If an = 3 — 4n, then show

Given that, nth term of the series is an = 3 – 4n                  …(i)

Put $n=1, \quad a_{1}=3-4(1)=3-4=-1$

Put $n=2, \quad a_{2}=3-4(2)=3-8=-5$

Put $n=3, \quad a_{3}=3-4(3)=3-12=-9$

Put $n=4, \quad a_{4}=3-4(4)=3-16=-13$

So, the series becomes $-1,-5,-9,-13, \ldots$

We see that,

$a_{2}-a_{1}=-5-(-1)=-5+1=-4$

$a_{3}-a_{2}=-9-(-5)=-9+5=-4$

$a_{4}-a_{2}=-13-(-9)=-13+9=-4$

i.e. $\quad a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=\ldots=-4$

Since, the each successive term of the series has the same difference. So, it forms an Al?. We know that, sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore$ Sum of 20 terms of the AP, $S_{20}=\frac{20}{2}[2(-1)+(20-1)(-4)]$

$=10(-2+(19)(-4))=10(-2-76)$

$=10 \times-78=-780$

the required sum of 20 terms i.e., S20 is – 780.