JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 14)
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
100
$$100{1 \over 2}$$
$$50{1 \over 4}$$
50
Explanation
T10 = a + 9d = $${1 \over {20}}$$ ....(1)
T20 = a + 19d = $${1 \over {10}}$$ .....(2)
Equation (2) – (1)
10d = $${1 \over {10}}$$ - $${1 \over {20}}$$
$$ \Rightarrow $$ d = $${1 \over {200}}$$
a + $${9 \over {200}}$$ = $${1 \over {20}}$$
$$ \Rightarrow $$ a = $${1 \over {200}}$$
S200 = $${{200} \over 2}\left[ {{2 \over {200}} + \left( {200 - 1} \right) \times {1 \over {200}}} \right]$$
= $$100\left[ {{2 \over {200}} + {{199} \over {200}}} \right]$$
= $${{201} \over 2}$$ = $$100{1 \over 2}$$
T20 = a + 19d = $${1 \over {10}}$$ .....(2)
Equation (2) – (1)
10d = $${1 \over {10}}$$ - $${1 \over {20}}$$
$$ \Rightarrow $$ d = $${1 \over {200}}$$
a + $${9 \over {200}}$$ = $${1 \over {20}}$$
$$ \Rightarrow $$ a = $${1 \over {200}}$$
S200 = $${{200} \over 2}\left[ {{2 \over {200}} + \left( {200 - 1} \right) \times {1 \over {200}}} \right]$$
= $$100\left[ {{2 \over {200}} + {{199} \over {200}}} \right]$$
= $${{201} \over 2}$$ = $$100{1 \over 2}$$
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