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s-12-30-s-8-4-a-3-




Question Number 120122 by Khalmohmmad last updated on 29/Oct/20
 { ((s_(12) =30)),((s_8 =4)) :}   a_3 =?
$$\begin{cases}{{s}_{\mathrm{12}} =\mathrm{30}}\\{{s}_{\mathrm{8}} =\mathrm{4}}\end{cases}\:\:\:{a}_{\mathrm{3}} =? \\ $$
Commented by Dwaipayan Shikari last updated on 29/Oct/20
a_3 =−1
$${a}_{\mathrm{3}} =−\mathrm{1} \\ $$
Commented by Khalmohmmad last updated on 29/Oct/20
solving method   pl
$${solving}\:{method}\:\:\:{pl} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 29/Oct/20
Considering A.P  ((12)/2)(a+a+(12−1)d)=30⇒2a+11d=5  (8/2)(a+a+(8−1)d)=4⇒2a+7d=1  d=1       a=−3  T_3 =a+(3−1)1=−3+2=−1
$${Considering}\:{A}.{P} \\ $$$$\frac{\mathrm{12}}{\mathrm{2}}\left({a}+{a}+\left(\mathrm{12}−\mathrm{1}\right){d}\right)=\mathrm{30}\Rightarrow\mathrm{2}{a}+\mathrm{11}{d}=\mathrm{5} \\ $$$$\frac{\mathrm{8}}{\mathrm{2}}\left({a}+{a}+\left(\mathrm{8}−\mathrm{1}\right){d}\right)=\mathrm{4}\Rightarrow\mathrm{2}{a}+\mathrm{7}{d}=\mathrm{1} \\ $$$${d}=\mathrm{1}\:\:\:\:\:\:\:{a}=−\mathrm{3} \\ $$$${T}_{\mathrm{3}} ={a}+\left(\mathrm{3}−\mathrm{1}\right)\mathrm{1}=−\mathrm{3}+\mathrm{2}=−\mathrm{1} \\ $$

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