(a)OR(b)

In an A.P., $\displaystyle 15{ }^{\text {th }}$ term exceeds the $\displaystyle 8{ }^{\text {th }}$ term by 21. If sum of first $\displaystyle 10$ terms is $\displaystyle 55$ , then form the A.P.

Let first term \(\displaystyle =\mathrm{a}\) and common difference \(\displaystyle =\mathrm{d}\) \[(a+14 d)=(a+7 d)+21 \] \(\displaystyle \Rightarrow \mathrm{d}=3\) Also, \(\displaystyle \mathrm{S}_{10}=55=\frac{10}{2}[2 \mathrm{a}+9 \times 3]\) \(\displaystyle \Rightarrow \mathrm{a}=-8\) ∴ A. P. is - $\displaystyle 8$, -$\displaystyle 5$, -2... \(\displaystyle \mathrm{S}_{\mathrm{n}}=2 \mathrm{n}^{2}+13 \mathrm{n}\) \[\mathrm{S}_{1}=\mathrm{a}_{1}=15 \] \(\displaystyle \mathrm{S}_{2}=\mathrm{a}_{1}+\mathrm{a}_{2}=34 \Rightarrow \mathrm{a}_{2}=19\) \(\displaystyle \Rightarrow \mathrm{d}=19-15=4\) \(\displaystyle \therefore \mathrm{a}_{\mathrm{n}}=15+(\mathrm{n}-1) \times 4=4 \mathrm{n}+11\) Hence \(\displaystyle \mathrm{a}_{10}=4 \times 10+11=51\)