Question

The sum of the first three terms of a G.P. is 39/10 and their product is 1 .

Find the terms and the common ratio.​

Answer 1

Answer:

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Let,

The first three terms of the G. P. be

• a/r , a , ar

Then,

• a/r + a + are = 39/10 → (1)
• a/r.a.ar = 1 => a³ = 1 :. a = 1 → (2)

Substituting in (1) we get

• 1/r + 1 + 1r = 39/10
• 1 + r + r² / r = 39/10
• 10 (1 + r + r²) = 39r
• 10r² - 29r + 10 = 0

Factorizing , we get

• (2r - 5) (5r - 2) = 0
• r = 5/2 (OR) r = 2/5

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We therefore have two geometric progressions

When ,

• r = 5/2

the first three terms are 2/5 , 1 , 5/2

.・゜゜・ and .・゜゜・

When,

• r = 2/5

the first three terms are 5/2 , 1 , 2/5 .

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Answer 2

Explanation:

let first three terms a/r ,a , ar

a/c to question,

sum of first three terms = 39/10

⇒a/r + a + ar = 39/10

⇒a[1/r + 1 + r] = 39/10 ........(1)

again, product of first three terms = 1

⇒a/r × a × ar = 1

⇒a³ = 1 = 1³

⇒a = 1 , putting in equation (1),

we get, 1[1/r + 1 + r] = 39/10

⇒(1 + r + r²)/r = 39/10

⇒10(1 + r + r²) = 39r

⇒10 + 10r + 10r² = 39r

⇒10r² - 29r + 10 = 0

⇒10r² - 25r - 4r + 10 = 0

⇒5r(2r - 5) - 2(2r - 5) = 0

⇒(5r - 2)(2r - 5) = 0

⇒r = 2/5 , 5/2

now, for a=1 and r = 2/5

three terms are ; 5/2, 1 , 2/5 

now , a= 1 and r= 5/2

three terms are ; 2/5,1,5/2