Question

three positive numbers form a Geometric progression.If the middle number of is increased by 8 the three numbers form an Arithmetic progression of the last number is also increased by 64 along with the previous increase in the middle number the resulting numbers form a Geometric progression then find the first term and the common ratio​

Answer 1

Answer:

Step-by-step explanation:

Let the three numbers be  

r

a

​  

,aandar

Now A/Q    2(a+8)=(  

r

a

​  

+ar)

Also A/Q    (a+8)  

2

=  

r

a

​  

×(ar+64)

Solving the above two equations :

From the 2nd equation a  

2

+64+16a=a  

2

+  

r

64a

​  

 

⇒4+a=  

r

4a

​  

 

⇒a=  

4−r

4r

​  

 

Substituting this in 1st equation

2(  

4−r

4r

​  

+8)=(  

4−r

4

​  

+  

4−r

4r  

2

 

​  

)

⇒16(4−r)=4r  

2

−8r+4

⇒r  

2

−2r+1=16−4r

⇒r  

2

+2r−15=0

⇒r=3,−5

anda=  

4−r

4r

​  

=12or  

9

−20

​  

 

but the second term is neglected because according to question the numbers are positive.

Hence common ratio=3 and  

first term =  

r

a

​  

=  

3

12

​  

=4