Find the number of terms in an AP whose first term is $3$, common difference is $2$, and total sum is $255$.

Step-by-step solution

1. Write the sum formula: $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.
2. Substitute $S_n=255$, $a=3$, $d=2$:
3. $255=\frac{n}{2}\left[2\times 3+(n-1)\times 2\right]=\frac{n}{2}\left[6+2n-2\right]=\frac{n}{2}(2n+4)$.
4. Simplify: $255=n\frac{(2n+4)}{2}=n(n+2)$.
5. Set up equation: $n^2+2n-255=0$.
6. Solve quadratic: discriminant $D=2^2+4\times 255=4+1020=1024$, $\sqrt{D}=32$.
7. Find positive root: $n=\frac{-2+32}{2}=15$.
8. Thus, the AP has $15$ terms.

Concept tested: finding number of terms from sum

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