Find the number of terms in the aritmetic progession 2,-9,-20,141
I assume you meant 2,-9,-20, ...,-141
since d = -9-2 = -11,
2+(n-1)(-11) = -141
-11(n-1) = -143
n-1 = 13
n=14
69
To find the number of terms in an arithmetic progression, you need to know the first term, the last term, and the common difference between consecutive terms.
In this arithmetic progression, the first term (a1) is 2, and the common difference (d) can be found by subtracting the second term from the first term or the third term from the second term:
d = -9 - 2 = -11 or d = -20 - (-9) = -11.
Now, we need to determine the last term (an) of the arithmetic progression. We know that the general formula for the nth term of an arithmetic progression is:
an = a1 + (n-1)d,
where n is the term number.
We can substitute the known values into this formula to find the last term:
141 = 2 + (n-1)(-11).
Simplifying the equation:
141 = 2 - 11n + 11
Combine like terms:
11n = 131
Dividing both sides of the equation by 11:
n = 11.
Therefore, there are 11 terms in the arithmetic progression 2, -9, -20, 141.