Question

s10=140 s9=120 t10=?​

Answer 1

Correct Question:

$\sf{In \ an \ AP, \ S_{10}=140, \ S_{9}=120} \\ \sf{Find \ 10^{th} \ term.}$

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Answer:

$\sf{The \ 10^{th} \ term \ is \ 20.}$

Given:

$\sf{\leadsto{S_{10}=140}} \\ \\ \sf{\leadsto{S_{9}=120}}$

To find:

$\sf{The \ 10^{th} \ term \ of \ the \ A.P.}$

Solution:

$\boxed{\sf{t_{n}=S_{n}-S_{n-1}}} \\ \\ \sf{\therefore{t_{10}=S_{10}-S_{9}}} \\ \\ \sf{\therefore{t_{10}=140-120}} \\ \\ \sf{\therefore{t_{10}=20}} \\ \\ \purple{\tt{\therefore{The \ the \ 10^{th} \ term \ is \ 20.}}}$

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$\sf{Long \ method} \\ \\ \boxed{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}} \\ \\ \sf{According \ to \ the \ first \ condition.} \\ \\ \sf{140=\dfrac{10}{2}[2a+9d]} \\ \\ \sf{\therefore{2a+9d=28...(1)}} \\ \\ \sf{According \ to \ the \ second \ condition.} \\ \\ \sf{120=\dfrac{9}{2}[2a+8d]} \\ \\ \sf{\therefore{2a+8d=\dfrac{80}{3}...(2)}}$

$\sf{Subtract \ equations \ (2) \ from \ equation \ (1), \ we \ get} \\ \\ \sf{d=28-\dfrac{80}{3}} \\ \\ \sf{d=\dfrac{84-80}{3}} \\ \\ \sf{\therefore{d=\dfrac{4}{3}}} \\ \\ \sf{Substitute \ d=\dfrac{4}{3} \ in \ equation \ (1), \ we \ get} \\ \\ \sf{2a+9\times\dfrac{4}{3}=28} \\ \\ \sf{\therefore{2a=16}} \\ \\ \sf{\therefore{a=8}} \\ \\ \boxed{\sf{t_{n}=a+(n-1)d}} \\ \\ \sf{\therefore{t_{10}=8+9\times\dfrac{4}{3}}} \\ \\ \sf{\therefore{t_{10}=8+12}} \\ \\ \sf{\therefore{t_{10}=20}} \\ \\ \purple{\tt{\therefore{The \ 10^{th} \ term \ is \ 20.}}}$