Q184 / 270 · NDA 2015 (Phase 2) · Mathematics

The $n$th term of an AP. is $\frac{3+n}{4}$, then the sum of first 105 terms is

  • A
    270
  • B
    735
  • C
    1409
  • Correct answer
    D
    1470

Solution

$\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\frac{3+\mathrm{n}}{4} \\ & \mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{T}_{\mathrm{n}} \\ & =\sum\left(\frac{3}{4}+\frac{\mathrm{n}}{4}\right) \\ & =\frac{3}{4} \mathrm{n}+\frac{1}{4} \times \frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ & =\frac{7}{8} \mathrm{n}+\frac{\mathrm{n}^2}{8} \\ & \mathrm{~S}_{105}=\frac{7}{8} \times 105+\frac{(105)^2}{8}=1470\end{aligned}$

Hence, the correct answer is option: 'D'.

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