Sum of Linear Number Sequence Calculator
Calculate the sum of arithmetic sequences with step-by-step solutions
Calculate Sum of Linear Sequence
Quick Examples
The first term of the sequence
Constant difference between consecutive terms
Total number of terms in the sequence
Calculation Results
Formula Used:
Sum = n/2 × (2a + d(n-1))
Sum = 10/2 × (2×1 + 1×9) = 55
Sequence Preview:
Step-by-Step Calculation
Given: Initial value a = 1
Common difference d = 1
Number of terms n = 10
Step 1: Find the final value
Final value = a + d × (n - 1)
Final value = 1 + 1 × (10 - 1)
Final value = 1 + 1 × 9 = 10
Step 2: Apply the sum formula
Sum = n/2 × (first term + last term)
Sum = 10/2 × (1 + 10)
Sum = 5 × 11 = 55
Alternative formula: Sum = n/2 × (2a + d(n-1))
Sum = 10/2 × (2×1 + 1×9)
Sum = 5 × (2 + 9) = 55
Example: Cloud Storage Billing
Problem
Alice uploads 5GB initially, then adds 2GB each month for 12 months.
At $1 per GB, how much storage will she have and what's the total cost?
Solution
Initial value: a = 5 GB
Common difference: d = 2 GB per month
Number of periods: n = 12 months
Final storage: 5 + 2×(12-1) = 5 + 22 = 27 GB
Total storage used: 12/2 × (5 + 27) = 6 × 32 = 192 GB
Total cost: 192 GB × $1 = $192
Key Formulas
Sum Formula
S = n/2 × (2a + d(n-1))
Or: S = n/2 × (first + last)
nth Term
aₙ = a + d(n-1)
Find any term in the sequence
Average Value
Average = Sum / n
Also: (first + last) / 2
Sequence Properties
Linear/Arithmetic
Constant difference between terms
Symmetry
Sum equals n × average value
Growth Rate
Linear increase/decrease
Applications
Finance, physics, statistics
Understanding Linear Number Sequences
What is a Linear Sequence?
A linear number sequence (also called an arithmetic sequence) is a sequence where each term differs from the previous term by the same constant amount, called the common difference.
Why Calculate the Sum?
- •Financial calculations (loans, savings)
- •Physics problems (motion, forces)
- •Statistics and data analysis
- •Business planning and forecasting
Sum Formula Derivation
For sequence: a, a+d, a+2d, ..., a+(n-1)d
Sum = a + (a+d) + (a+2d) + ... + (a+(n-1)d)
Sum = na + d(0+1+2+...+(n-1))
Sum = na + d × n(n-1)/2
Sum = n/2 × (2a + d(n-1))
Alternative: Sum = n/2 × (first term + last term)