FIND THE TERMS FROM SUM AND PRODUCT OF 3 CONSECUTIVE TERMS OF GP

In this section, we will learn how to find the terms from sum and product of 3 consecutive terms of a geometric progression.

We can consider the first three terms of a geometric progression as 

a/r, a, ar

Example 1 :

If the product of three consecutive terms in G.P is 216 the sum of their product in pairs is 156, find them.

Solution :

Let the first three terms are a/r, a, ar

Product of three terms  =  216

(a/r) ⋅ a ⋅ a r = 216

a³  =  6³  

a  =  6

Sum of their product in pairs  =  156

(a/r) ⋅ a  +  a ⋅ ar  +  ar ⋅ (a/r)  =  156

a² / r  +  a²  r  +  a²  =  156

a² [ (1/r) + r + 1 ]  =  156

a² [ (1+r²+r)/r]  =  156

a² (r²+r+1)/r  =  156

(6²/r)(r²+r+1)  =  156

(r²+r+1)/r  =  156/36

(r²+r+1)/r  =  13/3

3(r²+r+1)  =  13 r

3r² + 3r + 3 - 13r  =  0

3r² - 10r + 3  =  0

(3r - 1)(r - 3)  =  0

If a  =  6, then r  =  1/3

a/r  =  6/(1/3) ==> 18

a  =  6

ar  =  6(1/3)  ==> 2

If a  =  6, then r  =  1/3

a/r  =  6/3 ==> 2

a  =  6

ar  =  6(3)  ==> 18

Hence the required three terms are 18, 6 and 2 or 2, 6, 18.

Example 2 :

Find the first three consecutive terms in G.P whose sum is 7 and the sum of their reciprocals is 7/4.

Solution :

Let the first three terms are a/r, a, ar

Sum of three terms  =  7

a/r + a + a r  =  7

a [(1/r) + 1 + r]  =  7

[(1/r) + 1 + r]  =  7/a -----------(1)

Sum of their reciprocals  =  7/4

(r/a) + (1/a)  + (1/ar)  =  7/4

 (1/a)[r + 1 + (1/r)]  =  7/4

 (1/a)(7/a)  =  7/4

 7/a²  =  7/4

a²  =  4

a  =  2

By applying a  =  2 in the in (1), we get 

[(1/r) + 1 + r]  =  7/2

(1 + r + r²)/r  =  7/2

 2r² + 2r + 2  =  7r

 2r² + 2r - 7r + 2  =  0

 2r² - 5r + 2  =  0

(2r - 1) (r - 2)  =  0

If a  =  2 and r  =  1/2

a/r  =  2/(1/2)  ==>  4 

a  =  2

ar  =  2(1/2)  ==>  1

Therefore the three terms are  4, 2 , 1 or 1 , 2 , 4

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