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Question Number 127360 by bemath last updated on 29/Dec/20

Answered by liberty last updated on 29/Dec/20

let : a−7b,a−6b,a−5b,..., a+5b, a+6b, a+7b is AP  given condition → { ((3a−18b=−60; a−6b=−20 (the first 3 terms))),((3a+18b=84; a+6b = 28 (the last 3 terms))) :}  we get  { ((a=4)),((b=4)) :}. we want to compute the  sum of the middle 3 terms ⇒T_7 +T_8 +T_9    = (a−b)+a+(a+b) = 3a = 3×4=12

$${let}\::\:{a}−\mathrm{7}{b},{a}−\mathrm{6}{b},{a}−\mathrm{5}{b},...,\:{a}+\mathrm{5}{b},\:{a}+\mathrm{6}{b},\:{a}+\mathrm{7}{b}\:{is}\:{AP} \\ $$$${given}\:{condition}\:\rightarrow\begin{cases}{\mathrm{3}{a}−\mathrm{18}{b}=−\mathrm{60};\:{a}−\mathrm{6}{b}=−\mathrm{20}\:\left({the}\:{first}\:\mathrm{3}\:{terms}\right)}\\{\mathrm{3}{a}+\mathrm{18}{b}=\mathrm{84};\:{a}+\mathrm{6}{b}\:=\:\mathrm{28}\:\left({the}\:{last}\:\mathrm{3}\:{terms}\right)}\end{cases} \\ $$$${we}\:{get}\:\begin{cases}{{a}=\mathrm{4}}\\{{b}=\mathrm{4}}\end{cases}.\:{we}\:{want}\:{to}\:{compute}\:{the} \\ $$$${sum}\:{of}\:{the}\:{middle}\:\mathrm{3}\:{terms}\:\Rightarrow{T}_{\mathrm{7}} +{T}_{\mathrm{8}} +{T}_{\mathrm{9}} \\ $$$$\:=\:\left({a}−{b}\right)+{a}+\left({a}+{b}\right)\:=\:\mathrm{3}{a}\:=\:\mathrm{3}×\mathrm{4}=\mathrm{12} \\ $$$$ \\ $$

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