Find the common difference and write the next four terms of each of the following arithmetic progressions:

Solution:

In the given problem, we need to find the common difference and the next four terms of the given A.P.

(i) $1,-2,-5,-8, \ldots$

Here, first term $\left(a_{1}\right)=1$

 

Common difference $(d)=a_{2}-a_{1}$

$=-2-1$

 

$=-3$

Now, we need to find the next four terms of the given A.P

That is we need to find $a_{5}, a_{6}, a_{7}, a_{8,}$

So, using the formula $a_{n}=a+(n-1) d$

Substituting $n=5,6,7,8$ in the above formula

Substituting n = 5, we get

$a_{5}=1+(5-1)(-3)$

$a_{5}=1-12$

 

$a_{5}=-11$

Substituting n = 6, we get

$a_{6}=1+(6-1)(-3)$

$a_{6}=1-15$

 

$a_{6}=-14$

Substituting n = 7, we get

$a_{7}=1+(7-1)(-3)$

$a_{7}=1-18$

 

$a_{7}=-17$

Substituting n = 8, we get

$a_{8}=1+(8-1)(-3)$

$a_{8}=1-21$

 

$a_{3}=-20$

Therefore, the common difference is $d=-3$ and the next four terms are $-11,-14,-17,-20$

(ii) $0,-3,-6,-9, \ldots$

Here, first term $\left(a_{1}\right)=0$

 

Common difference $(d)=a_{2}-a_{1}$

$=-3-0$

 

$=-3$

Now, we need to find the next four terms of the given A.P

That is we need to find $a_{5}, a_{6}, a_{7}, a_{8}$.

So, using the formula $a_{n}=a+(n-1) d$

Substituting $n=5,6,7,8$ in the above formula

 

Substituting $n=5$, we get

$a_{5}=0+(5-1)(-3)$

$a_{5}=0-12$

 

$a_{5}=-12$

Substituting n = 6, we get

$a_{6}=0+(6-1)(-3)$

$a_{6}=0-15$

 

$a_{6}=-15$

Substituting n = 7, we get

$a_{7}=0+(7-1)(-3)$

$a_{7}=0-18$

 

$a_{7}=-18$

Substituting n = 8, we get

$a_{y}=0+(8-1)(-3)$

$a_{8}=0-21$

 

$a_{3}=-21$

Therefore, the common difference is $d=-3$ and the next four terms are $-12,-15,-18,-21$

(iii) $-1, \frac{1}{4}, \frac{3}{2}, \ldots$

Here, first term (a1) =−1

Common difference $(d)=a_{2}-a_{1}$

$=\frac{1}{4}-(-1)$

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

$=\frac{5}{4}$

Now, we need to find the next four terms of the given A.P

That is we need to find $a_{4}, a_{5}, a_{6}, a_{7}$

 

So, using the formula $a_{n}=a+(n-1) d$

Substituting $n=4,5,6,7$ in the above formula

 

Substituting $n=4$, we get

$a_{4}=-1+(4-1)\left(\frac{5}{4}\right)$

$a_{4}=-1+\frac{15}{4}$

$a_{4}=\frac{-4+15}{4}$

$a_{4}=\frac{11}{4}$

Substituting n = 5, we get

$a_{5}=-1+(5-1)\left(\frac{5}{4}\right)$

$a_{5}=-1+5$

$a_{5}=4$

Substituting n = 6, we get

$a_{6}=-1+(6-1)\left(\frac{5}{4}\right)$

$a_{6}=-1+\frac{25}{4}$

$a_{6}=\frac{-4+25}{4}$

$a_{6}=\frac{21}{4}$

Substituting n = 7, we get

$a_{7}=-1+(7-1)\left(\frac{5}{4}\right)$

$a_{7}=-1+\frac{30}{4}$

$a_{7}=\frac{-4+30}{4}$

$a_{7}=\frac{26}{4}$

Therefore, the common difference is $d=\frac{5}{4}$ and the next four terms are $\frac{11}{4}, 4, \frac{21}{4}, \frac{26}{4}$

(iv) $-1,-\frac{5}{6},-\frac{2}{3}, \ldots$

Here, first term $\left(a_{1}\right)=-1$

 

Common difference $(d)=a_{2}-a_{1}$

$=-\frac{5}{6}-(-1)$

$=\frac{-5+6}{6}$

$=\frac{1}{6}$

Now, we need to find the next four terms of the given A.P

That is we need to find $a_{4}, a_{5}, a_{6}, a_{7}$

So, using the formula $a_{n}=a+(n-1) d$

Substituting $n=4,5,6,7$ in the above formula

 

Substituting $n=4$, we get

$a_{4}=-1+(4-1)\left(\frac{1}{6}\right)$

$a_{4}=-1+\left(\frac{1}{2}\right)$

 

$a_{4}=\frac{-2+1}{2}=\frac{-1}{2}$

Substituting n = 5, we get

$a_{5}=-1+(5-1)\left(\frac{1}{6}\right)$

$a_{5}=-1+\frac{2}{3}$

$a_{5}=\frac{-3+2}{3}$

 

$a_{5}=-\frac{1}{3}$

Substituting n = 6, we get

$a_{6}=-1+(6-1)\left(\frac{1}{6}\right)$

$a_{6}=-1+\frac{5}{6}$

$a_{6}=\frac{-6+5}{6}$

 

$a_{6}=-\frac{1}{6}$

Substituting n = 7, we get

$a_{7}=-1+(7-1)\left(\frac{1}{6}\right)$

$a_{7}=-1+1$

 

$a_{7}=0$

Therefore, the common difference is $d=\frac{1}{6}$ and the next four terms are $-\frac{1}{2},-\frac{1}{3},-\frac{1}{6}, 0$