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The $n$ th term of an AP. is $\frac{3+\mathrm{n}}{4}$, then the sum of first 105 terms is
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1470
$\mathrm{T}_{\mathrm{n}}=\frac{3+\mathrm{n}}{4}$
$\mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{T}_{\mathrm{n}}=\sum\left(\frac{3}{4}+\frac{\mathrm{n}}{4}\right)$
$=\frac{3}{4} \mathrm{n}+\frac{1}{4} \times \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{7}{8} \mathrm{n}+\frac{\mathrm{n}^{2}}{8}$
$\mathrm{~S}_{105}=\frac{7}{8} \times 105+\frac{(105)^{2}}{8}=1470$
$\mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{T}_{\mathrm{n}}=\sum\left(\frac{3}{4}+\frac{\mathrm{n}}{4}\right)$
$=\frac{3}{4} \mathrm{n}+\frac{1}{4} \times \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{7}{8} \mathrm{n}+\frac{\mathrm{n}^{2}}{8}$
$\mathrm{~S}_{105}=\frac{7}{8} \times 105+\frac{(105)^{2}}{8}=1470$
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