First find the common ratio r
r= (x+10)/(x+2) = (x+32)/(x+10)
Cross-multiplying: (x+10)(x+10) = (x+2)(x+32)
Solving this, x=2
Hence 1st term is: x+2 = 4
2nd term: x+10 = 12
Hence Common Ratio r = 12/4 = 3
1st term = 4
Common Ratio r = 3
X=2. First 3terms are 4,12,36
T_1=a=x+2 ------(1)
T_2=ar=x+10------(2)
T_3=〖ar〗^2=x+34------(3)
Now
((3))/((2)
)→〖ar〗^2/ar=(x+34)/(x+10)→r=(x+34)/(x+10)------(4)
((2))/((3))→ar/a=(x+10)/(x+2) →
r=(x+10)/(x+2)------(5)
Now (4) =(5)
(x+34)/(x+10)=(x+10)/(x+2)→(x+34)(x+2)=(x+10)(x+10)
x^2+36x+68=x^2+20x+100
36x-20x=100-68
16x=32
x=2
T_1=x+2=2+2=4
r=(x+10)/(x+2)=(2+10)/(2+2)=12/4=3
கருத்திடுகை
First find the common ratio r
r= (x+10)/(x+2) = (x+32)/(x+10)
Cross-multiplying: (x+10)(x+10) = (x+2)(x+32)
Solving this, x=2
Hence 1st term is: x+2 = 4
2nd term: x+10 = 12
Hence Common Ratio r = 12/4 = 3
1st term = 4
Common Ratio r = 3
X=2. First 3terms are 4,12,36
T_1=a=x+2 ------(1)
T_2=ar=x+10------(2)
T_3=〖ar〗^2=x+34------(3)
Now
((3))/((2) )→〖ar〗^2/ar=(x+34)/(x+10)→r=(x+34)/(x+10)------(4)
((2))/((3))→ar/a=(x+10)/(x+2) → r=(x+10)/(x+2)------(5)
Now (4) =(5)
(x+34)/(x+10)=(x+10)/(x+2)→(x+34)(x+2)=(x+10)(x+10)
x^2+36x+68=x^2+20x+100
36x-20x=100-68
16x=32
x=2
T_1=x+2=2+2=4
r=(x+10)/(x+2)=(2+10)/(2+2)=12/4=3