(A)OR(B)

The ratio of the $\displaystyle 11^{\text {th }}$ term to the $\displaystyle 18^{\text {th }}$ term of an A.P. is $\displaystyle 2$ : 3. Find the ratio of the $\displaystyle 5^{\text {th }}$ term to the $\displaystyle 21$ $\displaystyle { }^{\text {st }}$ term. Also, find the ratio of the sum of first $\displaystyle 5$ terms to the sum of first $\displaystyle 21$ terms.

\[\begin{aligned} & \frac{a+10 \mathrm{~d}}{\mathrm{a}+17 \mathrm{~d}}=\frac{2}{3} \\ & 3 \mathrm{a}+30 \mathrm{~d}=2 \mathrm{a}+34 \mathrm{~d} \Rightarrow \mathrm{a}=4 \mathrm{~d} \\ & \text { Therefore, } \frac{a+4 \mathrm{~d}}{\mathrm{a}+20 \mathrm{~d}}=\frac{4 d+4 \mathrm{~d}}{4 \mathrm{~d}+20 \mathrm{~d}}=\frac{8 d}{24 d}=\frac{1}{3} \\ & \frac{S_{5}}{S_{21}}=\frac{\dfrac{5}{2}[2 a+4 \mathrm{~d}]}{\dfrac{21}{2}[2 a+20 \mathrm{~d}]}=\frac{5[8 d+4 \mathrm{~d}]}{21[8 d+20 \mathrm{~d}]} \\ & =\frac{5 \times 12 \mathrm{~d}}{21 \times 28 \mathrm{~d}}=\frac{5}{49} \text { or } S_{5}: S_{21}=5: 49 \end{aligned} \]

\(\displaystyle \mathrm{S}_{6}=36 \Rightarrow \frac{6}{2}[2 \mathrm{a}+5 \mathrm{~d}]=36\)

\[\Rightarrow 2 \mathrm{a}+5 \mathrm{~d}=12 \quad \text {----------- (1) } \]

\(\displaystyle \mathrm{S}_{16}=256 \Rightarrow \frac{16}{2}[2 \mathrm{a}+15 \mathrm{~d}]=256\)

\[\Rightarrow 2 \mathrm{a}+15 \mathrm{~d}=32 \quad \text {------------ (2) } \]

Solving ($\displaystyle 1$) and ($\displaystyle 2$)

\[\begin{aligned} d & =2 \\ a & =1 \\ S_{10} & =\frac{10}{2}[2(1)+9(2)] \\ & =100 \end{aligned} \]