Arithmetic sequence of 5 terms - calculate sum and nth term

    Every linear pattern starts with two numbers: the first term and the common difference. Enter both plus the position n to get the nth term, the general formula, and the sum of the first n terms using Gauss's method.

    An arithmetic sequence with 5 terms is a sequence where the difference between consecutive terms is constant. The sum of 5 terms is calculated as S = n/2 x (a1 + an). Enter the first term and common difference, and the calculator will list all 5 terms with their sum and general formula.

    Parameters

    Enter data for calculations

    The starting value of the sequence

    The constant difference between consecutive terms

    Which term to calculate

    Form progress0 / 3 fields

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    Every linear pattern boils down to two numbers

    An arithmetic sequence adds the same value - the common difference d - to each term. Give the calculator the first term a1, the difference d, and a position n, and it returns the nth term plus the sum of all terms from a1 to an. The sum uses the Gauss formula: pair the first and last terms, multiply by the count, divide by two.

    Quick start
    Enter a1 = 2, d = 5, n = 10. Result: a10 = 47. Sum S10 = 245. Formula: an = 2 + (n-1) x 5.

    Core formulas

    Nth term: an = a1 + (n - 1) x d

    Sum of n terms: Sn = n x (a1 + an) / 2

    Alternative sum: Sn = n x (2a1 + (n-1)d) / 2

    Common difference: d = an - an-1 (constant for all pairs)

    Reference table - common sequences

    Sequence a1 d a10 S10
    Natural numbers (1, 2, 3...) 1 1 10 55
    Odd numbers (1, 3, 5...) 1 2 19 100
    Multiples of 7 (7, 14, 21...) 7 7 70 385
    Countdown (100, 90, 80...) 100 -10 10 550
    Negative start (-5, -2, 1, 4...) -5 3 22 85

    Practical examples

    Saving plan: You save $200 in month 1 and increase by $50 each month. After 12 months: a12 = 200 + 11 x 50 = $750. Total saved: S12 = 12 x (200 + 750) / 2 = $5,700.
    An increasing savings plan is an arithmetic sequence
    Gauss's trick: Sum of 1 to 100? a1 = 1, d = 1, n = 100, a100 = 100. S100 = 100 x (1 + 100) / 2 = 5,050. Legend says young Gauss solved this in seconds while his classmates added one by one.
    The pairing trick: 1+100 = 2+99 = 3+98 = ... = 101, fifty pairs
    Seats in a theater: Row 1 has 20 seats, each next row adds 2. Row 15: a15 = 20 + 14 x 2 = 48 seats. Total seats in 15 rows: S15 = 15 x (20 + 48) / 2 = 510.
    Architecture often uses arithmetic spacing
    Depreciation: A machine depreciates by $3,000/year from $45,000. After year 8: a8 = 45000 + 7 x (-3000) = $24,000. Value hits zero at n = 16 (fully depreciated).
    Straight-line depreciation = arithmetic sequence with negative d
    Staircase: First step at 18 cm, each next 18 cm higher. Step 14 is at: a14 = 18 + 13 x 18 = 252 cm (2.52 m). Total rise for a standard floor height.
    Constant step height = constant common difference

    Arithmetic vs geometric - comparison

    Property Arithmetic (this calculator) Geometric
    Pattern Add d each time Multiply by q each time
    Formula an = a1 + (n-1)d an = a1 x qn-1
    Growth Linear (steady) Exponential (accelerating)
    Graph Straight line Exponential curve
    Example Salary +$500/year Investment x1.08/year

    FAQ

    What is the common difference?
    The common difference d is the constant value added to each term to get the next. In the sequence 5, 11, 17, 23... the difference is d = 11 - 5 = 6. If d is negative, the sequence decreases. If d = 0, every term equals a1.
    How did Gauss sum 1 to 100?
    He paired terms from opposite ends: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101... That gives 50 pairs, each summing to 101. Total: 50 x 101 = 5,050. This is exactly the formula S = n(a1 + an)/2. It works for any arithmetic sequence, not just natural numbers.
    Can the common difference be a decimal?
    Yes. The sequence 1.5, 1.8, 2.1, 2.4... has d = 0.3. All formulas work identically with decimals, fractions, or even irrational differences. The calculator handles any real-number inputs.
    How do I find d if I only know two terms?
    If you know am and ak (terms at positions m and k): d = (am - ak) / (m - k). Example: a3 = 11, a7 = 27. Then d = (27 - 11) / (7 - 3) = 16/4 = 4. Work backward to find a1: a1 = a3 - 2d = 11 - 8 = 3.
    When is an arithmetic sequence useful in real life?
    Whenever something changes by a fixed amount per period: straight-line depreciation ($5,000/year), fixed salary raises ($500/year), stacking objects (each row adds 2 seats), counting steps (constant riser height), or phone data plans (add 2 GB per tier). If the change is proportional rather than fixed, use a geometric sequence instead.
    What happens if n is very large?
    The nth term grows linearly: an = a1 + (n-1)d. For d = 1, a1000000 = 1,000,000. The sum grows quadratically: Sn is roughly n2d/2 for large n. The calculator handles very large n without issues - it uses direct formulas, not iteration.

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    Natalia Skrzek

    Reviewed by: Natalia Skrzek