Answer:
Terms are 6, 8, 10 and 27, 8, -11
Step-by-step explanation:
Let the terms of AP be,
a,a+d,a+2d
Sum of three terms in AS is 24
S₃= a+a+d+a+2d= 3a+3d = 24
a+d = 8 —-(1)
When 1, 6, and 18 are added to them, the terms become
a+1, a+d+6=14, a+2d+18 = 26+d
If they are in GP, then the ratios of two consecutive terms should be same.
(26+d)/14 = 14 /(a+1)
(26+d)/14 = 14 /[(8-d)+1]
d²+17d-38 =0
d²+19d-2d-38 =0
d(d+19)-2(d+19)=0
d+19= or d-2 =0
d=-19 or d =2
Case 1: d=-19
a= 8-(-19) = 27
Terms are 27, 8, -11
Verification:
27+1= 28 ; 8+6=14 ; -11+18 = 7
7/14 =14/28 =1/2; Hence they ae in GP
Case 2: d=2
a = 8–2 =6
Terms are 6, 8, 10
Verification:
6+1= 7 ; 8+6=14 ; 10+18 =28
28/14 =14/7 =2 ; Hence they ae in GP
Ans: Terms are 6, 8, 10 and 27, 8, -11