In an AP, the first is 2 and the sum of first five terms is one - fourth of the sum of next five terms.show that its 20th term is -112 and the sum of its frist 20 terms is -1100.

Explanation

we have,
`(T_(1)+T_(2)+T_(3)+T_(4)+T_(5) )=1/4(T_(6)+T_(7)+T_(8)+T_(9)+T_(10))`
`=1/4.[(T_(1)+T_(2)+......+T_(5)+T_(6)+.......+T_(10))-(T_(1) +T_(2) +T_(3) +T_(4)+T_(5))]`
` Rightarrow S_(4) = 1/4 . ( S_(10) -S_(5)) ` Where ` S_(n)` denotes the sum of first n terms.
`4S_(5) =S_(10) -S_(5)`
` Rightarrow 5S_(10) =S_(10)`
` Rightarrow 5 xx 5/2 xx [ 2xx2 + ( 5-1) d] = 10/2 xx [ 2 xx 2 + (10 -1) d]`
` 5d = -30 Rightarrow d=-6`
a =2 and d=-6
` T_(20) = [ 2 + (20-1) xx (-6) ] = -112`
` and S_(20) = 20/2 xx [ 2 xx 2+( 20-1) xx (-6)] = -1100`