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Question Number 124093 by Don08q last updated on 30/Nov/20

The 7th term of an A.P. is 3p + 5q    and the 18th term is 19(2q − p). Find  the first three terms of the sequence.

$$\mathrm{The}\:\mathrm{7}{th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}.\:\mathrm{is}\:\mathrm{3}{p}\:+\:\mathrm{5}{q}\:\: \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{18}{th}\:\mathrm{term}\:\mathrm{is}\:\mathrm{19}\left(\mathrm{2}{q}\:−\:{p}\right).\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{first}\:{three}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence}. \\ $$

Answered by liberty last updated on 30/Nov/20

T_7  = 3p+5q ; T_(18) = 38q−19p   d = ((T_(18) −T_7 )/(18−7)) = ((38q−19p−3p−5q)/(11))=((33q−22p)/(11))=3q−2p  T_1  = T_7 +(1−7)d = 3p+5q−6(3q−2p)  T_1 = 3p+5q+12p−18q = 15p−13q  The first three terms of the sequence   T_1 = 15p−13q  T_2 = 15p−13q+3q−2p=13p−10q   T_3 =13p−10q+3q−2p=11p−7q

$${T}_{\mathrm{7}} \:=\:\mathrm{3}{p}+\mathrm{5}{q}\:;\:{T}_{\mathrm{18}} =\:\mathrm{38}{q}−\mathrm{19}{p}\: \\ $$$${d}\:=\:\frac{{T}_{\mathrm{18}} −{T}_{\mathrm{7}} }{\mathrm{18}−\mathrm{7}}\:=\:\frac{\mathrm{38}{q}−\mathrm{19}{p}−\mathrm{3}{p}−\mathrm{5}{q}}{\mathrm{11}}=\frac{\mathrm{33}{q}−\mathrm{22}{p}}{\mathrm{11}}=\mathrm{3}{q}−\mathrm{2}{p} \\ $$$${T}_{\mathrm{1}} \:=\:{T}_{\mathrm{7}} +\left(\mathrm{1}−\mathrm{7}\right){d}\:=\:\mathrm{3}{p}+\mathrm{5}{q}−\mathrm{6}\left(\mathrm{3}{q}−\mathrm{2}{p}\right) \\ $$$${T}_{\mathrm{1}} =\:\mathrm{3}{p}+\mathrm{5}{q}+\mathrm{12}{p}−\mathrm{18}{q}\:=\:\mathrm{15}{p}−\mathrm{13}{q} \\ $$$${The}\:{first}\:{three}\:{terms}\:{of}\:{the}\:{sequence}\: \\ $$$${T}_{\mathrm{1}} =\:\mathrm{15}{p}−\mathrm{13}{q} \\ $$$${T}_{\mathrm{2}} =\:\mathrm{15}{p}−\mathrm{13}{q}+\mathrm{3}{q}−\mathrm{2}{p}=\mathrm{13}{p}−\mathrm{10}{q}\: \\ $$$${T}_{\mathrm{3}} =\mathrm{13}{p}−\mathrm{10}{q}+\mathrm{3}{q}−\mathrm{2}{p}=\mathrm{11}{p}−\mathrm{7}{q}\: \\ $$

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