Visual SolutionPYQ 2023 · Apr 13 Shift 2Tricky

Question

Let

a_1, a_2, a_3, \ldots

be a G.P. of increasing positive numbers. Let the sum of its

6^{\text{th}}

and

8^{\text{th}}

terms be 2 and the product of its

3^{\text{rd}}

and

5^{\text{th}}

terms be

\frac{1}{9}

. Then

6(a_2 + a_4)(a_4 + a_6)

is equal to

(A)

3

(B)

3\sqrt{3}

(C)

2

(D)

2\sqrt{2}

Solution Path

From

ar^3 = \frac{1}{3}

and

r^2 = 2

a_2 + a_4 = \frac{1}{2}

a_4 + a_6 = 1

. Answer:

6 \cdot \frac{1}{2} \cdot 1 = 3

01Question Setup

1/4

GP of increasing positive numbers.

a_6 + a_8 = 2

a_3 \cdot a_5 = \frac{1}{9}

. Find

6(a_2 + a_4)(a_4 + a_6)

6(a_2 + a_4)(a_4 + a_6) = \;?

02Find a and r

2/4

From

a^2r^6 = \frac{1}{9}

, get

ar^3 = \frac{1}{3}

. Substituting into the sum condition gives

r^4 + r^2 - 6 = 0

, so

r^2 = 2

r^2 = 2

03Compute ExpressionKEY INSIGHT

3/4

a_2 + a_4 = \frac{1}{2}

and

a_4 + a_6 = 1

. So

6 \cdot \frac{1}{2} \cdot 1 = 3

6 \cdot \dfrac{1}{2} \cdot 1 = 3

04Final Answer

4/4

6(a_2 + a_4)(a_4 + a_6) = 3

\boxed{3}

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