Question

he 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

Let the first term be 'a'

Let the common difference be 'r'

The geometric progression sequence is as below

a, ar, ar², ar³,..........,ar^(n-1)

The first three terms are a, ar, ar²

Given,

a(ar)(ar²) = 1

=> a³r³ = 1

=> (ar)³ = 1

=> ar = 1

a+ar+ar² = 39/10

=> (1/r) + 1 + (1/r)r² = 39/10

=> 1/r + 1 + r = 39/10

=> 1/r + r = 39/10 - 1

=> r²+1/r = 39-10/10

=> r²+1/r = 29/10

=> 10r²+10 = 29r

=> 10r²-29r+10 = 0

Applying the formula to find from quadratic equation, we get r = 5/2  and r = 2/5

Condition 1 (r = 5/2)

ar = 1 => a = 1/r = 2/5

First term = a = 2/5

Second term = ar = 1

Third term = ar² = 1(5/2) = 5/2

Condition 2 (r = 2/5)

ar = 1 => a = 1/r = 5/2
 
First term = a = 5/2

Second term = ar = 1

Third term = ar² = 1(2/5) = 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

Substituting a = 1 in the (1) equation 

a/r + a + ar = 39/10

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

= 2/5                     = 5/2

a = 1                    a = 1

ar = 1*(5/2)        ar = 1*(2/5)

= 5/2                      = 2/5

The three terms are (2/5), 1, (5/2)  or (5/2), 1, (2/5)

Answer.