Question

)Find four consecutive terms in an A.P. such that the sum of the middle two terms is 18 and the product of the two end terms is 45.(3M)​

Answer 1

Answer:

let 4 terms of A.P. are (a-d), a, (a+d), (a+2d)

sum of second and third terms = 2a+d = 18 or a = 9-(d/2) ........(1)

product of first and fourth terms are (a-d)(a+2d) = 45 .............(2)

let us substitute value of a from eqn.(1) in eqn (2) , [ 9 -(3/2)d ] [ 9 + (3/2)d ] = 81 - (9/4)d2 = 45 or d2 = 16 or d = 4

Let us substitute d value in eqn.(1) to get value of a. we get a = 7

4 terms of A.P. are 3, 7, 11, 15

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