Question
Find the common difference and write the next four terms of the following arithmetic progressions:
$0,-3,-6,-9, \ldots . .$
$0,-3,-6,-9, \ldots . .$
✓
Answer
Here, first term $\left(a_1\right)=0$
$\text { Common difference }(d)=a_2-a_1$
$=-3-0$
$=-3$
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
$a_8=0+(8-1)(-3)$
$a_8=0-21$
$a_8=-21$
Therefore, the common difference is $d=-3$ and the next four terms are $-12,-15,-18,-21$.
$\text { Common difference }(d)=a_2-a_1$
$=-3-0$
$=-3$
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
$a_8=0+(8-1)(-3)$
$a_8=0-21$
$a_8=-21$
Therefore, the common difference is $d=-3$ and the next four terms are $-12,-15,-18,-21$.
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