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JEE Mains · Maths · STD 11 - 8. sequence and series

જો \({A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ..... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}\)  અને \(B_n \,= 1 - A_n\) હોય તો \(p\) ની ન્યુનત્તમ અયુગ્મ કિમત મેળવો કે જેથી બધા \(n \geq p\) \({B_n} > {A_n}\) માટે થાય 

  1. A \(5\)
  2. B \(7\)
  3. C \(11\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(7\)

Step-by-step Solution

Detailed explanation

\({A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ...... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}\) Which is a \(G.P.\) with \(a = \frac{3}{4}'r = \frac{{ - 3}}{4}\) and number of terms \(=n\)…
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