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  • Let S1 be the sum of first 2n terms of an arthimetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If  is

Let S1 be the sum of first 2n terms of an arthimetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S_2 - S_1) is 1000, then the sum of the first 6n terms of the arthimetic progression is equal to:
Option: 1 1000
Option: 2 3000
Option: 3 5000
Option: 4 7000

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best_answer

\\\mathrm{S}_{2 \mathrm{n}}=\frac{2 \mathrm{n}}{2}[2 \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}]\\\\\mathrm{S}_{4 \mathrm{n}}=\frac{4 \mathrm{n}}{2}[2 \mathrm{a}+(4 \mathrm{n}-1) \mathrm{d}]

\\\Rightarrow \mathrm{S}_{2}-\mathrm{S}_{1}=\frac{4 \mathrm{n}}{2}[2 \mathrm{a}+(4 \mathrm{n}-1) \mathrm{d}]-\frac{2 \mathrm{n}}{2}[2 \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}]\\\begin{array}{l} \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=2 n a+n d[8 n-2-2 n+1] \end{array}

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

Posted by

himanshu.meshram

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