Find the number of terms in the aritmetic progession 2,-9,-20,-141
the common difference is -11 , so there appear to be 10 terms "missing" between -20 and -141
a = 2, d = -11
term(n) = a + (n-1)d
-141 = 2 + (n-1)(-11)
-143 = -11n + 11
11n = 154
n = 14
there are 14 terms in the sequence.
(check by listing them, there are not that many)
To find the number of terms in an arithmetic progression, we need to identify the common difference between consecutive terms.
Here, we observe that the common difference is changing between -7 and -11 for consecutive terms: -9 - 2 = -11, -20 - (-9) = -11, and -141 - (-20) = -121.
Now, let's find the common difference: -7, -11, -7, -11.
Since we have alternating common differences, we can conclude that there are two different common differences in this sequence: -7 and -11.
To determine the number of terms, we can use the formula for the nth term of an arithmetic sequence:
tn = a + (n - 1)d
Where tn represents the nth term, a is the first term, n is the position of the term, and d is the common difference.
Let's first solve for n when d = -7:
-141 = 2 + (n - 1)(-7)
Simplifying the equation:
-141 = 2 - 7(n - 1)
Expanding and bringing terms to one side:
-141 = 2 - 7n + 7
Rearranging terms:
-148 = -7n
Dividing by -7:
n = 21
The obtained value of n when d = -7 is 21.
Now let's solve for n when d = -11:
-141 = 2 + (n - 1)(-11)
Simplifying the equation:
-141 = 2 - 11(n - 1)
Expanding and bringing terms to one side:
-141 = 2 - 11n + 11
Rearranging terms:
-152 = -11n
Dividing by -11:
n = 14
The obtained value of n when d = -11 is 14.
Since we have two possible values for n, we can conclude that the number of terms in the arithmetic progression is the sum of the values of n, i.e., 21 + 14 = 35.
Therefore, there are 35 terms in the arithmetic progression 2, -9, -20, -141.