In an A.P., the sum of first ten terms is -150 and the sum of its next ten terms is -550. Find the A.P.

Given: In an A.P., the sum of first ten terms is -150 and the sum of its next ten terms is -550.

What to do: To find the A.P.

Solution: Here given:

Sum of first ten terms $S_{10}=-150$

Sum of next ten terms$=-550$

Sum of first 20 terms, S_{20} =-150-550=-700$

Let us say first term of A.P.$=a$

Common difference$=d$

We know that sum of n terms of an A.P. $S_{n} =\frac{n}{2}[ 2a+( n-1) d]$

sum of first 10 terms $S_{10} =\frac{10}{2}[ 2a+( 10-1) d]$

$-150=5( 2a+9d) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( here\ S_{10} =-150\ and\ n=10\ )$

$\ \Rightarrow \ 2a+9d=-30\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc ..\ ( 1)$

Sum of first 20 terms $S_{20} =\frac{20}{2}[ 2a+( 20-1) d]$

$-700=10( 2a+19d) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( here\ S_{20} =-700\ and\ n=20)$

$\Rightarrow 2a+19d=-70\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \ \ ( 2)$

Subtracting $( 1)$ from $( 2)$

$2a+19d-2a-9d=-70+30$

$10d=-40$

$d=\frac{-40}{4}$

$\Rightarrow d=-4$

On substituting the value of $d=-4$ in equation$( 2)$

$\Rightarrow 2a-76=-70$

$\Rightarrow 2a=-70+76=6$

$\Rightarrow a=3$

After solving $( 1)$  and $( 2)$

We get first term of the A.P.$=3$

Common difference$=-4$

Therefore the A.P. is $3,\ -1,\ -5,\ -9,\ -13,\ \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc $