3 Tutor System
Starting just at 265/hour

# 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