Question

Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) $2,4,8,16,\dots$
(ii) $2,\frac{5}{2},3,\frac{7}{2},\dots$
(iii)$-1.2,-3.2,-5.2,-7.2\dots$
$\left[5\right]$

Answer 1

If the difference between any two consecutive terms of a series is a constant value, then the series is an A.P

(i) Given, the series is $2,4,8,16,\dots$
Here,
$a_2-a_1=4-2=2$
$a_3-a_2=8-4=4a_4-a_3=16-8=8$

Note that the difference between any two consecutive terms of the series is not a constant value.
Hence, the given series is not an AP.
$\left[1\right]$

(ii) Given, the series is :
$2,\frac{5}{2},3,\frac{7}{2}\dots$
Here,
$a_2-a_1=\frac{5}{2}-2=\frac{1}{2}{a}_3-a_2=3-\frac{5}{2}=\frac{1}{2}{a}_4-a_3=\frac{7}{2}-3=\frac{1}{2}$
​​​​​​​$\left[1\right]$
Note that the difference between any two consecutive terms of the series is a constant value.
Therefore, $d=\frac{1}{2}$ and the given numbers are in A.P.
The next three terms are:
$a_5=\frac{7}{2}+\frac{1}{2}=4a_6=4+\frac{1}{2}=\frac{9}{2}{a}_7=\frac{9}{2}+\frac{1}{2}=5$
​​​​​​​$\left[1.5\right]$

(iii) $-1.2,-3.2,-5.2,-7.2\dots \dots$
Here,
$a_2-a_1=(-3.2)-(-1.2)=-2a_3-a_2=(-5.2)-(-3.2)=-2a_4-a_3=(-7.2)-(-5.2)=-2$
$\Rightarrow a_{n+1}-a_n$ is same for all n.
Therefore, d=-2 and the given numbers are in A.P.
Next three terms are:
$a_5=-7.2-2=-9.2a_6=-9.2-2=-11.2a_7=-11.2-2=-13.2$
​​​​​​​$\left[1.5\right]$