Sum of the First 10 Terms

**sharkman** · November 13th 2009, 01:04 PM

Find the sum of the first 10 terms if s_2 = 356 and a_8 = 83

**skeeter** · November 13th 2009, 01:51 PM

Originally Posted by sharkman

Find the sum of the first 10 terms if s_2 = 356 and a_8 = 83

it would help if we knew what kind of series this is ...

**e^(i*pi)** · November 13th 2009, 01:58 PM

Originally Posted by sharkman

Find the sum of the first 10 terms if s_2 = 356 and a_8 = 83

What kind of sequence is it? (Arithmetic has a common difference between terms and geometric has a common ratio between terms)

**sharkman** · November 15th 2009, 09:26 PM

Originally Posted by skeeter

it would help if we knew what kind of series this is ...

Originally Posted by e^(i*pi)

What kind of sequence is it? (Arithmetic has a common difference between terms and geometric has a common ratio between terms)

It's an arithmetic sequence.

Can you help me now?

**HallsofIvy** · November 16th 2009, 04:05 AM

Not a whole lot! My question now is why you have $s_2$ and $a_8$ ? You are giving us one term from each of two sequences!

Or did your finger just slip on the key board and it is $s_2$ and $s_8$ or $a_2$ and $a_8$ ?

Knowing that $a_2= 356$ and $a_8= 83$ tells us that a+ 2d= 356 and a+ 8d= 83. Subtracting the first equation from the second, 6d= 273 so d= 45.5. Now, a+2d= a+ 91= 356 so a= 356- 91= 265. You could, now calculate the first ten terms and add.

But arthmetic sequences have the very nice property that the "average" of an n consecutive terms is the same as the average of the first and last terms in the sequence. Here, the first term is 265 and the tenth term is [tex]a_{10}= 265- 10(45.5)= 265- 455= -190. The average of those two numbers is (265-190)/2= 75/2= 37.5 and that is the average of the first 10 numbers.

**sharkman** · November 16th 2009, 05:21 AM

Originally Posted by HallsofIvy

Not a whole lot! My question now is why you have $s_2$ and $a_8$ ? You are giving us one term from each of two sequences!

Or did your finger just slip on the key board and it is $s_2$ and $s_8$ or $a_2$ and $a_8$ ?

Knowing that $a_2= 356$ and $a_8= 83$ tells us that a+ 2d= 356 and a+ 8d= 83. Subtracting the first equation from the second, 6d= 273 so d= 45.5. Now, a+2d= a+ 91= 356 so a= 356- 91= 265. You could, now calculate the first ten terms and add.

Are you saying that I now must do the math and calculate the first 10 terms and then add? Can you please show me how this is done?

But arthmetic sequences have the very nice property that the "average" of an n consecutive terms is the same as the average of the first and last terms in the sequence. Here, the first term is 265 and the tenth term is [tex]a_{10}= 265- 10(45.5)= 265- 455= -190. The average of those two numbers is (265-190)/2= 75/2= 37.5 and that is the average of the first 10 numbers.

What do you mean by the symbol [tex]a{10}?

**HallsofIvy** · November 17th 2009, 05:18 AM

I meant $a_{10}$ - your a_10.

**sammy28** · November 20th 2009, 12:19 AM

I would use the fact that

$U_{n}=a+(n-1)d$

and

$S_{n}=\frac{n}{2}\left[2a+(n-1)d\right]$

where a is the first term and d is the common difference.

Ideally you should know the proofs behind these.
you can solve the two linear equations simultaneously for a and d

$356=2a+d$

$83=a+7d$

and hence solve for $S_{10}$

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