The numbers p,10 and q are 3 consecutive terms of an arithmetic progression .the numbers p,6 and q are 3 consecutive terms of a geometric progression .by first forming two equations in p and q show that p^2-20p+36=0

Question

The numbers p,10 and q are 3 consecutive terms of an arithmetic progression .the numbers p,6 and q are 3 consecutive terms of a geometric progression .by first forming two equations in p and q show that p^2-20p+36=0
Hence find the values of p and q for which the geometric progression converges
Help please

Reiny · Accepted Answer

from the first:
10-p = q-10
q = 20-p

from the second:
pq = 36

by substitution,
p(20-p) = 36
20p - p^2 = 36
p^2 - 20p + 36 = 0 ----> thus shown

(p - 18)(p - 2) = 0
p = 18 or p = 2
then q = 36/18 = 2 or p = 36/2 = 18

for p = 18, q = 2
the AS is 18, 10 and 2
the GS is 18, 6, 2 -----> which converges, r = 1/2

for p = 2, q = 18
the AS is 2, 10, 18
the GS is 2, 6, 18, ---> diverges , r > 1

Nakiv · Answer

i want specifically for P,6andQ