Let S_1 be the sum of first 2n terms of an arithmetic progression. Let S_2 be the sum of first 4n terms of the same arithmetic progression. If ( S_2 - S_1 ) is 1000 , n then the sum of the first 6n terms of the arithmetic progression is equal to

Correct option Is (B) 3000 S_2 n=(2 n / 2)[2 a+(2 n-1) d] \ S_4 n=(4 n / 2)[2 a+(4 n-1) d] \ => S_2-S_1=(4 n / 2)[2 a+(4 n-1) d] \ -(2 n / 2)[2 a+(2 n-1) d] \ =4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n \ =2 n a+n d[8 n-2-2 n+1] \ => 2 n a+n d[6 n-1]=1000 \ 2 a+(6 n-1) d=(1000 / n) Now, S_6 n=frac6 n2[2 a+(6 n-1) d] =3 n * frac1000 n=3000

Let S_{1} be the sum of first 2n terms of an arithmet... | Sequence & Series (Mathematics)

Correct option Is (B) 3000

$\begin{aligned} & S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d] \\ & S_{4 n}=\frac{4 n}{2}[2 a+(4 n-1) d] \\ & \Rightarrow S_2-S_1=\frac{4 n}{2}[2 a+(4 n-1) d] \\ & \quad-\frac{2 n}{2}[2 a+(2 n-1) d] \\ & =4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n \\ & =2 n a+n d[8 n-2-2 n+1] \\ & \Rightarrow 2 n a+n d[6 n-1]=1000 \\ & 2 a+(6 n-1) d=\frac{1000}{n} \end{aligned}$

Now, $\mathrm{S}_{6 \mathrm{n}}=\frac{6 \mathrm{n}}{2}[2 \mathrm{a}+(6 \mathrm{n}-1) \mathrm{d}]$ $=3 \mathrm{n} \cdot \frac{1000}{\mathrm{n}}=3000$

Explanation

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