Question

the sum of three terms of a geometric sequence is 39/10 and their product is 1. find the common ratio and their terms

Answer 1

When
a = the first term
r = the common ratio

then the first 3 terms can be written as;

a, ar, ar²

The sum of the first 3 terms = 39/10

So we have; 

a+ar+ar² =39/10 ………(i) 

The product of the first 3 terms = 1

So we have;

 a*ar*ar² = 1……(ii) 

Simplifying this, we get

a³*r³=1 

(ar)³=1 

 
ar=1 

a = 1/r……(iii) 

Substituting this in (i), we get 

a+ar+ar²=39/10 

1/r+1/r *r +1/r*r²=39/10 

1/r+1+r=39/10 1/r + r/r +r²/r = 39/10

(1+r+r²)=39/10 * r 

10(1+r+r²)=39r 

10r²+10r+10=39r 

10r^2-29r+10= 0 

Use the quadratic formula to find the value of ‘r’ . 

r= -(-29)+-√((-29)^2-4*10*10) / 2*10 

r= 29+-√(441) / 20 

r=29+-21 / 20 

So, r= 29+21 / 20 

or r=29-21/20 

r= 5/2 or r=2/5

a = 1/r……(iii)

when r = 5/2, a = 2/5

and when r = 2/5, a= 5/2

the first 3 terms can be

2/5, 1, 5/2

OR

5/2, 1, 2/5

Answer 2

Solution :-

Let the first three terms be a/r, a and ar

Sum of three terms = 39/10

Product of three terms = 1

⇒ (a/r) + a + ar = 39/10 ...............(1)

⇒ (a/r)*a*ar = 1

⇒ (a*a*a*r)/r = 1

⇒ a³ = 1

a = ∛1

a = 1

Substituting a = 1 in (1)

(a/r + a + ar) = 39/10

⇒ (1/r + 1 + 1*r) = 39/10

⇒ (1 + r + r²)/r = 39/10

⇒ 10(1 + r + r²) = 39r

⇒ 10 + 10r + 10r² = 39r

⇒ 10r² + 10r - 39r + 10 = 0

⇒ 10r² - 29r + 10 = 0

Solving it we get.

⇒ (2r - 5)  (5r - 2) = 0

⇒ 2r = 5     or  5r = 2

⇒ r = 5/2    or r = 2/5
 
a/r = 1/(5/2)  or          a/r = 1/(2/5)

a/r = 2/5       or          a/r = 5/2    

a = 1                         a = 1

ar = 1*(5/2)              ar = 1*(2/5)

ar = 5/2                    ar = 2/5

So, three terms are (2/5), 1, (5/2) or (5/2), 1. (2/5)

Answer.