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

In summary, the conversation discusses solving problems related to arithmetic sequences. The first problem involves finding the sum of the first 20 terms given the 20th term and the sum of the 6th to 10th term. The second problem involves finding the 50th term of a sequence given the sum of 50 terms and the sum of 49 terms. The third problem involves proving a formula for a term in the sequence. Techniques such as finding the mean of the first n terms and using simultaneous equations are utilized in solving these problems.
  • #1
UnD
21
0
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 [itex]T_n = S_n - S_n_-_1[/itex]
I have no idea how to do that. Thanks
 
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  • #2
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 = [tex]5(2a + 9d) - \frac{5}{2}(2a + 4d) = 235[/tex]
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 [itex]T_n = S_n - S_n_-_1[/itex]
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,
 
  • #3
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.
 
  • #4
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.
 

What is an arithmetic series?

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. For example, the series 2, 5, 8, 11, 14 is an arithmetic series with a difference of 3 between each term.

What is the formula for finding the sum of the first 20 terms of an arithmetic series?

The formula for finding the sum of the first n terms of an arithmetic series is Sn = (n/2)(a1 + an), where n is the number of terms, a1 is the first term, and an is the last term. For this specific series, the formula would be S20 = (20/2)(131 + 235) = 3660.

How many terms are in the given arithmetic series?

There are 20 terms in the given series. This can be determined by counting the number of terms or by using the formula n = (an - a1)/d + 1, where n is the number of terms, an is the last term, a1 is the first term, and d is the difference between terms.

What is the sum of the first 10 terms of this arithmetic series?

The sum of the first 10 terms of this series would be S10 = (10/2)(131 + 175) = 1530.

How can the sum of the first 20 terms of an arithmetic series be used in real life?

The sum of the first 20 terms of an arithmetic series can be used in financial calculations, such as finding the total cost of a loan or investment with a fixed interest rate. It can also be used in physics and engineering to calculate the total distance or displacement of an object with a constant velocity.

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