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, \({{5}\over{2}}\) , 3, \({{7}\over{2}}\)…
(iii) \(-1.2, -3.2, -5.2, -7.2…\)
(iv) \(-10, -6, -2, 2…\)
(v) \(3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}\)…
(vi) \(0.2, 0.22, 0.222, 0.2222…\)
(vii) \(0, -4, -8, -12…\)
(viii) \({{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}\)…
(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) \(1^2, 5^2, 7^2, 73\)…

(i) 2, 4, 8, 16…

It is not an AP because difference between consecutive terms is not equal.

4 – 2 \( \ne \) 8 - 4


(ii) 2, \({{5}\over{2}}\) , 3, \({{7}\over{2}}\)…

It is an AP because difference between consecutive terms is equal.

\(\Rightarrow {{5}\over{2}} - 2 = 3 - {{5}\over{2}} = {{1}\over{2}}\)

Common difference (d) = \({{1}\over{2}}\)
Fifth term = \({{7}\over{2}} + {{1}\over{2}} = 4\)
Sixth term = 4 + \({{1}\over{2}} = {{9}\over{2}}\)
Seventh term = \({{9}\over{2}} + {{1}\over{2}} = 5\)

Therefore, next three terms are 4, \({{9}\over{2}}\) and 5.


(iii)-1.2, -3.2, -5.2, -7.2…

It is an AP because difference between consecutive terms is equal.

-3.2 - (-1.2)
= -5.2 - (-3.2)
= -7.2 - (-5.2) = -2

Common difference (d) = -2

Fifth term = -7.2 – 2 = -9.2
Sixth term = -9.2 – 2 = -11.2
Seventh term = -11.2 – 2 = -13.2

Therefore, next three terms are -9.2, -11.2 and -13.2


(iv) -10, -6, -2, 2…

It is an AP because difference between consecutive terms is equal.

-6 - (-10) = -2 - (-6)
= 2 - (-2) = 4

Common difference (d) = 4

Fifth term = 2 + 4 = 6
Sixth term = 6 + 4 = 10
Seventh term = 10 + 4 = 14

Therefore, next three terms are 6, 10 and 14


(v) 3 ,\(3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}\)…

It is an AP because difference between consecutive terms is equal.

= \(3 + \sqrt{2} - 3 \)
= 3 + 2\sqrt{2} - 3 + \sqrt{2}\)
\( = \sqrt{2}\)

Common difference (d) = \(\sqrt{2}\)

Fifth term
= \(3 + 3\sqrt{2} + \sqrt{2} \)
\( = 3 + 4\sqrt{2}\)
Sixth term
= \(3 + 4\sqrt{2} + \sqrt{2} \)
\( = 3 + 5\sqrt{2}\)
Seventh term
= \(3 + 5\sqrt{2} + \sqrt{2}\)
\( = 3 + 6\sqrt{2}\)

Therefore, next three terms are \( 3 + 4\sqrt{2}, 3 + 5\sqrt{2}. 3 + 6\sqrt{2}\)


(vi) 0.2, 0.22, 0.222, 0.2222…

It is not an AP because difference between consecutive terms is not equal.

0.22 - 0.2 \( \ne \) 0.222 - 0.22


(vii) 0, -4, -8, -12…

It is an AP because difference between consecutive terms is equal.

-4 – 0 = -8 - (-4)
= -12 - (-8) = -4

Common difference (d) = -4

Fifth term = -12 – 4 =-16
Sixth term = -16 – 4 = -20
Seventh term = -20 – 4 = -24

Therefore, next three terms are -16, -20 and -24


(viii) \({{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}\)…

It is an AP because difference between consecutive terms is equal.

=\({{-1}\over{2}} - {{-1}\over{2}} = {{-1}\over{2}} - {{-1}\over{2}} = 0\)

Common difference (d) = 0

Fifth term = \({{-1}\over{2}} + 0 = {{-1}\over{2}}\)
Sixth term = \({{-1}\over{2}} + 0 = {{-1}\over{2}}\)
Seventh term =\({{-1}\over{2}} + 0 = {{-1}\over{2}}\)

Therefore, next three terms are \({{-1}\over{2}},{{-1}\over{2}},{{-1}\over{2}}\)

(ix) 1, 3, 9, 27…

It is not an AP because difference between consecutive terms is not equal.

3 – 1 \( \ne \) 9 - 3


(x) a, 2a, 3a, 4a…

It is an AP because difference between consecutive terms is equal.

2a – a
= 3a - 2a
= 4a - 3a = a

Common difference (d) = a

Fifth term = 4a + a = 5a
Sixth term = 5a + a = 6a
Seventh term = 6a + a = 7a

Therefore, next three terms are 5a, 6a and 7a


(xi) \(a , a^2 , a^3 , a^4 \)…

It is not an AP because difference between consecutive terms is not equal

\(a^2 - a \ne a^3 - a^2\)


(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \)…

It is an AP because difference between consecutive terms is equal.

=\(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\)

Common difference (d) = \(\sqrt{2}\)

Fifth term = \(4\sqrt{2} + \sqrt{2} = 5\sqrt{2}\)
Sixth term = \(5\sqrt{2} + \sqrt{2} = 6\sqrt{2}\)
Seventh term = \(6\sqrt{2} + \sqrt{2} = 7\sqrt{2}\)

Therefore, next three terms are \(5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}\)


(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12} \)…

It is not an AP because difference between consecutive terms is not equal.

= \(\sqrt{6} - \sqrt{3} \ne \sqrt{9} - \sqrt{6}\)


(xiv) \(1^2, 3^2, 5^2, 7^2\)…

It is not an AP because difference between consecutive terms is not equal.

= \(3^2 - 1^2 \ne 5^2 - 3^2\)


(xv) \(1^2, 5^2, 7^2, 73\)…
1, 25, 49, 73…

It is an AP because difference between consecutive terms is equal.

= \( 5^2 - 1^2 \)
\(= 7^2 - 5^2\)
\( = 73 - 7^2\)
\( = 24\)

Common difference (d) = 24

Fifth term = 73 + 24 = 97
Sixth term = 97 + 24 = 121
Seventh term = 121 + 24 = 145

Therefore, next three terms are 97, 121 and 145