JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 14)
Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 $$-$$ S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :
7000
1000
3000
5000
Explanation
S1 = $${{2n} \over 2}$$[2a + (2n $$-$$ 1)d]
S2 = $${{4n} \over 2}$$[2a + (4n $$-$$ 1)d]
(where a = T1 and d is common difference)
S2 $$-$$ S1$$ \Rightarrow $$ 2n[2a + (4n $$-$$ 1)d] $$-$$ n[2a + (2n $$-$$ 1)d] = 1000
$$ \Rightarrow $$ n[2a + d(8n $$-$$ 2 $$-$$ 2n + 1)] = 1000
$$ \Rightarrow $$ n[2a + (6n $$-$$ 1)d] = 1000
S6 = $${{6n} \over 2}$$[2a + (6n $$-$$ 1)d] = 3(S2 $$-$$ S1) = 3000
S2 = $${{4n} \over 2}$$[2a + (4n $$-$$ 1)d]
(where a = T1 and d is common difference)
S2 $$-$$ S1$$ \Rightarrow $$ 2n[2a + (4n $$-$$ 1)d] $$-$$ n[2a + (2n $$-$$ 1)d] = 1000
$$ \Rightarrow $$ n[2a + d(8n $$-$$ 2 $$-$$ 2n + 1)] = 1000
$$ \Rightarrow $$ n[2a + (6n $$-$$ 1)d] = 1000
S6 = $${{6n} \over 2}$$[2a + (6n $$-$$ 1)d] = 3(S2 $$-$$ S1) = 3000
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