Question:
For the AP -3, – 7, – 11,… can we find directly a30 – a20 without actually finding a30 and
a20? Give reason for your answer.
Solution:
True
$\because n$th term of an AP, $a_{n}=a+(n-1) d$
$\therefore \quad a_{30}=a+(30-1) d=a+29 d$
and $\quad a_{20}=a+(20-1) d=a+19 d$ $\ldots(\mathrm{i})$
Now, $\quad a_{30}-a_{20}=(a+29 d)-(a+19 d)=10 d$
and from given AP common difference, $d=-7-(-3)=-7+3$
$=-4$
$\therefore \quad a_{30}-a_{20}=10(-4)=-40$ [from Eq. (i)]