Menu Close

if-a-b-c-d-e-are-in-AP-find-the-value-of-a-4b-6c-4d-e-please-help-i-do-not-know-where-to-begin-




Question Number 99575 by MWSuSon last updated on 21/Jun/20
if a,b,c,d,e are in AP. find the value  of a−4b+6c−4d+e.  please help, i do not know where to  begin.
$$\mathrm{if}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{a}−\mathrm{4b}+\mathrm{6c}−\mathrm{4d}+\mathrm{e}. \\ $$$$\mathrm{please}\:\mathrm{help},\:\mathrm{i}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{where}\:\mathrm{to} \\ $$$$\mathrm{begin}. \\ $$
Answered by mathmax by abdo last updated on 21/Jun/20
take a try first ...
$$\mathrm{take}\:\mathrm{a}\:\mathrm{try}\:\mathrm{first}\:… \\ $$
Commented by MWSuSon last updated on 21/Jun/20
okay sir, although I've tried, but your comment gives me a little assurance that it looks easier than I think.
Commented by mathmax by abdo last updated on 21/Jun/20
perhaps ...
$$\mathrm{perhaps}\:… \\ $$
Commented by MWSuSon last updated on 22/Jun/20
a,b,c,d,e  b=a+k  c=a+2k  d=a+3k  e=a+4k  a−4b+6c−4d+e  (a+e)+6c−4(b+d)  =(2a+4k)+6a+12k−4(2a+4k)  =0.  Thank you sir, you were right, i did  not give it a try.
$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e} \\ $$$$\mathrm{b}=\mathrm{a}+\mathrm{k} \\ $$$$\mathrm{c}=\mathrm{a}+\mathrm{2k} \\ $$$$\mathrm{d}=\mathrm{a}+\mathrm{3k} \\ $$$$\mathrm{e}=\mathrm{a}+\mathrm{4k} \\ $$$$\mathrm{a}−\mathrm{4b}+\mathrm{6c}−\mathrm{4d}+\mathrm{e} \\ $$$$\left(\mathrm{a}+\mathrm{e}\right)+\mathrm{6c}−\mathrm{4}\left(\mathrm{b}+\mathrm{d}\right) \\ $$$$=\left(\mathrm{2a}+\mathrm{4k}\right)+\mathrm{6a}+\mathrm{12k}−\mathrm{4}\left(\mathrm{2a}+\mathrm{4k}\right) \\ $$$$=\mathrm{0}. \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{you}\:\mathrm{were}\:\mathrm{right},\:\mathrm{i}\:\mathrm{did} \\ $$$$\mathrm{not}\:\mathrm{give}\:\mathrm{it}\:\mathrm{a}\:\mathrm{try}. \\ $$
Answered by mathmax by abdo last updated on 22/Jun/20
b =a+r ,c =a+2r ,d =a+3r and e =a+4r ⇒  a−4b+6c−4d +e =a−4a−4r +6a+12r −4a−12r+ a+4r  =0
$$\mathrm{b}\:=\mathrm{a}+\mathrm{r}\:,\mathrm{c}\:=\mathrm{a}+\mathrm{2r}\:,\mathrm{d}\:=\mathrm{a}+\mathrm{3r}\:\mathrm{and}\:\mathrm{e}\:=\mathrm{a}+\mathrm{4r}\:\Rightarrow \\ $$$$\mathrm{a}−\mathrm{4b}+\mathrm{6c}−\mathrm{4d}\:+\mathrm{e}\:=\mathrm{a}−\mathrm{4a}−\mathrm{4r}\:+\mathrm{6a}+\mathrm{12r}\:−\mathrm{4a}−\mathrm{12r}+\:\mathrm{a}+\mathrm{4r} \\ $$$$=\mathrm{0}\:\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *