Question

Question:

-16, 8, -4, 2, ….., A.M and G.M of pth and qth terms are roots of 4x²-9x+5=0
then p+q = ?



Answer: 10.00​

Answer 1

Answer:

4x2 - 9x + 5 = 0

⇒ x = 1, 5/4

Now given 5/4 =tp+tq2,tp+tq2, tp tq where

tr = -16 (-1/2)r-1

so 5/4 = -8[(-1/2)p-1+ (-1/2)q-1]

1 = 256(-1/2)p+q-2

⇒ 2p+q-2 = (-1)p+q-2 28

hence p + q =10

Step-by-step explanation:

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Answer 2

ANSWER:

• The value of p + q = 10.

GIVEN:

• Series is -16, 8, -4, 2……

• A.M and G.M of pth and qth terms are roots of 4x² - 9x + 5 = 0.

TO FIND:

• The value of p + q.

EXPLANATION:

$\boxed{ \bold{ \large{ \red{A.M = \dfrac{T_p + T_q}{2}}}}} \\ \\ \\ \boxed{ \bold{ \large{ \blue{ G.M = \sqrt{T_pT_q}}}}} \\ \\ \\ \sf \orange{\dashrightarrow Also \ A.M \geq G.M.} \\ \\ \\\leadsto\sf 4x^{2} - 9x + 5 = 0. \\ \\ \\ \leadsto\sf 4x(x) - 4x - 5x + 5 = 0 \\ \\ \\ \leadsto\sf 4x(x - 1) - 5x(x - 1) = 0 \\ \\ \\ \leadsto\sf (x - 1)(4x - 5) = 0 \\ \\ \\ \sf\leadsto x = 1 \ and \ x = \dfrac{ 5}{4} \\ \\ \\ \sf G.M = 1(A.M \geq G.M)$

$\boxed{ \bold{ \large{ \red{T_n= ar^{n-1}}}}} \\ \\ \\ \leadsto \sf a = - 16 \ and \ r = - \dfrac{1}{2} \\ \\ \\ \leadsto \sf T_p= - 16 \bigg(- \dfrac{1}{2} \bigg)^{p-1} \\ \\ \\ \leadsto \sf T_q= - 16 \bigg(- \dfrac{1}{2} \bigg)^{q-1} \\ \\ \\ \sf \leadsto \sqrt{256 \bigg(- \dfrac{1}{2} \bigg)^{p +q -2} } = 1 \\ \\ \\ \sf \leadsto 16 \sqrt{ \bigg(- \dfrac{1}{2} \bigg)^{p +q -2} } = 1 \\ \\ \\ \sf \green{ \star \ Squaring \ on \:both \ sides.} \\ \\ \\ \sf \leadsto \bigg(- \dfrac{1}{2} \bigg)^{p +q -2} = \dfrac{1}{256} \\ \\ \\ \sf \leadsto 2^{p +q -2} = {2}^{8} \\ \\ \\ \sf \pink{\star\ Bases \ are \ equal \ so, \ equate \ the\ powers.}\\ \\ \\ \sf \leadsto p +q -2 = 8 \\ \\ \\ \sf \leadsto p +q = 10$

The value of p + q = 10.