Question

Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

$$\left( i \right)\;2,4,8,16,.... \\\ \left( {ii} \right)\;2,\dfrac{5}{2},3,\dfrac{7}{2},... \\\ \left( {iii} \right)\; - 1.2, - 3.2, - 5.2, - 7.2,... \\\ \left( {iv} \right)\; - 10, - 6, - 2,2,... \\\ \left( v \right)\;3,3 + \sqrt 2 ,3 + 2\sqrt 2 ,3 + 3\sqrt 2 \\\ \left( {vi} \right)\;0.2,0.22,0.222,0.2222,.... \\\ \left( {vii} \right)\;0, - 4, - 8, - 12,... \\\ \left( {viii} \right)\; - \dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2},... \\\ \left( {ix} \right)\;1,3,9,27 \\\ \left( x \right)\;a,2a,3a,4a,... \\\$$

Answer 1

A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same.
Check if differences between consecutive elements are the same or not. If all differences are the same, Arithmetic Progression is possible.
An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……
a, (a+d), (a+2d), (a+3d),….. where a = first term, d = common difference.
a,b,c are said to be in AP if the common difference between any two consecutive number of the series is same ie b−a=c−b⇒2b=a+c

Complete step-by-step answer:
Step 1: find the difference between the two consecutive two numbers of the series. If the difference between consecutive numbers is the same, it is AP else it is not. If it is AP then we move to step 2.
Step 2: Calculate the common difference i.e. the difference between two consecutive numbers.
Step 3: Add the common difference to the last number in the series to find the next number in the series. Similarly find the next two numbers.
(i)$2,4,8,16,....$
Difference between consecutive numbers = 2, 4, 8 which is not equal.
It is not in AP, as the difference between consecutive terms is different.
(ii) $2,\dfrac{5}{2},3,\dfrac{7}{2},...$
Difference between consecutive numbers $= \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2}$ which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step$d = \dfrac{1}{2}$.
Next three consecutive terms:
$\dfrac{7}{2} + \dfrac{1}{2} = \dfrac{8}{2} = 4;\;4 + \dfrac{1}{2} = \dfrac{9}{2};\;\dfrac{9}{2} + \dfrac{1}{2} = \dfrac{{10}}{2} = 5$
(iii) $- 1.2, - 3.2, - 5.2, - 7.2,...$
Difference between consecutive numbers $= - 2.0, - 2.0, - 2.0$ which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step $d = - 2.0$.
Next three consecutive terms:
$- 7.2 - 2.0 = - 9.2;\; - 9.2 - 2.0 = - 11.2;\; - 11.2 - 2.0 = - 13.2$
(iv) $- 10, - 6, - 2,2,...$
Difference between consecutive numbers $= 4,4,4$ which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step $d = 4$.
Next three consecutive terms:
$2 + 4 = 6;\;6 + 4 = 10\;;\;10 + 4 = 14\;$
(v) $3,3 + \sqrt 2 ,3 + 2\sqrt 2 ,3 + 3\sqrt 2$…..
Difference between consecutive numbers $= \sqrt 2 ,\sqrt 2 ,\sqrt 2$ which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step $d = \sqrt 2$.
Next three consecutive terms:
$3 + 3\sqrt 2 + \sqrt 2 = 3 + 4\sqrt 2 ;\;3 + 4\sqrt 2 + \sqrt 2 = 3 + 5\sqrt 2 \;;\;3 + 5\sqrt 2 + \sqrt 2 = 3 + 6\sqrt 2$
(vi) $0.2,0.22,0.222,0.2222,....$
Difference between consecutive numbers = 0.02, 0.002, 0.0002 which is not equal.
It is not in AP, as the difference between consecutive terms is different.
(vii) $0, - 4, - 8, - 12,...$
Difference between consecutive numbers = -4,-4, -4 which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step d=-4.
Next three consecutive terms:
$- 12 - 4 = - 16;\; - 16 - 4 = - 20;\; - 20 - 4 = - 24$
(viii) $- \dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2},...$
Difference between consecutive numbers = 0, 0, 0 which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step d=0.
Next three consecutive terms:
$- \dfrac{1}{2} + 0 = - \dfrac{1}{2};\; - \dfrac{1}{2} + 0 = - \dfrac{1}{2};\; - \dfrac{1}{2} + 0 = - \dfrac{1}{2}$
(ix) $1,3,9,27$
Difference between consecutive numbers = 2, 6, 18 which is not equal.
It is not in AP, as the difference between consecutive terms is different.
(x) $a,2a,3a,4a,...$
Difference between consecutive numbers = a, a, a which is not equal.
It is an AP, as the difference between consecutive terms is the same.
Difference we found in the last step d=a.
Next three consecutive terms:
$4a + a = 5a;\;5a + a = 6a;\;6a + a = 7a$

Note: To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a−d),a,(a+d)
4 terms: (a−3d),(a−d),(a+d),(a+3d)
5 terms: (a−2d),(a−d),a,(a+d),(a+2d)
$t_n=S_n−S_{n−1}$
If each term of an AP is increased, decreased, multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.
In an AP, the sum of terms equidistant from beginning and end will be constant.