First slide

Arithmetic progression

Question

$Let a_{1}, a_{2}, a_{3} \dots be an A.P such that \frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{3}}{q^{3}}, p \neq q then \frac{a_{6}}{a_{21}} is$

Moderate

A

$\frac{41}{11}$
B

$\frac{31}{121}$
C

$\frac{11}{41}$
D

$\frac{121}{1861}$

Solution

$\begin{array}{l} take p=2,q=1 then the given equation becomes= \frac{a_{1} + a_{2}}{a_{1}} = 8 \Rightarrow \frac{a_{1} + (a_{1} + d)}{a_{1}} = 8 \\ \Rightarrow d = 6 a_{1} \\ Now \frac{a_{6}}{a_{21}} = \frac{a_{1} + 5 d}{a_{1} + 20 d} = \frac{a_{1} + 5 \times 6 a_{1}}{a_{1} + 20 \times 6 a_{1}} = \frac{1 + 30}{1 + 120} = \frac{31}{121} (n' t h t e r m i n A . P = a + (n - 1) d) \end{array}$

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