In an AP the sum of …

According to the question,the sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550
Let a be the first term and d be the common difference of the given AP.
Then, we have
S₁₀=-150
{tex}\Rightarrow \frac { 10 } { 2 } [ 2 a + 9 d ] = - 150{/tex}
{tex}\Rightarrow{/tex}5[2a+9d]=-150
{tex}\Rightarrow{/tex}2a+9d=-30...(i)
Clearly, the sum of first 20 terms =-150+(-550)=-700
{tex}\therefore{/tex}S₂₀=-700
{tex}\Rightarrow \frac { 20 } { 2 } [ 2 a + 19 d ] = - 700{/tex}
{tex}\Rightarrow{/tex}10[2a+19d]=-700
{tex}\Rightarrow{/tex}2a+19d=-70...(iii)
Subtracting (i) from (ii), we get
10d=-40
{tex}\Rightarrow{/tex}d=-4
{tex}\Rightarrow{/tex}2a=-30-9(-4)=-30+36=6
{tex}\Rightarrow{/tex}a=3
Thus, we have 
First term=a=3
Second term= a+d=3+2(-4)=-1
Third term=a+2d=3+2(-4)=3-8=-5
Fourth term=a+3d=3+3(-4)=3-12=-9
Thus, the given AP is 3,-1,-5,-9,....