# What is T = (2007a1 + a2006)/(a2503) if S26 = 4S13? Sn is sum of n terms of arithmetic progresion?

Asked on by greenbel

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should write the summation of terms, `S_26` and `S_13` , using the first term `a_1` and the common difference, `d` , such that:

`S_13 = ((a_1 + a_13)*13)/2`

Using the general term identity, yields:

`a_13 = a_1 + 12d`

Replacing `a_1 + 12d` for `a_13` yields:

`S_13 = ((a_1 + a_1 + 12d)*13)/2 => S_13 = (2a_1 + 12d)*13/2`

Factoring out 2 yields:

`S_13 = (2(a_1 + 6d)*13)/2`

Reducing duplicate terms yields:

`S_13 = (a_1 + 6d)*13`

`S_26 = ((a_1 + a_26)*26)/2 => S_26 = (a_1 + a_26)*13`

Replacing `(a_1 + a_26)*13` for `S_26` and` (a_1 + 6d)*13` for `S_13` , yields:

`S_26 = 4S_13 => (a_1 + a_26)*13 = 4*(a_1 + 6d)*13`

Reducing duplicate terms yields:

`a_1 + a_26 = 4*(a_1 + 6d) => a_1 + a_1 + 25d = 4a_1 + 24d`

`2a_1 + 25d = 4a_1 + 24d => 25d - 24d = 4a_1 - 2a_1`

`d = 2a_1`

You need to evaluate T such that:

`T = (2007a_1 + a_1 + 2005d)/(a_1 + 2502d)`

Replacing `2a_1` for` d` yields:

`T = (2007a_1 + a_1 + 2005*2a_1)/(a_1 + 2502*2a_1)`

Factoring out `a_1` yields:

`T = (a_1(2008 + 2*2005))/(a_1(1 + 2502*2))`

Reducing duplicate terms yields:

`T = 6018/5005`

Hence, evaluating T, using the information provided by the problem, yields `T = 6018/5005.`

We’ve answered 319,175 questions. We can answer yours, too.