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a-n-numbers-series-If-S-16-S-13-S-106-S-103-Find-3a-3-4a-4-5a-5-2a-12-




Question Number 208218 by hardmath last updated on 07/Jun/24
a_n   numbers series  If  S_(16)  − S_(13)   =  S_(106)  − S_(103)   Find:    ((3a_3  + 4a_4  + 5a_5 )/(2a_(12) ))  =  ?
$$\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:\:\mathrm{numbers}\:\mathrm{series} \\ $$$$\mathrm{If}\:\:\mathrm{S}_{\mathrm{16}} \:−\:\mathrm{S}_{\mathrm{13}} \:\:=\:\:\mathrm{S}_{\mathrm{106}} \:−\:\mathrm{S}_{\mathrm{103}} \\ $$$$\mathrm{Find}:\:\:\:\:\frac{\mathrm{3a}_{\mathrm{3}} \:+\:\mathrm{4a}_{\mathrm{4}} \:+\:\mathrm{5a}_{\mathrm{5}} }{\mathrm{2a}_{\mathrm{12}} }\:\:=\:\:? \\ $$
Commented by mr W last updated on 07/Jun/24
this is not the first time when i ask  you what you mean with “numbers  series”, e.g. in Q207930.
$${this}\:{is}\:{not}\:{the}\:{first}\:{time}\:{when}\:{i}\:{ask} \\ $$$${you}\:{what}\:{you}\:{mean}\:{with}\:“{numbers} \\ $$$${series}'',\:{e}.{g}.\:{in}\:{Q}\mathrm{207930}. \\ $$
Commented by hardmath last updated on 07/Jun/24
  For example, there is a numerical series and a geometric series
$$ \\ $$For example, there is a numerical series and a geometric series
Commented by mr W last updated on 08/Jun/24
then you mean arithmetic series.
$${then}\:{you}\:{mean}\:{arithmetic}\:{series}. \\ $$
Commented by hardmath last updated on 08/Jun/24
yes dear professor
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by zamin2001 last updated on 07/Jun/24
S_(16) −S_(13) =a_(16) +a_(15) +a_(14) =3a_1 +42d ★  S_(106) −S_(103) =a_(106) +a_(105) +a_(104) =3a_1 +312d■   ★ =■ ⇒ d=0 ⇒ constant numerical series    ((3a_3  + 4a_4  + 5a_5 )/(2a_(12) ))  =  ((3C+4C+5C)/(2C))=((12C)/(2C))=6
$${S}_{\mathrm{16}} −{S}_{\mathrm{13}} ={a}_{\mathrm{16}} +{a}_{\mathrm{15}} +{a}_{\mathrm{14}} =\mathrm{3}{a}_{\mathrm{1}} +\mathrm{42}{d}\:\bigstar \\ $$$${S}_{\mathrm{106}} −{S}_{\mathrm{103}} ={a}_{\mathrm{106}} +{a}_{\mathrm{105}} +{a}_{\mathrm{104}} =\mathrm{3}{a}_{\mathrm{1}} +\mathrm{312}{d}\blacksquare \\ $$$$\:\bigstar\:=\blacksquare\:\Rightarrow\:{d}=\mathrm{0}\:\Rightarrow\:{constant}\:{numerical}\:{series} \\ $$$$\:\:\frac{\mathrm{3a}_{\mathrm{3}} \:+\:\mathrm{4a}_{\mathrm{4}} \:+\:\mathrm{5a}_{\mathrm{5}} }{\mathrm{2a}_{\mathrm{12}} }\:\:=\:\:\frac{\mathrm{3}{C}+\mathrm{4}{C}+\mathrm{5}{C}}{\mathrm{2}{C}}=\frac{\mathrm{12}{C}}{\mathrm{2}{C}}=\mathrm{6} \\ $$

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