Series Calculator
Calculate the sum of arithmetic and geometric series. Enter the first term, common difference or ratio, and number of terms.
What is the difference between arithmetic and geometric series?
Arithmetic series: constant difference between terms (e.g., 2, 5, 8, 11 with d=3). Sum = n/2 * (2a + (n-1)d). Geometric series: constant ratio between terms (e.g., 2, 6, 18, 54 with r=3). Sum = a(1-r^n)/(1-r). Example: 1+2+3+4+5 = 15 (arithmetic), 1+2+4+8+16 = 31 (geometric).
How do you calculate the sum of an arithmetic series?
Formula: S_n = n/2 * (2a + (n-1)d) or S_n = n/2 * (first + last), where n=number of terms, a=first term, d=common difference. Example: Sum of 2, 5, 8, 11, 14 (n=5, a=2, d=3): S_5 = 5/2 * (2*2 + 4*3) = 5/2 * 16 = 40.
How do you calculate the sum of a geometric series?
Formula: S_n = a(1-r^n)/(1-r) for r≠1, where a=first term, r=common ratio, n=number of terms. For r=1, S_n=na. Example: Sum of 3, 6, 12, 24, 48 (a=3, r=2, n=5): S_5 = 3(1-2^5)/(1-2) = 3(-31)/(-1) = 93. Infinite geometric series (|r|<1): S = a/(1-r).
What is the formula for the nth term of a series?
Arithmetic: a_n = a_1 + (n-1)d. Geometric: a_n = a_1 * r^(n-1). Example: For arithmetic 5, 8, 11, 14... (d=3), 10th term = 5 + 9*3 = 32. For geometric 2, 6, 18... (r=3), 5th term = 2 * 3^4 = 162.
Can a geometric series have an infinite sum?
Yes, if |r| < 1 (ratio between -1 and 1), the infinite geometric series converges to S = a/(1-r). Example: 1 + 1/2 + 1/4 + 1/8 + ... (a=1, r=1/2) converges to 1/(1-0.5) = 2. If |r| >= 1, the series diverges (no finite sum).