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??


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

$i f t h e m e a n o f t h e d a t a 1, 2, 3, 4, . . ., 100$ $i s 50.5 t h e n w h a t i s t h e m e a n o f t h e$ $d a t a 5, 7, 9, 11, . . ., 203 ? ?$
Commented by Rasheed.Sindhi last updated on 31/Jul/22

$\frac{5 + 203}{2} = \frac{208}{2} = 104$
Commented by Best1 last updated on 31/Jul/22

$e x p l a i n p l e a s e$
Commented by Rasheed.Sindhi last updated on 31/Jul/22

$I f a i s a f i r s t t e r m a n d l i s l a s t t e r m$ $o f a n A P, t h e m e a n (m) i s :$ $m = \frac{a + l}{2}$ $m = \frac{5 + 203}{2} = \frac{208}{2} = 104$
Commented by cortano1 last updated on 31/Jul/22
$mean = (S_n /n) = ((((100)/2)(1+100))/(100)) = (1/2)×101=50.5 then mean of data : 5,7,9,...,203 { ((203=5+(n−1)2⇒n=100)),((mean=((((100)/2)(5+203))/(100))= 104)) :}$
$m e a n = \frac{S_{n}}{n} = \frac{\frac{100}{2} (1 + 100)}{100} = \frac{1}{2} \times 101 = 50.5$ $t h e n m e a n o f d a t a : 5, 7, 9, . . ., 203$ ${\begin{cases} 203 = 5 + (n - 1) 2 \Rightarrow n = 100 \\ m e a n = \frac{\frac{100}{2} (5 + 203)}{100} = 104 \end{cases}$
Commented by Best1 last updated on 31/Jul/22

$t h a n k s u v e r y m u c h$
Commented by peter frank last updated on 01/Aug/22

$thank$
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