Question

The sum of the first n terms of an arithmetic series is n/2(3n-5). If the second and fourth terms of the arithmetic series are the second and the third terms of a geometric series respectively, find the sum of the first eleven terms of this geometric series.​

Answer 1

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:

$a + (a+d) + (a+2d) + \cdots + (a+(n-1)d) = \dfrac{n(3n-5)}2$

We can simplify the left side as

$\displaystyle \sum_{i=1}^n (a+(i-1)d) = (a-d)\sum_{i=1}^n1 + d\sum_{i=1}^ni = an+\dfrac{dn(n-1)}2$

so that

$an+\dfrac{dn(n-1)}2 = \dfrac{n(3n-5)}2$

or

$a+\dfrac{d(n-1)}2 = \dfrac{3n-5}2$

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

$\begin{cases}a+d = br \\\\ a+3d = br^2\end{cases}$

and it follows that

$\dfrac{br^2}{br} = r = \dfrac{a+3d}{a+d}$

From the earlier result, we then have

$n=7 \implies a+\dfrac{d(7-1)}2 = a+3d = \dfrac{3\cdot7-5}2 = 8$

and

$n=2 \implies a+\dfrac{d(2-1)}2 = a+d = \dfrac{3\cdot2-5}2 = \dfrac12$

so that

$r = \dfrac8{\frac12} = 16$

and since the second arithmetic and geometric terms are both 1/2, this means that

$br=16b=\dfrac12 \implies b = \dfrac1{32}$

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 :

$S = \dfrac1{32}\cdot\dfrac{1-\dfrac1{2^{11}}}{1-\dfrac12} = \boxed{\dfrac{2047}{32768}}$