = (-8) - (-4) = -4
a4 - a3 = (-12) - (-8) = -4
⇒ an+1 - an is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a5 = -12 - 4 = -16
a6 = -16 - 4 = -20
a7 = -20 - 4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ....
Here,
a2 - a1 = (-1/2) - (-1/2) = 0
a3 - a2 = (-1/2) - (-1/2) = 0
a4 - a3 = (-1/2) - (-1/2) = 0
⇒ an+1 - an is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a5 = (-1/2) - 0 = -1/2
a6 = (-1/2) - 0 = -1/2
a7 = (-1/2) - 0 = -1/2
(ix) 1, 3, 9, 27 …
Here,
a2 - a1 = 3 - 1 = 2
a3 - a2 = 9 - 3 = 6
a4 - a3 = 27 - 9 = 18
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(x) a, 2a, 3a, 4a …
Here,
a2 - a1 = 2a - a = a
a3 - a2 = 3a - 2a = a
a4 - a3 = 4a - 3a = a
⇒ an+1 - an is same every time.
Therefore, d = a and 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 …
Here,
a2 - a1 = a2 - a = (a - 1)
a3 - a2 = a3 - a2 = a2 (a - 1)
a4 - a3 = a4 - a3 = a3(a - 1)
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(xii) √2, √8, √18, √32 ...
Here,
a2 - a1 = √8 - √2 = 2√2 - √2 = √2
a3 - a2 = √18 - √8 = 3√2 - 2√2 = √2
a4 - a3 = 4√2 - 3√2 = √2
⇒ an+1 - an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a5 = √32 + √2 = 4√2 + √2 = 5√2 = √50
a6 = 5√2 +√2 = 6√2 = √72
a7 = 6√2 + √2 = 7√2 = √98
(xiii) √3, √6, √9, √12 ...
Here,
a2 - a1 = √6 - √3 = √3 × 2 -√3 = √3(√2 - 1)
a3 - a2 = √9 - √6 = 3 - √6 = √3(√3 - √2)
a4 - a3 = √12 - √9 = 2√3 - √3 × 3 = √3(2 - √3)
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(xiv) 12, 32, 52, 72 …
Or, 1, 9, 25, 49 …..
Here,
a2 − a1 = 9 − 1 = 8
a3 − a2 = 25 − 9 = 16
a4 − a3 = 49 − 25 = 24
⇒ an+1 - an is not the same every time.
Therefore, the given numbers are forming an A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2 − a1 = 25 − 1 = 24
a3 − a2 = 49 − 25 = 24
a4 − a3 = 73 − 49 = 24
i.e., ak+1 − ak is same every time.
⇒ an+1 - an is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a5 = 73+ 24 = 97
a6 = 97 + 24 = 121
a7 = 121 + 24 = 145 .
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