anonymous
  • anonymous
The fifth term of an arithmetic progression is 28 and the tenth term in 58. The sum of all the terms in this progression is 444. How many terms are there? In the previous question the answer shows that the first term is 4 and the common difference is 6. I also do have the answers available if needed. Please help me! I don't know what to do to answer this. If you could just set me on the right track I could try to solve it myself.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Nnesha
  • Nnesha
to find common difference \[\huge\rm \frac{ a_{10} -a_5 }{ 10-5 }\] difference between both terms and the number of terms
Nnesha
  • Nnesha
a_10 = 58 and a_5 =28 so subtract \[\huge\rm \frac{ 58-28 }{ 10-5 }\]
anonymous
  • anonymous
Ok, so you get the common difference of 6.

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Nnesha
  • Nnesha
yes right now we to find number of the terms use the sum formula for arithmetic\[S_n =\frac{ n(a_1 +a_n) }{ 2 }\]
Nnesha
  • Nnesha
and i just noticed we don't have the first term hm
anonymous
  • anonymous
yes we do- its 4
Nnesha
  • Nnesha
we can use the formula to find term \[\huge\rm a_n =(r)^{n-k}\]
Nnesha
  • Nnesha
opps sorry i was looking at the wrong page thats for geometric sequence
Nnesha
  • Nnesha
alright so to find 1st term \[\huge\rm a_5 = a_1 +(n-1)d\] d= 6 and we are using 5th term to find first one so n would be 5 \[\huge\rm 28=a_1+(5-1)6\] solve for a_1
Nnesha
  • Nnesha
btw you can use 10th term doesn't matter u will get the same answer :=)
Nnesha
  • Nnesha
oh!! you already found it yes it's 4
Nnesha
  • Nnesha
can i see the answer for last part little confused on that part
anonymous
  • anonymous
The last part? You mean the first question that I didn't include?
Nnesha
  • Nnesha
`How many terms are there? ` i assumed previous question was the same like this one bec we got same value for d and a_1
anonymous
  • anonymous
The entire question is: 7. The fifth term of an arithmetic progression is 28 and the tenth term is 58 (i) Find the first term and the common difference. (ii) The sum of all the terms in this progression is 444. How many terms are there? Answers to 7. (i) common difference=6 first term=4
anonymous
  • anonymous
So I got help yesterday for the first one and now I m stuck on how to find the answers for (ii)
Nnesha
  • Nnesha
ohh i see..
Nnesha
  • Nnesha
alright so we should use \[\huge\rm s_n =\frac{ n(a_1-a_n )}{ 2 }\] a_1 is 4 replace a_n with the a_1 equation which is a_1 = 4+(n-1)6 then solve for n s_n=144
Nnesha
  • Nnesha
444**
anonymous
  • anonymous
Ok I will try that quickly
Nnesha
  • Nnesha
try to work on it ive to go to eat something BRB
anonymous
  • anonymous
sure
anonymous
  • anonymous
Ok I also actually have to be away for a little while. Will come back to this as soon as I can.
Nnesha
  • Nnesha
alright tag me when u r ready :=)
Nnesha
  • Nnesha
ie found this http://math.stackexchange.com/questions/709276/arithmetic-sequence-find-term-given-sum-of-terms-a1-and-d you should get quadratic equation hm
anonymous
  • anonymous
I am back now @Nnesha , sorry for taking up so much of your time!
anonymous
  • anonymous
Anyway the quadtratic I ended up with was \[-7n^{2}+2n +888\] which looks a bit odd. Did I go wrong somewhere?
anonymous
  • anonymous
Trying to solve it now
Nnesha
  • Nnesha
888 ?O_*
anonymous
  • anonymous
Yeah that didn't work
anonymous
  • anonymous
I can send a picture of my working out
Nnesha
  • Nnesha
ohh it supposed to be a_1 `+` a_n
Nnesha
  • Nnesha
sorry about taht ..
anonymous
  • anonymous
What do you mean?
Nnesha
  • Nnesha
alright so we should use \[\huge\rm444 =\frac{ n(4+(4+(n-1)6 )}{ 2 }\] a_1 is 4 replace a_n with the a_1 equation which is a_1 = 4+(n-1)6
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha alright so we should use \[\huge\rm s_n =\frac{ n(a_1-a_n )}{ 2 }\] a_1 is 4 replace a_n with the a_1 equation which is a_1 = 4+(n-1)6 then solve for n s_n=144 \(\color{blue}{\text{End of Quote}}\) there supposed to be plus sign
anonymous
  • anonymous
Oooooh ok let me try that quickly
Nnesha
  • Nnesha
i'll try it let's see what we get
anonymous
  • anonymous
Now the quadratic I get is \[n ^{2} + \frac{ 1 }{ 3 } -148=0\]
Nnesha
  • Nnesha
1/3 ? how did you get that ? o.O
anonymous
  • anonymous
Oh dear. Umm I had -6n^2 -2n +888 and I divided everything by -6
Nnesha
  • Nnesha
ohh so 1/3n*
anonymous
  • anonymous
Oh yes, sorry
Nnesha
  • Nnesha
\[n ^{2} + \frac{ 1 }{ 3 }n -148=0\] can you solve for n ?
anonymous
  • anonymous
Yup doing that now-almost done
Nnesha
  • Nnesha
alright let me know what you get
anonymous
  • anonymous
Ok done- the positive value I got was 12.00333333. Do I round that down to 12?
Nnesha
  • Nnesha
looks right \[ 444 =\frac{ n(8+6n-6) }{ 2 } ~~~~~= \frac{ n(6n+2) }{ 2 }\] i did it differently but got same answer so i guess 12 is right :=)
Nnesha
  • Nnesha
good job! :=)
anonymous
  • anonymous
Thank you! Again I am sorry for taking up so much of your time. I really appreciate your help. Hope you have a nice day :)
Nnesha
  • Nnesha
np & you too!

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