Question

Question 2 (ii)
Verify that each of the following is an AP and then write its next three terms.
$5,\frac{14}{3},\frac{13}{3},4,\cdots$

Answer 1

$Here,a_1=5,a_2=\frac{14}{3},a_3=\frac{13}{3}and{a}_4=4$

$a_2-a_1=\frac{14}{3}-5=\frac{14-15}{3}=\frac{-1}{3},$
$a_3-a_2=\frac{13}{3}-\frac{14}{3}=-\frac{1}{3},$
$a_4-a_3=4-\frac{13}{3}=\frac{12-13}{3}=\frac{-1}{3}$

$\therefore a_2-a_1=a_3-a_2=a_4-a_3$

Since the difference between successive terms are same, it forms an AP.
The next three terms are,
$a_5=a_1+4d=5+4(-\frac{1}{3})=5-\frac{4}{3}=\frac{11}{3}$
$[\because a_n=a_1+(n-1\left)d\right]$
$a_6=a_1+5d=5+5(-\frac{1}{3})=5-\frac{5}{3}=\frac{10}{3}$
$a_7=a_1+6d=5+6(-\frac{1}{3})=5-2=3$