Question

K is a positive integer such that 36+k,300+k,596+k are the square of three consecutive terms of an ap.Find k

Answer 1

Answer:

Step-by-step explanation:

The  three expressions are squares of consecutive terms of an AP ,then  

√(36 + k), √(300 + k), and √(596 + k)  

hence √(300 +k) - √(36 +k) = √(596 + k) - √(300 + k)  

[2 √(300 + k) = √(596 + k) + √(36 + k)]^2  

4(300 + k) = 596 + k + 2 √[(596 + k)(36 + k)] + 36 + k  

1200 + 4k = 632 + 2k + 2 √[(596 + k)(36 + k)]  

{568 + 2k = 2 √[(596 + k)(36 + k)] } / 2  

{284 + k = √[(596 + k)(36 + k)]}^2  

80656 + 568k + k^2 = 21456 + 632k + k^2  

59200 = 64k  

k = 925  

Thus, k = 925