Question

how many terms of AP 20,58/3,56/3 should be taken to get the sum as 300

Answer 1

Answer:

AP : 20. 19 1/3 ,18 2/3................

19 1/3 and 18 2/3 are Mixed Numbers :=19 1/3= 19 + (1/ 3) =58 / 3

= 18 2/3 =18+(2 /3) = 56/ 3

So AP :20, 58 / 3 , 56 / 3..............

First term (a) = 20

Common Difference (d) = (58 / 3) - 20 = (-2 / 3)

Sum of n term (S n) = 300

Apply :Sum of n term (S n) = [n/2] *(2a +(n-1)*d )

300 =[n/2] * ( 2*{20} +(n-1)*{-2 / 3} )

300 *2= n (40 + [- 2 / 3] n- [-2 / 3] )

600 = n (40 *3 - 2 n +2) / 3

1800 = n (120 +2 - 2n )

1800 = n (122 - 2n)

Divide the whole equation by "2"

900 = n ( 61 - n )

n2 - 61n + 900 = 0

n2 +(- 61n) + 900 = 0

Apply Splitting the middle term :900 = -36 *-25

n2 +(-36n - 25n) + 900 = 0

n ( n - 36)-25 (n - 36) = 0

So :(n-36) * (n - 25) = 0

= ( n -36 )= 0OR( n - 25