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Question Number 17187 by virus last updated on 02/Jul/17

	Answered by Tinkutara last updated on 02/Jul/17

	$\frac{a_{1} + a_{2} + . . . + a_{p}}{a_{1} + a_{2} + . . . + a_{q}} = \frac{\frac{p}{2} [2 a_{1} + (p - 1) d]}{\frac{q}{2} [2 a_{1} + (q - 1) d]}$ $= \frac{p [2 a_{1} + (p - 1) d]}{q [2 a_{1} + (q - 1) d]} = \frac{p^{2}}{q^{2}} (Given)$ $2 a_{1} q + (p - 1) d q = 2 a_{1} p + (q - 1) d p$ $2 a_{1} (p - q) = d (p q - q - p q + p)$ $2 a_{1} = d \Rightarrow a_{1} = \frac{d}{2}$ $Now \frac{a_{6}}{a_{12}} = \frac{a_{1} + 5 d}{a_{1} + 11 d} = \frac{\frac{11 d}{2}}{\frac{23 d}{2}} = \frac{11}{23}$
	Answered by RasheedSoomro last updated on 03/Jul/17

	$AnOther way$ $\frac{a_{1} + a_{2} + . . . a_{p}}{a_{1} + a_{2} + . . . a_{q}} = \frac{p^{2}}{q^{2}}, \frac{a_{6}}{a_{12}} = ?$ $[W e require \frac{a_{1} + 5 d}{a_{1} + 11 d}]$ $\frac{\frac{p}{2} [2 a_{1} + (p - 1) d]}{\frac{q}{2} [2 a_{1} + (q - 1) d]} = \frac{p^{2}}{q^{2}}$ $\frac{a_{1} + (\frac{p - 1}{2}) d}{a_{1} + (\frac{q - 1}{2}) d} = \frac{p}{q}$ $Now let \frac{p - 1}{2} = 5 \Rightarrow p = 11$ $and \frac{q - 1}{2} = 11 \Rightarrow q = 23$ $S o$ $\frac{a_{6}}{a_{12}} = \frac{a_{1} + 5 d}{a_{1} + 11 d} = \frac{11}{23}$
	Answered by myintkhaing last updated on 03/Jul/17

	$another way$ $Let a_{1} + a_{2} + a_{3} + . . . + a_{p} = S_{p} = {kp}^{2} and$ $a_{1} + a_{2} + a_{3} + . . . + a_{q} = S_{q} = {kq}^{2}$ $So, S_{n} = {kn}^{2}$ $Thus$ $a_{6} = S_{6} - S_{5} = 11 k and a_{12} = S_{12} - S_{11} = 23 k$ $∴ \frac{a_{6}}{a_{12}} = \frac{11}{23}$
	Commented by virus last updated on 15/Jul/17

	$y e s t h a n k y o u$
	Commented by RasheedSoomro last updated on 03/Jul/17

	$More efficient way than that of mine!$
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