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Question Number 174376 by Best1 last updated on 31/Jul/22

if the mean of the data 1,2,3,4,..., 100 is 50.5 then what is the mean of the data 5,7,9,11,...,203??

$${if}\:{the}\:{mean}\:{of}\:{the}\:{data}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},...,\:\mathrm{100} \\ $$$$\:{is}\:\mathrm{50}.\mathrm{5}\:{then}\:{what}\:{is}\:{the}\:{mean}\:{of}\:{the}\: \\ $$$${data}\:\mathrm{5},\mathrm{7},\mathrm{9},\mathrm{11},...,\mathrm{203}?? \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 31/Jul/22

((5+203)/2)=((208)/2)=104

$$\frac{\mathrm{5}+\mathrm{203}}{\mathrm{2}}=\frac{\mathrm{208}}{\mathrm{2}}=\mathrm{104} \\ $$

Commented by Best1 last updated on 31/Jul/22

explain please

$${explain}\:{please} \\ $$

Commented by Rasheed.Sindhi last updated on 31/Jul/22

If a is a first term and l is last term of an AP, the mean(m) is : m=((a+l)/2) m=((5+203)/2)=((208)/2)=104

$${If}\:\boldsymbol{{a}}\:{is}\:{a}\:{first}\:{term}\:{and}\:\boldsymbol{{l}}\:{is}\:{last}\:{term} \\ $$$${of}\:{an}\:{AP},\:{the}\:{mean}\left(\boldsymbol{{m}}\right)\:{is}\:: \\ $$$$\:\:\:\:\:\:\boldsymbol{{m}}=\frac{\boldsymbol{{a}}+\boldsymbol{{l}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\boldsymbol{{m}}=\frac{\mathrm{5}+\mathrm{203}}{\mathrm{2}}=\frac{\mathrm{208}}{\mathrm{2}}=\mathrm{104} \\ $$$$ \\ $$

Commented by cortano1 last updated on 31/Jul/22

mean = (S_n /n) = ((((100)/2)(1+100))/(100)) = (1/2)×101=50.5 then mean of data : 5,7,9,...,203 { ((203=5+(n−1)2⇒n=100)),((mean=((((100)/2)(5+203))/(100))= 104)) :}

$${mean}\:=\:\frac{{S}_{{n}} }{{n}}\:=\:\frac{\frac{\mathrm{100}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{100}\right)}{\mathrm{100}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{101}=\mathrm{50}.\mathrm{5} \\ $$$${then}\:{mean}\:{of}\:{data}\::\:\mathrm{5},\mathrm{7},\mathrm{9},...,\mathrm{203} \\ $$$$\:\begin{cases}{\mathrm{203}=\mathrm{5}+\left({n}−\mathrm{1}\right)\mathrm{2}\Rightarrow{n}=\mathrm{100}}\\{{mean}=\frac{\frac{\mathrm{100}}{\mathrm{2}}\left(\mathrm{5}+\mathrm{203}\right)}{\mathrm{100}}=\:\mathrm{104}}\end{cases} \\ $$

Commented by Best1 last updated on 31/Jul/22

thanks u very much

$${thanks}\:{u}\:\:{very}\:{much} \\ $$

Commented by peter frank last updated on 01/Aug/22

thank

$$\mathrm{thank} \\ $$

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