If `a_(n)=3-4n`, then show that `a_(1),a_(2),a_(3), …` form an AP. Also, find `S_(20)`.

Explanation

Given that, `n`th term of the series is `a_(n)=3-4n " " `…(i)
Put`n=1, " " a_(1)=3-4(1)=3-4= -1`
Put`n=2, " " a_(2)=3-4(2)= 3-8=-5`
Put`n=3, " " a_(3)=3-4(3)= 3-12=-9`
Put`n=4, " " a_(4)=3-4(4)=3-16= -13`
So the series becomes `-1,-5,-9,-13, …`
We see that,
`a_(2)-a_(1)=-5-(-1)=-5+1=-4`
`a_(3)-a_(2)=-9-(-5)=-9+5= -4`
`a_(4)-a_(3)= -13-(-9)=-13+9= -4`
` "i.e., " a_(2)-a_(1)=a_(3)-a_(2)=a_(4)-a_(3)=...= -4`
Since, the each successive term of the series has the same difference. So, it forms an AP.
We know that, sun of n terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]`
` :. ` Sum of 20 terms of the AP,`S_(20)=(20)/(2)[2(-1)+(20-1)(-4)]`
`S_(20)=10(-2+(19)(-4))=10(-2-76)`
` " " =10xx-78= -780`
Hence, the required sum of 20 terms i.e., `S_(20)` is `-780`.