Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

The correct answer is (c) 11/41

(a1+a2+…ap.)/(a1+a2+…aq) = p2/q2

(p/2)[(2a1+(p-1)d)] / (q/2)(2a1+(q-1)d) = p2/q2

(2a1+(p-1)d) / (2a1+(q-1)d) = p/q

Cross multiplying

q(2a1+(p-1)d) = (2a1+(q-1)d)p

2a1(p-q) = (pq-q-qp+p)d

2a1(p-q) = (p-q)d

d = 2a1

a6/a21 = (a1+5d)/(a1+20d)

= (a1+10a1)/(a1+40a1)

= 11a1 / 41a1

= 11/41

Answer 2

EXPLANATION.

a₁, a₂, a₃, . . . . . be terms of an A.P.

$\implies \dfrac{a_{1} + a_{2} + . . . . . + a_{p}}{a_{1} + a_{2} + . . . . . + a_{q}} = \dfrac{p^{2} }{q^{2} } , \ \ p \ne q$

As we know that,

$\implies \dfrac{\dfrac{p}{2} \bigg[ 2a_{1} + (p - 1) d \bigg]}{\dfrac{q}{2} \bigg[ 2a_{1} + (q - 1) d \bigg]} \ = \dfrac{p^{2} }{q^{2} }$

⇒ (2a₁ + (p - 1)d)/(2a₁ + (q - 1)d) = p/q.

⇒ [2a₁ + (p - 1)d]q = [2a₁ + (q - 1)d]p.

⇒ [2a₁ + (pd - d)]q = [2a₁ + (qd - d)]p.

⇒ 2a₁q + pqd - dq = 2a₁p + pqd - dp.

⇒ 2a₁q + pqd - dq - 2a₁p - pqd + dp.

⇒ 2a₁(q - p) = d[(q - 1)p - (p - 1)q].

⇒ 2a₁(q - p) = d(q - p).

⇒ d = 2a₁.

⇒ a₆/a₂₁  = a₁ + 5d/a₁ + 20d.

⇒ a₁ + 5(2a₁)/(a₁ + 20(2a₁)) = 11a₁/41a₁ = 11/41.

Option [C] is correct answer.