Let a1-a2-a3- be an A-P- such that a 1 - a 2 - a p a 1 - a 2 - a 3 - a q - p 3 q 3 -p- q- Then a 6 a 21 is equal to-

The correct option is B 121 1681 a_1 + a_2 + ...... + a_pa_1 + a_2 + a_3 + ...... + a_q = p^3q^3 ⇒p2[ 2a + ( p 1)d]q2[ 2a + ( q 1)d] = p^3q ...

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Sum of n Terms

Let a1,a2,a...

Question

Let $a_{1}, a_{2}, a_{3}, .$ be an $A . P .$ such that $\frac{a_{1} + a_{2} + . . . . + a_{p}}{a_{1} + a_{2} + a_{3} + . . . . + a_{q}} = \frac{p^{3}}{q^{3}}; p \neq q$ . Then $\frac{a_{6}}{a_{21}}$ is equal to:

\frac{121}{1861}

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\frac{11}{41}

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\frac{121}{1681}

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\frac{41}{11}

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a_{1}, a_{2}, a_{3} \dots

I f \frac{a_{1} + a_{2} + \dots a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q,

\frac{a_{6}}{a_{21}}

a_{1}, a_{2}, a_{3} \dots

I f \frac{a_{1} + a_{2} + \dots a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q,

\frac{a_{6}}{a_{21}}

a_{1}, a_{2}, a_{3}, . . . .

\frac{a_{1} + a_{2} + . . . . + a_{p}}{a_{1} + a_{2} + . . . . + a_{q}} = \frac{p^{2}}{q^{2}}

p \neq q

\frac{a_{6}}{a_{21}}

a_{1}, a_{2}, a_{3} . . . .

\frac{a_{1} + a_{2} + . . . . + a_{p}}{a_{1} + a_{2} + . . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q

\frac{a_{6}}{a_{21}}

a_{1}, a_{2}, a_{3}, \dots

\frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{2}}{q^{2}} (p \neq q)

\frac{a_{6}}{a_{21}}

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