Let <em>a</em> be the first term in the arithmetic sequence. Since it's arithmetic, consecutive terms in the sequence differ by a constant <em>d</em>, so the sequence is
<em>a</em>, <em>a</em> + <em>d</em>, <em>a</em> + 2<em>d</em>, <em>a</em> + 3<em>d</em>, …
with the <em>n</em>-th term, <em>a</em> + (<em>n</em> - 1)<em>d</em>.
The sum of the first <em>n</em> terms of this sequence is given:
![a + (a+d) + (a+2d) + \cdots + (a+(n-1)d) = \dfrac{n(3n-5)}2](https://tex.z-dn.net/?f=a%20%2B%20%28a%2Bd%29%20%2B%20%28a%2B2d%29%20%2B%20%5Ccdots%20%2B%20%28a%2B%28n-1%29d%29%20%3D%20%5Cdfrac%7Bn%283n-5%29%7D2)
We can simplify the left side as
![\displaystyle \sum_{i=1}^n (a+(i-1)d) = (a-d)\sum_{i=1}^n1 + d\sum_{i=1}^ni = an+\dfrac{dn(n-1)}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bi%3D1%7D%5En%20%28a%2B%28i-1%29d%29%20%3D%20%28a-d%29%5Csum_%7Bi%3D1%7D%5En1%20%2B%20d%5Csum_%7Bi%3D1%7D%5Eni%20%3D%20an%2B%5Cdfrac%7Bdn%28n-1%29%7D2)
so that
![an+\dfrac{dn(n-1)}2 = \dfrac{n(3n-5)}2](https://tex.z-dn.net/?f=an%2B%5Cdfrac%7Bdn%28n-1%29%7D2%20%3D%20%5Cdfrac%7Bn%283n-5%29%7D2)
or
![a+\dfrac{d(n-1)}2 = \dfrac{3n-5}2](https://tex.z-dn.net/?f=a%2B%5Cdfrac%7Bd%28n-1%29%7D2%20%3D%20%5Cdfrac%7B3n-5%7D2)
Let <em>b</em> be the first term in the geometric sequence. Consecutive terms in this sequence are scaled by a fixed factor <em>r</em>, so the sequence is
<em>b</em>, <em>br</em>, <em>br</em> ², <em>br</em> ³, …
with <em>n</em>-th term <em>br</em> ⁿ⁻¹.
The second arithmetic term is equal to the second geometric term, and the fourth arithmetic term is equal to the third geometric term, so
![\begin{cases}a+d = br \\\\ a+3d = br^2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da%2Bd%20%3D%20br%20%5C%5C%5C%5C%20a%2B3d%20%3D%20br%5E2%5Cend%7Bcases%7D)
and it follows that
![\dfrac{br^2}{br} = r = \dfrac{a+3d}{a+d}](https://tex.z-dn.net/?f=%5Cdfrac%7Bbr%5E2%7D%7Bbr%7D%20%3D%20r%20%3D%20%5Cdfrac%7Ba%2B3d%7D%7Ba%2Bd%7D)
From the earlier result, we then have
![n=7 \implies a+\dfrac{d(7-1)}2 = a+3d = \dfrac{3\cdot7-5}2 = 8](https://tex.z-dn.net/?f=n%3D7%20%5Cimplies%20a%2B%5Cdfrac%7Bd%287-1%29%7D2%20%3D%20a%2B3d%20%3D%20%5Cdfrac%7B3%5Ccdot7-5%7D2%20%3D%208)
and
![n=2 \implies a+\dfrac{d(2-1)}2 = a+d = \dfrac{3\cdot2-5}2 = \dfrac12](https://tex.z-dn.net/?f=n%3D2%20%5Cimplies%20a%2B%5Cdfrac%7Bd%282-1%29%7D2%20%3D%20a%2Bd%20%3D%20%5Cdfrac%7B3%5Ccdot2-5%7D2%20%3D%20%5Cdfrac12)
so that
![r = \dfrac8{\frac12} = 16](https://tex.z-dn.net/?f=r%20%3D%20%5Cdfrac8%7B%5Cfrac12%7D%20%3D%2016)
and since the second arithmetic and geometric terms are both 1/2, this means that
![br=16b=\dfrac12 \implies b = \dfrac1{32}](https://tex.z-dn.net/?f=br%3D16b%3D%5Cdfrac12%20%5Cimplies%20b%20%3D%20%5Cdfrac1%7B32%7D)
The sum of the first 11 terms of the geometric sequence is
<em>S</em> = <em>b</em> + <em>br</em> + <em>br</em> ² + … + <em>br</em> ¹⁰
Multiply both sides by <em>r</em> :
<em>rS</em> = <em>br</em> + <em>br</em> ² + <em>br</em> ³ + … + <em>br</em> ¹¹
Subtract this from <em>S</em>, then solve for <em>S</em> :
<em>S</em> - <em>rS</em> = <em>b</em> - <em>br</em> ¹¹
(1 - <em>r</em> ) <em>S</em> = <em>b</em> (1 - <em>r</em> ¹¹)
<em>S</em> = <em>b</em> (1 - <em>r</em> ¹¹) / (1 - <em>r</em> )
Plug in <em>b</em> = 1/32 and <em>r</em> = 1/2 to get the sum :
![S = \dfrac1{32}\cdot\dfrac{1-\dfrac1{2^{11}}}{1-\dfrac12} = \boxed{\dfrac{2047}{32768}}](https://tex.z-dn.net/?f=S%20%3D%20%5Cdfrac1%7B32%7D%5Ccdot%5Cdfrac%7B1-%5Cdfrac1%7B2%5E%7B11%7D%7D%7D%7B1-%5Cdfrac12%7D%20%3D%20%5Cboxed%7B%5Cdfrac%7B2047%7D%7B32768%7D%7D)