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) = \frac{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+\frac{dn(n-1)}2`

so that

`an+\frac{dn(n-1)}2 = \frac{n(3n-5)}2`

or

`a+\frac{d(n-1)}2 = \frac{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

`(br^2)/(br) = r = (a+3d)/(a+d)`

From the earlier result, we then have

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

and

`n=2 \implies a+\frac{d(2-1)}2 = a+d = \frac{3\cdot2-5}2 = \frac12`

so that

`r = \frac8{\frac12} = 16`

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

`br=16b=\frac12 \implies b = \frac1{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 = \frac1{32}\cdot\frac{1-\frac1{2^(11)}}{1-\frac12} = \boxed{(2047)/(32768)}`