If the 10th term of an A.P. is ${1 \over {20}}$ and its 20th term is ${1 \over {10}}$, then the sum of its first 200 terms is
Solution
T<sub>10</sub> = a + 9d = ${1 \over {20}}$ ....(1)
<br><br>T<sub>20</sub> = a + 19d = ${1 \over {10}}$ .....(2)
<br><br>Equation (2) – (1)
<br><br>10d = ${1 \over {10}}$ - ${1 \over {20}}$
<br><br>$\Rightarrow$ d = ${1 \over {200}}$
<br><br>a + ${9 \over {200}}$ = ${1 \over {20}}$
<br><br>$\Rightarrow$ a = ${1 \over {200}}$
<br><br>S<sub>200</sub> = $${{200} \over 2}\left[ {{2 \over {200}} + \left( {200 - 1} \right) \times {1 \over {200}}} \right]$$
<br><br>= $100\left[ {{2 \over {200}} + {{199} \over {200}}} \right]$
<br><br>= ${{201} \over 2}$ = $100{1 \over 2}$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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