Find the sum of the first n terms of the sequence. The sequences are either arithmetic or geometric.
13. -1, 11, -121, ...; n = 9
A. 235,794,769
B. 429,981,696
C. -429,981,696
D. 196,495,641
E. -196,495,641
F. -235,794,769
I know that n = # of terms (9)
d = -11
a = -1
formula: n/-11(a + last term)
9/-11(-1 +?)
Also term 4 = 1331
But, the last term is too large (-214358881).
Am I doing something wrong?
* The sequence is geometric and the r=-11. Thank you for any help!
yes, you realized that the series was geometric rather than arithmetic
so you have
a=-1, r = -11, n = 9
you now need the formula for the sum of a geometric, not arithmetic like you stated.
sum(n) = a(r^n - 1)/(r-1)
= -1( (-11)^9 - 1)/(-11-1)
= -196495641
Based on the given sequence -1, 11, -121, ..., we can see that this is a geometric sequence because each term can be obtained by multiplying the previous term by a constant ratio.
To find the sum of the first n terms of a geometric sequence, we can use the formula:
Sn = (a * (r^n - 1))/(r - 1)
Where:
Sn is the sum of the first n terms
a is the first term
r is the common ratio
In this case, we have:
a = -1
r = -11
n = 9
So we can substitute these values into the formula:
Sn = (-1 * (-11^9 - 1))/(-11 - 1)
Now let's calculate the expression in the numerator first:
-11^9 = -2357947691 (Note that it's negative because both the base and the exponent are odd)
Now let's calculate the expression in the denominator:
-11 - 1 = -12
Sn = (-1 * (-2357947691 - 1))/(-12)
Sn = (-1 * -2357947692)/(-12)
Sn = 2357947692/12
Sn = 196495641
Therefore, the correct answer is D. 196,495,641.