Find the sum of all two-digit numbers that leave a remainder of 1 when divided by 4.

**Sₙ = $(n/2)\times$ [$2a+(n-1)d$]**

Or equivalently, **Sₙ = $(n/2)\times (a+l)$**, where $l$ is the LAST term (the n-th term).

**Why these are equal**: $l=a+(n-1)d$, so $a+l=2a+(n-1)d$.

**Example**: Sum of 1, 4, 7, …, 28.
First find $n:a=1,d=3,l=28$. From aₙ = 28: $1+3(n-1)=28$ → $n=10$.
Then S₁₀ = $(10/2)(1+28)=5\times 29=145$.

**Special case $(n$ natural numbers)**: Sₙ = $n(n+1)/2$.
**First $n$ odd numbers**: Sₙ = $n^2$.
**First $n$ even numbers**: Sₙ = $n(n+1)$.