Q.Find the 20th term of the Arithmetic Progression $3, 7, 11, 15, \ldots$

Identify the first term and common difference, then substitute into the nth-term formula.

$t_{20} = 3 + (20-1)(4) = 3 + 76 = 79$.

Here $a = 3$ and $d = 7 - 3 = 4$ (checked: $11 - 7 = 4$, $15 - 11 = 4$ — constant, so this is an AP).

Apply $t_n = a + (n-1)d$ with $n = 20$:

Independent check (repeated addition): starting from $3$, add $4$ a total of $19$ times: $3 + 19\times 4 = 3 + 76 = 79$ — same value.

The last term can also be reached by extending the pattern in blocks — e.g. the 10th term is $3 + 9(4) = 39$, and ten more steps of $4$ add $40$, giving $39 + 40 = 79$; the same answer reached in two stages.

Using $n$ instead of $(n-1)$ as the multiplier on $d$: the first term needs zero additions of $d$, so the multiplier is $(n-1) = 19$, not $20$. The wrong version gives $3 + 20(4) = 83$.