Question

The sum of n terms of an AP is 49 and that of first 17 terms is 289, find the sum of first n terms.

Answer 1

$$\begin{gathered} S_n=\left(\frac{5n^2}{2}+\frac{3n}{2}\right)=\frac{1}{2}\left(5n^2+3n\right)...\left(i\right) \\ \mathrm{Replacing}n\mathrm{by}(n-1)\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ {S}_{n-1}=\frac{1}{2}\times \left[5{\left(n-1\right)}^2+3\left(n-1\right)\right] \\ =\frac{1}{2}\times \left[5n^2-10n+5+3n-3]\right]=\frac{1}{2}\times [5n^2-7n+2] \\ \therefore T_n=S_n-S_{n-1} \\ =\frac{1}{2}\left(5n^2+3n\right)-\frac{1}{2}\times [5n^2-7n+2] \\ =\frac{1}{2}\left(10n-2\right)=5n-1...\left(ii\right) \end{gathered}$$


Putting n = 20 in (ii), we get:
T₂₀ = (5 ⨯ 20) - 1 = 99
Hence, the 20^th term is 99.