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 :
Solution
S<sub>1</sub> = ${{2n} \over 2}$[2a + (2n $-$ 1)d]<br><br>S<sub>2</sub> = ${{4n} \over 2}$[2a + (4n $-$ 1)d]<br><br>(where a = T<sub>1</sub> and d is common difference)<br><br>S<sub>2</sub> $-$ S<sub>1</sub>$\Rightarrow$ 2n[2a + (4n $-$ 1)d] $-$ n[2a + (2n $-$ 1)d] = 1000<br><br>$\Rightarrow$ n[2a + d(8n $-$ 2 $-$ 2n + 1)] = 1000<br><br>$\Rightarrow$ n[2a + (6n $-$ 1)d] = 1000<br><br>S<sub>6</sub> = ${{6n} \over 2}$[2a + (6n $-$ 1)d] = 3(S<sub>2</sub> $-$ S<sub>1</sub>) = 3000
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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