I don't know what mistake I'm making
Question : Let x1, x2 …., x100 be in an arithmetic progression, with x1 = 2 and their mean equal to 200. If yi = i(xi - i),1 ≤ i ≤ 100, then the mean of y1, y2, ….., y100 is
#Arithmetic Progression
opal granite
tame doveBOT
I don't know what mistake I'm making
Question : Let x1, x2 …., x100 be in an a...
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oak hull
@opal granite
do you want me to solve the question for you?
ill just tell that you that
this will give you more clarification on what your doing wrong
opal granite
yeah we can write xi= 2 + (i-1)4
from that
opal granite
do you want me to solve the question for you?
No it's alright
I was confused about the algebra
opal granite
Can I just plug in 20,000
so I can't plug 20000 in the place of xi because the summation sign doesn't distribute
frosty willow
@opal granite
Mean = sum of all terms / total number of terms
200 = ap sum/100
20000 = ap sum.
if you use the ap sum formula = n/2 [2a + nd - d]
You find the common difference d.
Then you find the general term of y series by putting the general term xi in it.
You should get 3i^2 - 2i.
Then use the formula summation n^2 = n(n+1)(2n+1)/6 and n = n(n+1)/2.
You would get 10049.5
opal granite
Got it