Let a be the first term in the arithmetic sequence. Since it's arithmetic, consecutive terms in the sequence differ by a constant d, so the sequence is
a, a + d, a + 2d, a + 3d, …
with the n-th term, a + (n - 1)d.
The sum of the first n terms of this sequence is given:
[tex]a + (a+d) + (a+2d) + \cdots + (a+(n-1)d) = \dfrac{n(3n-5)}2[/tex]
We can simplify the left side as
[tex]\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[/tex]
so that
[tex]an+\dfrac{dn(n-1)}2 = \dfrac{n(3n-5)}2[/tex]
or
[tex]a+\dfrac{d(n-1)}2 = \dfrac{3n-5}2[/tex]
Let b be the first term in the geometric sequence. Consecutive terms in this sequence are scaled by a fixed factor r, so the sequence is
b, br, br ², br ³, …
with n-th term br 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
[tex]\begin{cases}a+d = br \\\\ a+3d = br^2\end{cases}[/tex]
and it follows that
[tex]\dfrac{br^2}{br} = r = \dfrac{a+3d}{a+d}[/tex]
From the earlier result, we then have
[tex]n=7 \implies a+\dfrac{d(7-1)}2 = a+3d = \dfrac{3\cdot7-5}2 = 8[/tex]
and
[tex]n=2 \implies a+\dfrac{d(2-1)}2 = a+d = \dfrac{3\cdot2-5}2 = \dfrac12[/tex]
so that
[tex]r = \dfrac8{\frac12} = 16[/tex]
and since the second arithmetic and geometric terms are both 1/2, this means that
[tex]br=16b=\dfrac12 \implies b = \dfrac1{32}[/tex]
The sum of the first 11 terms of the geometric sequence is
S = b + br + br ² + … + br ¹⁰
Multiply both sides by r :
rS = br + br ² + br ³ + … + br ¹¹
Subtract this from S, then solve for S :
S - rS = b - br ¹¹
(1 - r ) S = b (1 - r ¹¹)
S = b (1 - r ¹¹) / (1 - r )
Plug in b = 1/32 and r = 1/2 to get the sum :
[tex]S = \dfrac1{32}\cdot\dfrac{1-\dfrac1{2^{11}}}{1-\dfrac12} = \boxed{\dfrac{2047}{32768}}[/tex]
