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

Answer 1

(i) 2, 4, 8, 16 … It can be observed that
$a_2 - a_1$ = 4 - 2 = 2
$a_3 - a_2$ = 8 - 4 = 4
$a_4 - a_3$= 16 - 8 = 8
i.e., $a_{k+1}$ - $a_k$ is not the same every time.
Therefore, the given numbers are not forming an A.P.

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(ii) 2, $\frac{5}{2}, 3, \frac{7}{2}$
It can be observed that
$a_2-a$ = $\frac{5}{2}-2 = \frac{1}{2}$
$a_3 - a_2 = 3- \frac{5}{2} = \frac{1}{2}$
$a_4 - a_3 = \frac{7}{2} - 3 = \frac{1}{2}$
i.e., $a_{k+1} - a_k$ is same every time.
Therefore, d = $\frac{1}{2}$ and the given numbers are in A.P.
Three more terms are
$a_5 = \frac{7}{2} + \frac{1}{2}$
$a_6 = 4 + \frac{1}{2} = \frac {9}{2}$
$a_7 = \frac{9}{2} + \frac{1}{2} = 5$

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(iii) - 1.2, - 3.2, - 5.2, - 7.2 …
It can be observed that
$a_2 - a_1$ = ( - 3.2) - ( - 1.2) = - 2
$a_3 - a_2$ = ( - 5.2) - ( - 3.2) = - 2
$a_4 - a_3$= ( - 7.2) - ( - 5.2) = - 2
i.e., $a_{k+1} - a_k$ is same every time.
Therefore, d = - 2 The given numbers are in A.P. Three more terms are
$a_5$ = - 7.2 - 2 = - 9.2
$a_6$= - 9.2 - 2 = - 11.2
$a_7$ = - 11.2 - 2 = - 13.2

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(iv) - 10, - 6, - 2, 2 …
It can be observed that
$a_2 - a_1$= ( - 6) - ( - 10) = 4
$a_3 - a_2$ = ( - 2) - ( - 6) = 4
$a_4 - a_3$= (2) - ( - 2) = 4 i.e., $a_{k+1} - a_k$ is same every time.
Therefore, d = 4 The given numbers are in A.P.
Three more terms are
$a_5$= 2 + 4 = 6
$a_6$ = 6 + 4 = 10
$a_7$= 10 + 4 = 14

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(v) 3, $3+\sqrt{2} , 3+ 2\sqrt{2} , 3 + 3 \sqrt{2}$, .....
It can be observed that
$a_2 - a1$ = $3 + \sqrt{2}-3 = \sqrt {2}$
$a_3 - a_2$= $3 + 2 \sqrt{2} - 3 - \sqrt{2}= \sqrt{2}$
$a_4 - a_3$= $3 + 3 \sqrt {2} - 3 - 2 \sqrt {2}= \sqrt{2}$
i.e., $a_{k+1}-a_k$ is same every time. Therefore, d =$\sqrt{2}$
The given numbers are in A.P.
Three more terms are
$a_5$ = $3 + 3 \sqrt {2} + \sqrt {2} = 3 + 4 \sqrt{2}$
$a_6$ = $3 + 4 \sqrt{2} + \sqrt{2} = 3 + 5 \sqrt{2}$
$a_7$ = $3 + 5 \sqrt{2} + \sqrt{2} = 3 + 6 \sqrt{2}$

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(vi) 0.2, 0.22, 0.222, 0.2222 ….
It can be observed that=
$a_2 - a_1$ = 0.22 - 0.2 = 0.02
$a_3 - a_2$ = 0.222 - 0.22 = 0.002
$a_4 - a_3$= 0.2222 - 0.222 = 0.0002
i.e., $a_{k+1}$ - $a_k$ is not the same every time.
Therefore, the given numbers are not in A.P.

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(vii) 0, - 4, - 8, - 12 …
It can be observed that
$a_2-a_1$ = ( - 4) - 0 = - 4
$a_3-a_2$ = ( - 8) - ( - 4) = - 4
$a_4 - a_3$ = ( - 12) - ( - 8) = - 4
i.e., $a_{k+1}-a_k$ 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

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

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

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(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

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