Let a_1 ,a_2 ,a_3 … be an A.P. If a_7=3, the product a_1a_4 is minimum and the sum of its first n terms is zero, then n !-4a_n(n+2) is equal to?

Question

Accepted Answer

Correct answer: A. (a) Solution a+6d=3 ...................(1) z=a (a+3d ) =(3-6x) (3-3d ) =18d^2-27d+9 Differentiating with respect to d ⇒36d-27=0 ⇒d= 34 , from (1 )a= -32, (Z=minimum) Now, S_n= n2(-3+(n-1) 34)=0 ⇒n=5 n !-4a_n(n+2)=120-4 (a_25 ) =120-4(a+(35-1)d) =120-4( -32+34. ( 34 )) =120-4 ( -6+1024 ) =120-96=24

$Let a_{1}, a_{2}, a_{3} \dots be an A.P. If a_{7} = 3, the product a_{1} a_{4} is minimum and the sum of its first n terms is zero, then n! - 4 a_{n (n + 2)} is equal to?$

Solution

More Sequence & Series PYQs