Let a_1 ,a_2 ,a_3 , …. be a G.P. of increasing positive numbers. If a_3a_5=729 and a_2+a_4= 1114, then 24(a_1+a_2+a_3) is equal to

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Correct answer: B. (b) We start with the given sequence a_1 ,a_2 ,a_3 , … of an increasing geometric progression (G.P.). Two key conditions are provided: a_3 a_5=729a_2+a_4= 1114 Calculating using given conditions: Using the sequence terms: a_3=a r^2 ,a_5=a r^4 Then: a_3 a_5=(a r^2)(a r^4)=a^2 r^6=729=272 It follows: a r^3=27 ⇒a_4=a r^3=27…..(i) Using the second condition: a_2+a_4= 1114 Substituting a_4=27 : a_2= 1114-27a_2=a r= 34 … … ( ii ) Solving for r and a : From equation (i) and (ii), we derive r^2 : r^2= 4 273=4 9=36 Since the G.P. is increasing, r=6. Solve for a : Substituting r=6 into a r= 34 : a 6= 34 ⇒a= 324= 18 Calculating the requested sum: The task is to find: 24(a_1+a_2+a_3)=24(a+ ar+ar^2) Substitute the values: =24 18(1+6+6^2)=3 (1+6+36)=3 43=129 Therefore, the value is 129.

$Let a_{1}, a_{2}, a_{3}, \dots . be a G.P. of increasing positive numbers. If a_{3} a_{5} = 729 and a_{2} + a_{4} = \frac{111}{4}, then 24 (a_{1} + a_{2} + a_{3}) is equal to$

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