the following problem refers to a geometric sequence:
Find S_10 for -1, -1/3, -1/9
r=-1/3/-1=1/3
a_1=-1
a_2=-1/3
a_n=a_1*r^n-1
a_n=-1*1/3^n-1
s_n=
Help me.
For a GP
sum(n) = a(1 - r^n)/(1-r)
so for yours ....
sum(10) = -1( 1 - (1/3)^10)/(1 - 1/3)
= -(1 - 1/59049)/(2/3)
= -(59048/59049)(3/2) = ....
To find S_10, we need to calculate the sum of the first 10 terms of the geometric sequence represented by -1, -1/3, -1/9.
To find the common ratio (r) of the geometric sequence, we divide any term by its preceding term. So, for this sequence, r = (-1/3) / (-1) = 1/3.
To find the formula for the nth term (a_n) of the geometric sequence, we start with the first term (a_1) and multiply it by the common ratio raised to the power of (n - 1). So, a_n = a_1 * r^(n-1).
In this case, a_1 = -1 and r = 1/3. Therefore, a_n = -1 * (1/3)^(n-1).
Finally, to find S_10, which represents the sum of the first 10 terms of the sequence, we use the formula:
S_n = a_1 * (1 - r^n) / (1 - r)
In this case, n = 10. So, substituting the values:
S_10 = -1 * (1 - (1/3)^10) / (1 - 1/3)
Now we can simplify the equation and calculate the value of S_10.