If the sum of three consecutive terms of GP are k-4, k, k+5. Find k

k = 20

Step-by-step explanation:

t1 = a = k-4

a = k - 4

t2 = ar = k

(k-4) r = k - 4

r = k / (k-4)

t3 = ar^2 = k + 5

(k - 4) (k/k - 4) ^2 = k+5

k^2/k-4 = k+5

k^2 = (k+5) (k-4)

k^2 = k^2 - 4k + 5k - 20

0 = k - 20

k = 20