Adding a common difference, or multiplying by a common ratio. Two patterns, two formulas.
Find the sum
What is the sum of the first 20 terms of 3, 7, 11, 15, …?
It is arithmetic with a = 3 and d = 4. The 20th term is 3 + 19(4) = 79, and Sₙ = n(a + l)2 = 20(3 + 79)2 = 10 × 82 = 820. A is the 20th TERM, not the sum — the classic mix-up. D forgets to halve.
Subtract consecutive terms first. If the differences match it is arithmetic; if not, divide instead and look for a constant ratio.
aₙ = a + (n − 1)d
the nth term first
a₃₀ = 4 + 29(3) = 91
29 steps, not 30
Sₙ = n(a + l)2
count times the average of the ends
S₃₀ = 30(4 + 91)2 = 1425
the sum
The formula works because pairing the first term with the last, the second with the second-last and so on gives the same total every time. Fifteen pairs of 95 is 1,425.
a = 3, r = 4, n = 5
name them
Sₙ = a(rⁿ − 1)r − 1
valid whenever r ≠ 1
= 3(4⁵ − 1)3 = 4⁵ − 1
4⁵ = 1024
= 1023
check: 3+12+48+192+768 ✓
The nth TERM uses arⁿ⁻¹, but the SUM of n terms uses rⁿ. Carrying the term formula's exponent into the sum formula is the most common error here, and it always gives an answer that is too small by roughly a factor of r.
Each term is a third of the last, so the running total closes in on 32 without ever passing it. Adding forever is fine as long as the pieces shrink fast enough.
The same two formulas, used from an unusual starting point.
Two terms given: a₃ = 10 and a₈ = 25 in an arithmetic sequence. Find a₁.
a₈ − a₃ = 5d
five steps from term 3 to term 8
25 − 10 = 15, so d = 3
divide by the number of steps
a₁ = a₃ − 2d = 10 − 6 = 4
two steps back
Count the steps between the given terms (8 − 3 = 5), not the terms themselves.
A negative ratio: 12 − 6 + 3 − … forever
a = 12, r = −612 = −12
the signs alternate
|r| = 12 < 1, so it converges
check before using the formula
S = 121 − (−12) = 1232 = 8
1 − r becomes 1 + 12
The minus in r turns 1 − r into 1 + 12. Dropping it gives 1212 = 24, which is wrong.
n as the unknown: how many terms of 2 + 5 + 8 + … add up to 77?
Sₙ = (n2)(2a + (n − 1)d) = (n2)(3n + 1)
a = 2, d = 3
(n2)(3n + 1) = 77, so 3n² + n − 154 = 0
clear the 2
(n − 7)(3n + 22) = 0, so n = 7
a count must be a positive whole number
Check: 2 + 5 + 8 + 11 + 14 + 17 + 20 = 77. The other root, −223, is not a number of terms.
0.27 + 0.0027 + 0.000027 + …
a = 0.27, r = 0.01
|r| < 1, so it converges
the formula applies
S = 0.271 − 0.01 = 0.270.99
tidy it
= 2799 = 311
the exact fraction
This is why every repeating decimal is rational. The repeating block is the first term, the ratio is a power of ten below one, and the formula always returns a fraction of whole numbers.
Tests ask for an nth term, a sum of n terms, or an infinite sum, often with the sum and the last term both among the choices. The fastest route: decide arithmetic or geometric from differences or ratios, then use n(a + l)2 or a(rⁿ − 1)r − 1, or a1 − r when |r| < 1. A 10-second check: write out the first three or four terms and see that your formula gives their total.
Find the sum
What is the sum of the infinite series 8 + 4 + 2 + 1 + …?
a = 8 and r = 12, and since |r| < 1 the series converges: S = 81 − 12 = 80.5 = 16. A is the sum of just the four terms shown; the series keeps going and creeps up on 16 without reaching it.
Hint: find r by dividing any term by the one before it, check that it is smaller than 1 in size, then use a1 − r.
