(A) The sum of first seven terms of an A.P. is 182. If its 4^th term and the 17^th term are in the ratio 1: 5, find the A.P.

Let a be the first term and d be the common difference.

\[\begin{aligned} & \quad \mathrm{S}_{7}=182 \Rightarrow \frac{7}{2}(2 \mathrm{a}+6 \mathrm{~d})=182 \\ & \Rightarrow 2 \mathrm{a}+6 \mathrm{~d}=\frac{182 \times 2}{7}=52 \\ & \mathrm{a}+3 \mathrm{~d}=26 \\ & \frac{a_{4}}{a_{17}}=\frac{1}{5} \Rightarrow \frac{a+3 \mathrm{~d}}{\mathrm{a}+16 \mathrm{~d}}=\frac{1}{5} \Rightarrow 5 \mathrm{a}+15 \mathrm{~d}=\mathrm{a}+16 \mathrm{~d} \\ & 4 \mathrm{a}=\mathrm{d} \end{aligned} \]

Solving (i) and (ii)

\[\begin{aligned} & \mathrm{a}=2 \text { and } \mathrm{d}=8 \\ & \therefore \mathrm{AP} \text { is } 2,10,18,26, \ldots . \end{aligned} \]

\(\displaystyle \mathrm{Sq}=63 \mathrm{q}-3 \mathrm{q}^{2}\)

\[\begin{aligned} & \therefore S_{1}=60 \Rightarrow a_{1}=60\left(1^{\text {st }} \text { term }\right) \\ & S_{2}=63(2)-3(2)^{2}=126-12=114 \\ & a_{1}+a_{2}=114 \Rightarrow a_{2}=114-60=54 \\ & d=a_{2}-a_{1}=54-60=-6 \\ & a_{p}=-60 \\ & 60+(p-1) d=-60 \\ & (p-1)(-6)=-120 \Rightarrow p=21 \\ & a_{11}=a+10 d=60+10(-6)=0 \end{aligned} \]