Let a_1, a_2, a_3, be terms of an A.P. If a_1 + a_2 + + a_p/a_1 + a_2 + + a_q = p^2/q^2 , p ≠ q , then a_6/a_21 equals: - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

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Let a_{1}, a_{2}, a_{3}, \dots be terms of an A.P. If \frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q, then \frac{a _{6}}{a _{21}} equals:

Solution & Step-by-step Explanation

Let S_{n} denote the sum of n terms. \frac{S _{p}}{S _{q}} = \frac{p ^{2}}{q ^{2}} . \frac{\frac{p}{2} [ 2 a + ( p - 1 ) d ]}{\frac{q}{2} [ 2 a + ( q - 1 ) d ]} = \frac{p ^{2}}{q ^{2}} ⟹ \frac{2 a + ( p - 1 ) d}{2 a + ( q - 1 ) d} = \frac{p}{q} To find \frac{a _{6}}{a _{21}} = \frac{a + 5 d}{a + 20 d} = \frac{2 a + 10 d}{2 a + 40 d} .Comparing with the sum ratio formula, we need p - 1 = 10 ⟹ p = 11 and q - 1 = 40 ⟹ q = 41 . \frac{a _{6}}{a _{21}} = \frac{p}{q} = \frac{11}{41} .

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Let a_{1}, a_{2}, a_{3}, \dots be terms of an A.P. If \frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q, then \frac{a _{6}}{a _{21}} equals:

41/11

7/2

2/7

11/41

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