Question

if the 4th and 9th term of ap are 8 and -4 then which term is 0​

Answer 1

Step-by-step explanation:

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Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{\frac{22}{3}^{nd} \ term \ is \ zero.}$

$\bold\orange{Given:}$

$\bold{In \ an \ A.P.,}$

$\bold{=>4th \ term \ is \ 8}$

$\bold{=>9th \ term \ is \ -4}$

$\bold\pink{To \ find:}$

$\bold{Which \ term \ is \ zero}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{tn=a+(n-1)d... formula}$

$\bold{8=a+(4-1)d}$

$\bold{8=a+3d}$

$\bold{i.e.a+3d=8...(1)}$

______________________________________

$\bold{t9=a+(9-1)d}$

$\bold{-4=a+8d}$

$\bold{i.e.a+8d=-4...(2)}$

$\bold{Subtract \ eq(1) \ from \ eq (2)}$

$\bold{a+8d=-4}$

$\bold{-}$

$\bold{a+3d=8}$

$\bold{5d=-12}$

$\bold{d=\frac{-12}{5}}$

$\bold{Substitute \ value \ of \ d \ in \ eq(1)}$

$\bold{a+3(\frac{-12}{5})=8}$

$\bold{a+(\frac{-36}{5})=8}$

$\bold{a=8+\frac{36}{5}}$

$\bold{a=\frac{36+40}{5}}$

$\bold{a=\frac{76}{5}}$

__________________________________

$\bold{Now,tn=0, \ a=\frac{76}{5} , d=\frac{-12}{5}}$

$\bold{tn=a+(n-1)d...formul}$

$\bold{0=\frac{76}{5}+(n-1)×(\frac{-12}{5})}$

$\bold{\tt{\therefore{(n-1)×(\frac{-12}{5})=\frac{-76}{5}}}}$

$\bold{(n-1)=\frac{-76}{5}×\frac({-5}{12})}$

$\bold{n-1=\frac{19}{3}}$

$\bold{n=\frac{19+3}{3}}$

$\bold{n=\frac{22}{3}}$

$\bold\purple{\tt{\therefore{\frac{22}{3}^{nd} \ term \ is \ zero.}}}$