Question:

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) 2,52,3,722, \frac{5}{2},3,\frac{7}{2}, . . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, 3+2,3+32,3+323 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2} . . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . . .
(vii) 0, – 4, – 8, –12, . . . .
(viii) 12,12,12,12\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, . . . .
(ix) 1, 3, 9, 27, . . . .
(x) a, 2a, 3a, 4a, . . . .
(xi) a, a2,a3,a4,a^2, a^3, a^4,  . . . .
(xii) 2,8,18,32\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32} . . . .
(xiii) 3,6,9,12\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12} . . . . .
(xiv) 12,32,52,721^2 , 3^2 , 5^2 , 7^2, . . . .
(xv) 12,52,72,731^2 , 5^2, 7^2, 7^3, . . . .

Updated On: Jul 28, 2024
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Solution and Explanation

(i) 2, 4, 8, 16 … It can be observed that
a2a1a_2 - a_1 = 4 - 2 = 2
a3a2a_3 - a_2 = 8 - 4 = 4
a4a3a_4 - a_3= 16 - 8 = 8
i.e., ak+1a_{k+1} - aka_k is not the same every time.
Therefore, the given numbers are not forming an A.P.


(ii) 2, 52,3,72\frac{5}{2}, 3, \frac{7}{2}
It can be observed that
a2aa_2-a = 522=12\frac{5}{2}-2 = \frac{1}{2}
a3a2=352=12a_3 - a_2 = 3- \frac{5}{2} = \frac{1}{2}
a4a3=723=12a_4 - a_3 = \frac{7}{2} - 3 = \frac{1}{2}
i.e., ak+1aka_{k+1} - a_k is same every time.
Therefore, d = 12\frac{1}{2} and the given numbers are in A.P.
Three more terms are
a5=72+12a_5 = \frac{7}{2} + \frac{1}{2}
a6=4+12=92a_6 = 4 + \frac{1}{2} = \frac {9}{2}
a7=92+12=5a_7 = \frac{9}{2} + \frac{1}{2} = 5


(iii) - 1.2, - 3.2, - 5.2, - 7.2 …
It can be observed that
a2a1a_2 - a_1 = ( - 3.2) - ( - 1.2) = - 2
a3a2a_3 - a_2 = ( - 5.2) - ( - 3.2) = - 2
a4a3a_4 - a_3= ( - 7.2) - ( - 5.2) = - 2
i.e., ak+1aka_{k+1} - a_k is same every time.
Therefore, d = - 2 The given numbers are in A.P. Three more terms are
a5a_5 = - 7.2 - 2 = - 9.2
a6a_6= - 9.2 - 2 = - 11.2
a7a_7 = - 11.2 - 2 = - 13.2


(iv) - 10, - 6, - 2, 2 …
It can be observed that
a2a1a_2 - a_1= ( - 6) - ( - 10) = 4
a3a2a_3 - a_2 = ( - 2) - ( - 6) = 4
a4a3a_4 - a_3= (2) - ( - 2) = 4 i.e., ak+1aka_{k+1} - a_k is same every time.
Therefore, d = 4 The given numbers are in A.P.
Three more terms are
a5a_5= 2 + 4 = 6
a6a_6 = 6 + 4 = 10
a7a_7= 10 + 4 = 14


(v) 3, 3+2,3+22,3+323+\sqrt{2} , 3+ 2\sqrt{2} , 3 + 3 \sqrt{2}, .....
It can be observed that 
a2a1a_2 - a1 = 3+23=23 + \sqrt{2}-3 = \sqrt {2}
a3a2a_3 - a_23+2232=23 + 2 \sqrt{2} - 3 - \sqrt{2}= \sqrt{2}
a4a3a_4 - a_33+32322=23 + 3 \sqrt {2} - 3 - 2 \sqrt {2}= \sqrt{2}
i.e., ak+1aka_{k+1}-a_k is same every time. Therefore, d =2\sqrt{2}
The given numbers are in A.P.
Three more terms are
a5a_5 = 3+32+2=3+423 + 3 \sqrt {2} + \sqrt {2} = 3 + 4 \sqrt{2}
a6a_6 = 3+42+2=3+523 + 4 \sqrt{2} + \sqrt{2} = 3 + 5 \sqrt{2}
a7a_7 = 3+52+2=3+623 + 5 \sqrt{2} + \sqrt{2} = 3 + 6 \sqrt{2}


(vi) 0.2, 0.22, 0.222, 0.2222 ….
It can be observed that=
a2a1a_2 - a_1 = 0.22 - 0.2 = 0.02
a3a2a_3 - a_2 = 0.222 - 0.22 = 0.002
a4a3a_4 - a_3= 0.2222 - 0.222 = 0.0002
i.e., ak+1a_{k+1} - aka_k 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
a2a1a_2-a_1 = ( - 4) - 0 = - 4
a3a2a_3-a_2 = ( - 8) - ( - 4) = - 4
a4a3a_4 - a_3 = ( - 12) - ( - 8) = - 4
i.e., ak+1aka_{k+1}-a_k is same every time.
Therefore, d = - 4, so the given numbers are in A.P.
Three more terms are 
a+(5 - 1)d= -16,
a+(6 - 1)d= -20,
a+(7 - 1)d= -24


(viii) It is in AP, with common difference 0, therefore next three terms will also be same as previous ones, i.e., 12-\frac{1}{2}.


(ix) 9 - 3 = 3 - 1
6 ≠ 2
a3 - a2 ≠ a2 - a1
Therefore, the given numbers are not in A.P.


(x) It is in AP with common difference d = 2a - a = a and first term is a,
Next three terms are
a + (5 - 1)d = 5a,
a + (6 - 1)d = 6a,
a + (7 - 1)d = 7a


(xi) It is not in AP, as the difference is not constant.
(xii) It is in AP with common difference d = 2\sqrt{2} and a = 2\sqrt2,
Next three terms are
a + (5 - 1)d = 52=505\sqrt{2}=\sqrt{50}
a + (6 - 1)d = 72\sqrt{72}
a + (7 - 1)d = 98\sqrt{98}
(xiii) It is not in AP as difference is not constant.
(xiv) It is not in AP as difference is not constant.
(xv) It is in AP with common difference d = 52 - 1 = 24 and a = 1
Next three terms are
a + (5 - 1)d = 97,
a + (6 - 1)d = 121,
a + (7 - 1)d =145

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