Question

Question 12 The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Class X1 - Maths -Sequences and Series Page 192

Answer 1

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

Answer 2

Answer:

The first term is a=1 and the common ratio are $r=\frac{2}{5},5$

Step-by-step explanation:

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

To find : The common ratio and the terms?

Solution :

Let the three term of a G.P series is $\frac{a}{r},a,ar$

Where, r is the common ratio

According to question,

The sum of first three terms of a G.P. is $\frac{39}{10}$

So, $\frac{a}{r}+a+ar=\frac{39}{10}$  ......(1)

The product of first three terms of a G.P. is 1.

$\frac{a}{r}\times a\times ar=1$  ......(2)

Solving equation (2)

$a^3=1$

$a=1$

Now, substitute the value of a in equation (1)

$\frac{a}{r}+a+ar=\frac{39}{10}$

$\frac{1}{r}+1+(1)r=\frac{39}{10}$

Taking LCM,

$\frac{1+r+r^2}{r}=\frac{39}{10}$

$10r^2+10r+10=39r$

$10r^2-29r+10=0$

$10r^2-25r-4r+10=0$

$5r(r-5)-2(r-5)=0$

$(5r-2)(r-5)=0$

$5r-2=0,r-5=0$

$r=\frac{2}{5},r=5$

Therefore, The first term is a=1 and the common ratio are $r=\frac{2}{5},5$