Enter the first term, common ratio, and N-th to calculate the N-th term and the sum of the geometric sequence.
A geometric sequence is a series of numbers where the ratio between any two consecutive terms is constant. This constant ratio is called the common ratio.
The N-th term of a geometric sequence can be expressed using the following formula:
\( a_n = a_1 \cdot r^{(n - 1)} \)
Where:
The sum of the first N terms of a geometric sequence can be calculated using the following formula (when the common ratio \( r \neq 1 \)):
\( S_n = \frac{a_1 \cdot (1 - r^n)}{1 - r} \)
When the common ratio is 1, the sum is:
\( S_n = n \cdot a_1 \)
Solution:
Calculate the 5th term:
\( a_5 = 3 \cdot 2^{(5 - 1)} = 3 \cdot 16 = 48 \)
Calculate the sum:
\( S_5 = \frac{3 \cdot (1 - 2^5)}{1 - 2} = \frac{3 \cdot (1 - 32)}{-1} = \frac{3 \cdot (-31)}{-1} = 93 \)
Solution:
Calculate the 6th term:
\( a_6 = 10 \cdot 5^{(6 - 1)} = 10 \cdot 3125 = 31250 \)
Calculate the sum:
\( S_6 = \frac{10 \cdot (1 - 5^6)}{1 - 5} = \frac{10 \cdot (1 - 15625)}{-4} = \frac{10 \cdot (-15624)}{-4} = 39060 \)
Solution:
Calculate the 8th term:
\( a_8 = 12 \cdot (-2)^{(8 - 1)} = 12 \cdot (-128) = -1536 \)
Calculate the sum:
\( S_8 = \frac{12 \cdot (1 - (-2)^8)}{1 - (-2)} = \frac{12 \cdot (1 - 256)}{3} = \frac{12 \cdot (-255)}{3} = -1020 \)