In an A.P., the sum of its first ten terms is -80 and the sum of its next ten terms is -280. Find the A.P.
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In an A.P., the sum of its first ten terms is -80 and the sum of its next ten terms is -280. Find the A.P.
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[tex]\sf\red{\underline{\underline{Answer:}}}[/tex]
[tex]\sf{The \ required \ A.P. \ is \ 1, \ -1, \ -3,...}[/tex]
[tex]\sf\orange{Given:}[/tex]
[tex]\sf{In \ an \ A.P.,}[/tex]
[tex]\sf{The \ sum \ of \ first \ 10 \ terms (S_{10})=-80}[/tex]
[tex]\sf{The \ sum \ of \ next \ 10 \ terms=-280}[/tex]
[tex]\sf\pink{To \ find:}[/tex]
[tex]\sf{The \ A.P.}[/tex]
[tex]\sf\green{\underline{\underline{Solution:}}}[/tex]
[tex]\boxed{\sf{S_{n}=\frac{n}{2}[2a+(n-1)d]}}[/tex]
[tex]\sf{According \ to \ the \ first \ condition}[/tex]
[tex]\sf{-80=\frac{10}{2}[2a+(10-1)d]}[/tex]
[tex]\sf{\therefore{5[2a+9d]=-80}}[/tex]
[tex]\sf{\therefore{2a+9d=\frac{-80}{5}}}[/tex]
[tex]\sf{\therefore{2a+9d=-16...(1)}}[/tex]
[tex]\sf{According \ to \ the \ second \ condition.}[/tex]
[tex]\sf{Sum \ of \ next \ 10 \ terms=S_{20}-S_{10}}[/tex]
[tex]\sf{-280=\{\frac{20}{2}[2a(20-1)d]\}-(-80)}[/tex]
[tex]\sf{\therefore{10[2a+19d]=-280+(-80)}}[/tex]
[tex]\sf{\therefore{10[2a+19d]=-360}}[/tex]
[tex]\sf{\therefore{2a+19d=\frac{-360}{10}}}[/tex]
[tex]\sf{\therefore{2a+19d=-36...(2)}}[/tex]
[tex]\sf{Subtract \ equation (2) \ from \ equation (1)}[/tex]
[tex]\sf{2a+9d=-16}[/tex]
[tex]\sf{-}[/tex]
[tex]\sf{2a+19d=-36}[/tex]
_________________
[tex]\sf{10d=-20}[/tex]
[tex]\sf{\therefore{d=\frac{-20}{10}}}[/tex]
[tex]\boxed{\sf{\therefore{d=-2}}}[/tex]
[tex]\sf{Substitute \ d=-2 \ in \ equation (1)}[/tex]
[tex]\sf{2a+9(-2)=-16}[/tex]
[tex]\sf{\therefore{2a-18=-16}}[/tex]
[tex]\sf{\therefore{2a=-16+18}}[/tex]
[tex]\sf{\therefore{2a=2}}[/tex]
[tex]\sf{\therefore{a=\frac{2}{2}}}[/tex]
[tex]\boxed{\sf{\therefore{a=1}}}[/tex]
[tex]\sf{The \ A.P. \ is}[/tex]
[tex]\sf{t_{1}=a=1,}[/tex]
[tex]\sf{t_{2}=a+d=1+(-2)=-1,}[/tex]
[tex]\sf{t_{3}=a+2d=1+2(-2)=-3.}[/tex]
[tex]\sf\purple{\tt{\therefore{The \ required \ A.P. \ is \ 1, \ -1, \ -3,...}}}[/tex]
Step-by-step explanation:
[tex]\huge\boxed {\red {\mathbb{\underline {\underbrace {Answer}}}}}[/tex]
[tex]<font color =purple >[/tex]
let the first term and d is the common differences of AP
= Sn=n/2{2a+(n-1)d}
now.............
=S10=10/2{2a+(10-1)d}
= -80=5(2a+9d)
= 2a+9d=-16.........(1)
again....
S20-S10=-280
=20/2{2a+19d}+80=-280
= 10{2a+19d}=-360
= 2a+19d=-36.........(2)
solve this two equation
then..
we found
d=-2
a=1
so...
AP=1,-1,-3,-5......