\(a=6, S_{\infty}=18, r=? \)
\(S_{\infty} =\frac{a}{1-r}=\frac{6}{1-r}=18 \)
\(6 =18-18 r \)
\(18 r =12 \)
\(r =\frac{2}{3} \)

\(\mathrm{a} =1, \quad \mathrm{~d}=6, \quad \mathrm{r}=\frac{1}{2} \)
\(\mathrm{~S}_{\infty} =\frac{a}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1+\mathrm{r})^{2}} \)
\(=\frac{1}{1-\frac{1}{2}}+\frac{6 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^{2}} \)
\(=2+(3 \times 4)=14 \)

\(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32} \ldots . . =\sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2} . \)
\(=\sqrt{2}[1+2+3+\ldots .] \)
\(S_{n} =\frac{\sqrt{2}[n(n+1)]}{2} \)
\(=\frac{n(n+1)}{\sqrt{2}} \)

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

\(n^{\text {th }} \text { term is } \frac{n^{2}-n+2}{2}\)

\((38)^{5} =(39-1)^{15} \)
\(=(39)^{15}-{ }^{15} C_{1}(39)^{14}+\ldots+{ }^{15} C_{14}(39)-1 \)
The remainder will be 12 because all the terms except the last is divisible by 39 and so by 13, -1 remains

\(\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{n}}{2}[2 a+(\mathrm{n}-1) \mathrm{d}] \)
\(\mathrm{S}_{\mathrm{n}-1} =\frac{\mathrm{n}-1}{2}[2 a+(\mathrm{n}-2) \mathrm{d}] \)
\(\mathrm{S}_{\mathrm{n}-2} =\frac{\mathrm{n}-2}{2}[2 a+(\mathrm{n}-3) \mathrm{d}] \)
\(\mathrm{S}_{\mathrm{n}}-2 \mathrm{~S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2} =2 a\left[\frac{\mathrm{n}-2(\mathrm{n}-1)}{2}+\frac{(\mathrm{n}-2)}{2}\right]+\frac{\mathrm{d}}{2}[\mathrm{n}(\mathrm{n}-1)-2(\mathrm{n}-1)(\mathrm{n}-2)+(\mathrm{n}-2)(\mathrm{n}-3)] \)
\(=2 a\left[\frac{\mathrm{n}-2 \mathrm{n}+2+\mathrm{n}-2}{2}\right]+\frac{\mathrm{d}}{2}\left[\mathrm{n}^{2}-\mathrm{n}-2 \mathrm{n}^{2}+6 \mathrm{n}-4+\mathrm{n}^{2}-5 \mathrm{n}+6\right] \)
\(= 0+\frac{\mathrm{d}}{2}(2)=\mathrm{d} \)

\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2 \sqrt{2}}, \ldots \ldots \text { form } a \text { HP. }\)

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)