Sum of First 20 Terms of Arithmetic Series: 131+235

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The discussion revolves around solving for the sum of the first 20 terms of an arithmetic series, given that the 20th term is 131 and the sum of the 6th to 10th terms is 235. Participants highlight the need for correct equations, noting that the sum of the 6th to 10th terms should be calculated as S10 - S5, rather than using incorrect formulations. There are also queries about finding the 50th term of another series based on given sums. The importance of understanding the mean of terms in arithmetic sequences is emphasized, as it can simplify finding specific term values. Ultimately, correct identification of the first term and the common difference is crucial for determining the sum of the first 20 terms.
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Lol it's quite easy but these questions r annoying me.
The 20th term of an arithmetic series if 131 and sum of the 6th to 10th term inclusive is 235, find sum is the first 20 terms
well
131= a + 19 d 5(2a+ 9d) -3(2a+5d)=235
well i just went on and solve simutalneosly and got the wrong answer for S(20), i don't know how to make it at hte butotm.
2) the sum of 50 terms of an arithmetic is 249 and sum of 49 terms is 233, find 50th term of the series.
249= 50a + 1225d and 233= 49/2 (2a+48d)
and well got the wrong answer aagain.
3) prove T_n = S_n - S_n_-_1
I have no idea how to do that. Thanks
 
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UnD said:
...
well
131= a + 19 d
5(2a+ 9d) -3(2a+5d)=235
well i just went on and solve simutalneosly and got the wrong answer for S(20), ...
I have bolded the wrong equation in your post. In fact the sum of the sixth term to the tenth term is:
S6 - 10 = S10 - S5 = 5(2a + 9d) - \frac{5}{2}(2a + 4d) = 235
not
S6 - 10 = S10 - S6
UnD said:
2) the sum of 50 terms of an arithmetic is 249 and sum of 49 terms is 233, find 50th term of the series.
249= 50a + 1225d
233= 49/2 (2a+48d)
You are complicating the problem...
S50 = a1 + a2 + a3 + ... + a49 + a50
S49 = a1 + a2 + a3 + ... + a49
So what's a50?

UnD said:
3) prove T_n = S_n - S_n_-_1
I have no idea how to do that. Thanks
I have no idea what Tn is... Is that the n-th term of the series?
Viet Dao,
 
your second equation is wrong. it should read 5a + 35d = 235; after that you can solve the system formed withthis equation et your first equation to find a and d
After that it should be easy to find the sum of the first 20 terms.
 
UnD said:
The 20th term of an arithmetic series if 131 and sum of the 6th to 10th term inclusive is 235, find sum is the first 20 terms
Concept of mean is often helpful in questions on arithmetic sequences.
Mean of n first terms of such sequence is
amean = Sn/n
If n is odd, amean is one of the terms of the sequence.
Which one?
Knowing that
a6 + a7 + a8 + a9 + a10 = 235
you can determine the value of a?.
Now you can find a1 from a? and a20 in one step.
a1 and a20 lead you straight to S20.
 

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