anonymous
  • anonymous
please help The 11th term of an arithmetic progression is 1. the sum of the first 10 terms is 120. find the 4th term :S formula for the... sum of terms=1/2n(a+l)=1/2n{2a+(n-1)d} thanks
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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yuki
  • yuki
okay
yuki
  • yuki
\[{11(a_1 + a_{11}) \over 2} = 120 + 1 \]
yuki
  • yuki
so\[{a_1 + 1} = 22\] thus a_1 = 21

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yuki
  • yuki
Then you can say\[a_{11} = a_1 + 10d\] \[a_11 = 1 = 21 + 10d\]
yuki
  • yuki
so d = -2
yuki
  • yuki
that means that the 4th term is 21+3(-2) = 15
yuki
  • yuki
:)
yuki
  • yuki
did it make sense ?
anonymous
  • anonymous
but you didnt even use 4 once :S??? and i dont really understand where 21+3(-2)=15 comes from :(
yuki
  • yuki
ok, \[a_n = a_1 + d(n-1)\] so\[a_4 = a_1 + (-2)(4-1) = 21 + 3(-2)\]
yuki
  • yuki
the idea of arithmetic sequence is that you are "Adding the same number for all sequence"
anonymous
  • anonymous
thanks :)
yuki
  • yuki
np :)
anonymous
  • anonymous
sorry but i first thought that i knew where 11(a1+a11)2=120+1 came from but i dont :S
yuki
  • yuki
Hope you read this.
yuki
  • yuki
the first ten numbers add up to 120, so 10(a1+a10) /2= 120 now if you add a11 to it, it is 120 + 1 but adding a11 to the sum from a1 to a10 is as same as adding from a1 to a11 so 11(a1 + a11)/2 = 121

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