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Question Number 175814 by Mortalis last updated on 07/Sep/22

Commented by Mortalis last updated on 07/Sep/22

Can anyone help me solve this question?

$${Can}\:{anyone}\:{help}\:{me}\:{solve}\:{this}\:{question}? \\ $$

Answered by Rasheed.Sindhi last updated on 07/Sep/22

S_n =(n/2)(a+l) ; a: first term, l=last term  S_p =(p/2)(a+l)..............(i)  S_p =((q−3)/2)(a+l) ; l=13.....(ii)  Comparing (i) & (ii):  p=q−3⇒q=p+3

$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left({a}+{l}\right)\:;\:{a}:\:{first}\:{term},\:{l}={last}\:{term} \\ $$$${S}_{{p}} =\frac{{p}}{\mathrm{2}}\left({a}+{l}\right)..............\left({i}\right) \\ $$$${S}_{{p}} =\frac{{q}−\mathrm{3}}{\mathrm{2}}\left({a}+{l}\right)\:;\:{l}=\mathrm{13}.....\left({ii}\right) \\ $$$${Comparing}\:\left({i}\right)\:\&\:\left({ii}\right): \\ $$$${p}={q}−\mathrm{3}\Rightarrow{q}={p}+\mathrm{3} \\ $$

Answered by Cesar1994 last updated on 07/Sep/22

a)  S_p =(p/2)(a+13)  ((q−3)/2)(a+13)=(p/2)(a+13)  ⇒q−3 = p  ⇒q=p+3

$$\left.{a}\right) \\ $$$${S}_{{p}} =\frac{{p}}{\mathrm{2}}\left({a}+\mathrm{13}\right) \\ $$$$\frac{{q}−\mathrm{3}}{\mathrm{2}}\left({a}+\mathrm{13}\right)=\frac{{p}}{\mathrm{2}}\left({a}+\mathrm{13}\right) \\ $$$$\Rightarrow{q}−\mathrm{3}\:=\:{p} \\ $$$$\Rightarrow{q}={p}+\mathrm{3} \\ $$

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