Question

Find the common difference and write the next four terms of each of the following arithmetic progression:
$-1,\frac{1}{4},\frac{3}{2},...$

Answer 1

Given A.P is

$-1,\frac{1}{4},\frac{3}{2},....$

the common difference of an A.P is given by $d=a_{n+1}-a_n$

by putting n=1 in above equation

$d=a_2-a_1=\left(\frac{1}{4}\right)-(-1)$

$d=\frac{1}{4}+1$

$d=\frac{5}{4}$

first term of this A.P is

$a_1=-1$

the nth term of this A.P is given by

$a_n=a_1+(n-1)d$

$⟹a_n=-1+(n-1)\left(\frac{5}{4}\right)$

$⟹a_n=-1+\frac{5}{4}n-\frac{5}{4}$

$⟹a_n=-\frac{9}{4}+\frac{5}{4}n\dots .eq\left(1\right)$

finding next four terms of A.P

1) for finding the 4th term of sequence $\left(a_4\right)$ put n=4 in eq(1)

$⟹a_4=\frac{-9}{4}+\frac{5}{4}\times 4$

$⟹a_4=\frac{-9}{4}+\frac{20}{4}$

$⟹a_4=\frac{11}{4}$

2) for finding the 5th term of sequence $\left(a_5\right)$ put n=5 in eq(1)

$⟹a_5=\frac{-9}{4}+\frac{5}{4}\times 5$

$⟹a_5=\frac{-9}{4}+\frac{25}{4}$

$⟹a_5=\frac{16}{4}$

$⟹a_5=4$

3) for finding the 6th term of sequence $\left(a_6\right)$ put n=6 in eq(1)

$⟹a_6=\frac{-9}{4}+\frac{5}{4}\times 6$

$⟹a_6=\frac{-9}{4}+\frac{30}{4}$

$⟹a_6=\frac{21}{4}$

4) for finding the 7th term of sequence $\left(a_7\right)$ put n=7 in eq(1)

$⟹a_7=\frac{-9}{4}+\frac{5}{4}\times 7$

$⟹a_7=\frac{-9}{4}+\frac{35}{4}$

$⟹a_7=\frac{26}{4}$

$⟹a_7=\frac{13}{2}$