Question
Find $a_{30} - a_{20}$_ for the A.P.
$-9, -14, -19, -24$, .....
$-9, -14, -19, -24$, .....
✓
Answer
Given,
$a_{30}-a_{20}=a+(30-1) d-(a+(20-1) d)$
$\left(\therefore a_n=a+(n-1) d\right)$
$=a+29 d-a-19 d$
$=10 d$
$-9,-14,-19,-24, \ldots .$
Common differecne (d) = Secomd term - First term
$=-14-(-9)$
$=-14+9$
$d=-5$
Then $a _{30}- a _{20}=10 d$
$=10 \times(-5)$
$=-50$
$a_{30}-a_{20}=a+(30-1) d-(a+(20-1) d)$
$\left(\therefore a_n=a+(n-1) d\right)$
$=a+29 d-a-19 d$
$=10 d$
$-9,-14,-19,-24, \ldots .$
Common differecne (d) = Secomd term - First term
$=-14-(-9)$
$=-14+9$
$d=-5$
Then $a _{30}- a _{20}=10 d$
$=10 \times(-5)$
$=-50$
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