Question

Find $a_{30}-a_{20}$ for the A.P $-9,-14,-19,-24,......$

Answer 1

Given A.P is

$-9,-14,-19,-24,....$

hence the common difference is given by $d=a_{n+1}-a_n$

by putting n=1 in above equation

$d=a_2-a_1=(-14)-(-9)$

$d=(-14+9)$

$d=-5$

first term of this A.P is

$a_1=-9$

the nth term of this A.P is given by

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

$⟹a_n=-9+(n-1)(-5)$

$⟹a_n=-9-5n+5$

$⟹a_n=-4-5n\dots .eq\left(1\right)$

1) for finding the 30th term of sequence $\left(a_{30}\right)$ put n=30 in eq(1)

$⟹a_{30}=-4-5\times 30$

$⟹a_{30}=-4-150$

$⟹a_{30}=-154$

2) for finding the 20th term of sequence $\left(a_{20}\right)$put n=20 in eq(1)

$⟹a_{20}=-4-5\times 20$

$⟹a_{20}=-4-100$

$⟹a_{20}=-104$

now,

$a_{30}-a_{20}=(-154)-(-104)$

$a_{30}-a_{20}=(-154+104)$

$a_{30}-a_2=-50$