Understanding Geometric Progression Properties

We are given a Geometric Progression (GP) with the following conditions:

• The sum of the first 13 terms ($S_{13}$) is equal to the sum of the first 11 terms ($S_{11}$).
• The sum of the first 15 terms ($S_{15}$) is 1200.

We need to find the 21^st term ($T_{21}$).

Deriving the Common Ratio (r)

The sum of the first $n$ terms of a GP is given by $S_n = a \frac{r^n - 1}{r - 1}$, where $a$ is the first term and $r$ is the common ratio. However, a simpler relationship exists between sums:

$S_n = S_{n-1} + T_n$. Therefore, $S_{13} = S_{11} + T_{12} + T_{13}$.

Given $S_{13} = S_{11}$, we have:

$S_{11} = S_{11} + T_{12} + T_{13}$

$T_{12} + T_{13} = 0$

In terms of $a$ and $r$: $a r^{11} + a r^{12} = 0$. Factoring out $a r^{11}$ (assuming $a \neq 0$ and $r \neq 0$ for a valid GP):

$a r^{11} (1 + r) = 0$

This implies $1 + r = 0$, so the common ratio $r = -1$.

Calculating the First Term (a)

Now we use the formula for the sum of a GP with $r = -1$. The formula simplifies when $r=-1$:

If $n$ is odd, $S_n = a \frac{(-1)^n - 1}{-1 - 1} = a \frac{-1 - 1}{-2} = a$.

If $n$ is even, $S_n = a \frac{(-1)^n - 1}{-1 - 1} = a \frac{1 - 1}{-2} = 0$.

We are given $S_{15} = 1200$. Since 15 is odd:

$S_{15} = a = 1200$.

So, the first term $a = 1200$.

Finding the 21st Term

The formula for the $n^{th}$ term of a GP is $T_n = a r^{n-1}$.

We need to find the 21^st term ($T_{21}$) with $a = 1200$ and $r = -1$:

$T_{21} = a r^{21-1} = a r^{20}$

$T_{21} = 1200 \times (-1)^{20}$

Since $(-1)^{20} = 1$ (any negative number raised to an even power is positive):

$T_{21} = 1200 \times 1 = 1200$.

Therefore, the 21^st term in the GP is 1200.