Question

Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference. $\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},,...$

Answer 1

Given sequence is,

$\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},...\dots$

first term of this A.P is $a_1=\frac{1}{2}$

second term of this A.P is $a_2=\frac{1}{4}$

third term of this A.P is $a_3=\frac{1}{6}$

the condition for an sequence to be an A.P is their must be a common difference $(i.e.,d=a_{n+1}-a_n)$

putting n=1 in above equation

$d=a_2-a_1=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$

putting n=2 in above equation

$d=a_3-a_2=\frac{1}{6}-\frac{1}{4}=-\frac{1}{12}$

as we can see we get two values of d , but in an A.P their must be a single value of d which means for this sequence we can't define a common difference hence this sequence not form an A.P