Q.Find the 10th term of the Arithmetic Progression $3, 7, 11, 15, \ldots$

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

$T_{10}=3+(10-1)(4)=3+36=39$.

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

Apply the general-term formula: $T_n=a+(n-1)d$. For $n=10$: $T_{10}=3+(10-1)(4)=3+9(4)=3+36=39$.

Check by direct listing: $3,7,11,15,19,23,27,31,35,39$ — counting these ten terms confirms the $10$th is indeed $39$.

Alternative — repeated addition: since each step adds $4$, the $10$th term is $9$ steps beyond the first term: $3 + 9\times4 = 39$ — the same computation viewed as repeated addition rather than the formula.

A common slip is using $n$ instead of $(n-1)$ as the number of times $d$ is added — the first term $T_1=a$ needs ZERO additions of $d$, so the multiplier on $d$ is always $(n-1)$, not $n$. Using $n$ here would wrongly give $3+10(4)=43$.