Question

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

Answer 1

GEOMETRIC PROGRESSION :
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)

a1 = a , r = a(n+1)/ an

General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........
The three consecutive terms in an GP may be taken as a/r,a,ar.

SOLUTION :
Let the first three terms are a/r,a,ar
GIVEN :
Product of three terms = 216
sum of their product in pairs = 156

Product of three terms = 216
(a/r) x a x a r = 216
a³ = 6³
a = 6

sum of their product in pairs = 156
(a/r) x a + (a) x (ar) + (a r) x (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
3 r² + 3 r + 3 - 13 r = 0
3 r² - 10 r + 3 = 0
[By middle term splitting]
3r² - 1r - 9r +3= 0
r(3r - 1) -3(3r -1)= 0
(3r -1) (r-3) = 0
3 r - 1 = 0 or r - 3 = 0
r = 1/3 or r = 3

a/r = 6/(1/3) or a/r = 6/3
a/r = 6x(3/1)       a/r= 2
a/r = 18

a = 6 or a = 6
ar = 6(1/3) or ar = 6(3)
ar = 2 or ar = 18

Hence, the three terms are 18,6,2 or 2,6,18

HOPE THIS WILL HELP YOU...

Answer 2

Solution :

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

in G.P

i ) Product of terms = 216

( a/r ) × a × ( ar ) = 216

=> a³ = 6³

Therefore ,

a = 6----( 1 )

ii ) sum of products in pairs = 156

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

=> a²/r + a²r + a² = 156

=> a² ( 1/r + r + 1 ) = 156

=> 6² [( 1 + r² + r )/r ]= 156

=> ( r² + r + 1 ) = 156r/36

=> r² + r + 1 = 13r/3

=> 3( r² + r + 1 ) = 13r

=> 3r² + 3r + 3 - 13r = 0

=> 3r² - 10r + 3 = 0

=> 3r² - 9r - 1r + 3 = 0

.

=> 3r( r - 3 ) - 1 ( r - 3 ) = 0

=> ( r - 3 )( 3r - 1 ) = 0

r - 3 = 0 or 3r - 1 = 0

r = 3 or r = 1/3

ii ) if a = 6 , r = 3

Required three terms are

6/3 , 6 , 6 × 3

2 , 6 , 18

iii )

If a = 6 , r = 1/3 then

Required 3 terms are

18 , 6 , 2

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