Question-17187

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} + \dots + a_{p}}{a_{1} + a_{2} + \dots + 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} + \dots a_{p}}{a_{1} + a_{2} + \dots 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} + \dots + a_{p} = S_{p} = {kp}^{2} and

a_{1} + a_{2} + a_{3} + \dots + 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!