anonymous
  • anonymous
Find S25 for 1/2 +1+ 3/2 + 2+
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is that S sum or 25 term?
mathstudent55
  • mathstudent55
Is this an arithmetic or a geometric series?
anonymous
  • anonymous
just like the last one we did. its S with 25 in the lower right coner and @mathstudent55 its arithmetric

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anonymous
  • anonymous
then apply a+24d
mathstudent55
  • mathstudent55
Great. The formula for the sum of an arithmetic series is: \(\Large S_n = \dfrac{n(a_1 + a_n)}{2} \) We have the first term. We need the 25th term.
anonymous
  • anonymous
is @GIL.ojei way correct too?
anonymous
  • anonymous
yap correct for sum. that was why i asked
mathstudent55
  • mathstudent55
Yes, he's correct to find the 25th term. \(\Large a_n = a_1 + (n - 1)d\)
anonymous
  • anonymous
okay. so I can use his way then?
mathstudent55
  • mathstudent55
First use the formula just above for finding the 25th term. Then use the formula for \(S_n\) to find the sum of the first 25 terms.
anonymous
  • anonymous
follow his way. mine is term and his is sum
anonymous
  • anonymous
i dont think i did the first one right bc i got 318.75 @mathstudent55 @GIL.ojei helpp? lol
mathstudent55
  • mathstudent55
You are looking for the sum of the first 25 terms of an arithmetic series. This is the formula you need to use. The general formula is below. \(\Large S_n = \dfrac{n(a_1 + a_n)}{2}\) We replace n with 25 to give you the exact formula you need in your case, to sum the first 25 terms. \(\Large S_{25} = \dfrac{n(a_1 + a_{25})}{2}\) Now you see that you need the first term, which you have, and the 25th term. For the 25th term, we use the formula below. This is the general formula you sue to find the nth term of a series. Since we need the 25th term, we replace n with 25. \(\Large a_{25} = a_1 + (25 - 1)d\) Form the first two given terms, 1/2 and 1, we find d, the common difference: d = 1 - 1/2 = 1/2 = 0.5 Now we can find the 25th term: \(\Large a_{25} = 0.5 + (25 - 1)0.5\) \(\Large a_{25} = 12.5\) Now we can go back to the sum formula using the first and 25th terms: \(\Large S_{25} = \dfrac{25(0.5 + 12.5)}{2}\) \(\Large S_{25} = 162.5\)
jim_thompson5910
  • jim_thompson5910
Look at the formulas mathstudent55 wrote out, you will plug \(\Large {\color{red}{a_n=a_1+d(n-1)}}\) into the \(\Large S_n\) formula to get... \[\Large S_n = \dfrac{n(a_1 + a_n)}{2}\] \[\Large S_n = \dfrac{n(a_1 + {\color{red}{a_n}})}{2}\] \[\Large S_n = \dfrac{n(a_1 + {\color{red}{a_1 + d(n-1)}})}{2}\] \[\Large S_n = \dfrac{n(2a_1 + d(n-1))}{2}\]
anonymous
  • anonymous
i get it. thank you! sorry for the late responce, i was getting food.

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