If the 10^{text{th}} term of an A.P. is frac{ | vedclass

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(B) The $n^{	ext{th}}$ term of an $A$.$P$. is given by $T_n = a + (n-1)d$.Given $T_{10} = a + 9d = \frac{1}{20} \quad \dots(i)$Given $T_{20} = a + 19

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$100 \frac{1}{2}$

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(B) The $n^{	ext{th}}$ term of an $A$.$P$. is given by $T_n = a + (n-1)d$.Given $T_{10} = a + 9d = \frac{1}{20} \quad \dots(i)$Given $T_{20} = a + 19d = \frac{1}{10} \quad \dots(ii)$Subtracting equation $(i)$ from $(ii)$:$(a + 19d) - (a + 9d) = \frac{1}{10} - \frac{1}{20}$$10d = \frac{2-1}{20} = \frac{1}{20}$$d = \frac{1}{200}$Substituting $d$ in equation $(i)$:$a + 9(\frac{1}{200}) = \frac{1}{20}$$a = \frac{10}{200} - \frac{9}{200} = \frac{1}{200}$Now,the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.For $n = 200$:$S_{200} = \frac{200}{2} [2(\frac{1}{200}) + (200-1)(\frac{1}{200})]$$S_{200} = 100 [\frac{2}{200} + \frac{199}{200}]$$S_{200} = 100 [\frac{201}{200}] = \frac{201}{2} = 100 \frac{1}{2}$.

If the $10^{\text{th}}$ term of an $A$.$P$. is $\frac{1}{20}$ and its $20^{\text{th}}$ term is $\frac{1}{10}$,then the sum of its first $200$ terms is

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