Question Number 167851 by otchereabdullai@gmail.com last updated on 27/Mar/22
$$\:\:\mathrm{A}\:\mathrm{particular}\:\mathrm{A}.\mathrm{P}\:\mathrm{has}\:\mathrm{a}\:\mathrm{positive}\: \\ $$$$\:\:\mathrm{common}\:\mathrm{difference}\:\mathrm{and}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{for}\:\mathrm{any}\:\mathrm{three}\:\mathrm{adjacent}\:\mathrm{terms},\:\mathrm{three} \\ $$$$\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{squares}\:\mathrm{exceed} \\ $$$$\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{by}\:\mathrm{37}.\mathrm{5}\:.\:\mathrm{Find} \\ $$$$\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}.\: \\ $$
Answered by mr W last updated on 27/Mar/22
![say the terms are a−d, a, a+d 3[(a−d)^2 +a^2 +(a+d)^2 ]−(a−d+a+a+d)^2 =37.5 3(3a^2 +2d^2 )−(3a)^2 =((75)/2) d^2 =((75)/(12))=((25)/4) ⇒d=(5/2)](Q167854.png)
$${say}\:{the}\:{terms}\:{are}\:{a}−{d},\:{a},\:{a}+{d} \\ $$$$\mathrm{3}\left[\left({a}−{d}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} +\left({a}+{d}\right)^{\mathrm{2}} \right]−\left({a}−{d}+{a}+{a}+{d}\right)^{\mathrm{2}} =\mathrm{37}.\mathrm{5} \\ $$$$\mathrm{3}\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{d}^{\mathrm{2}} \right)−\left(\mathrm{3}{a}\right)^{\mathrm{2}} =\frac{\mathrm{75}}{\mathrm{2}} \\ $$$${d}^{\mathrm{2}} =\frac{\mathrm{75}}{\mathrm{12}}=\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$\Rightarrow{d}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Mar/22
$$\mathrm{Infact}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{and}\:\mathrm{give}\:\mathrm{you}\: \\ $$$$\mathrm{long}\:\mathrm{life}\:\mathrm{because}\:\mathrm{you}\:\mathrm{are}\:\mathrm{beneficial}\: \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{world}!\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}! \\ $$
Commented by peter frank last updated on 27/Mar/22
$$\mathrm{thanks} \\ $$