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  • The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n.

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n.

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\\ $Given: a = 8, d\textsubscript{1} $= 20 for first AP\\ b = - 30, d\textsubscript{2} = 8 $for second AP\\$ $ Given $ S\textsubscript{n} = S\textsubscript{2n}\\ \frac{\mathrm{n}}{2}\left[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}_{1}\right]=\frac{2 \mathrm{n}}{2}\left[2 \mathrm{~b}+(2 \mathrm{n}-1) \mathrm{d}_{2}\right] \\ \frac{\mathrm{n}}{2}[2(8)+(\mathrm{n}-1) 20]=\mathrm{n}[2(-30)+(2 \mathrm{n}-1)(8)] \\ \frac{\mathrm{n}}{2}[16+20 \mathrm{n}-20]=\mathrm{n}[-60+16 \mathrm{n}-8] \\ \frac{\mathrm{n}}{2}[20 \mathrm{n}-4]=\mathrm{n}[-68+16 \mathrm{n}] \\ \frac{2(10 \mathrm{n}-2)}{2}=-68+16 \mathrm{n} 10n - 16n = - 68 + 2\\ -6n = - 66\\ n = \frac{+66}{+6}=11 \\ n = 11\\

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