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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 $
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