The sum of 8 terms of an AP is 160 while the sum of 20 terms is 880. Find the 23rd term
Let's denote the first term of the arithmetic progression as $a$, and the common difference as $d$.
The sum of an arithmetic progression can be calculated using the formula $\text{Sum} = \frac{n}{2}(2a + (n-1)d)$, where $n$ is the number of terms.
Using this formula, we can set up two equations:
For the sum of 8 terms:
$160 = \frac{8}{2}(2a + 7d)$
$80 = 2a + 7d$ \[ \text{(equation 1)} \]
For the sum of 20 terms:
$880 = \frac{20}{2}(2a + 19d)$
$44 = 2a + 19d$ \[ \text{(equation 2)} \]
We can solve these equations simultaneously to find $a$ and $d$:
Multiplying equation 1 by 19 and equation 2 by 7, we get:
$19(80) = 19(2a + 7d)$
$1440 = 38a + 133d$ \[ \text{(equation 3)} \]
$7(44) = 7(2a + 19d)$
$308 = 14a + 133d$ \[ \text{(equation 4)} \]
Subtracting equation 4 from equation 3, we have:
$(1440 - 308) = (38a + 133d) - (14a + 133d)$
$1132 = 24a$
$a = \frac{1132}{24} = 47$
Substituting the value of $a$ into equation 1:
$80 = 2(47) + 7d$
$80 = 94 + 7d$
$-14 = 7d$
$d = -2$
So, the arithmetic progression has a first term, $a$, of 47, and a common difference, $d$, of -2.
To find the 23rd term, we use the formula $a + (n-1)d$:
$23$rd term $= 47 + (23-1)(-2)$
$= 47 + 22(-2)$
$= 47 - 44$
$= \boxed{3}$