8. The sum of three terms of a geometric sequence is 39\10 and their product is 1. Find the
common ratio and the terms
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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:
Sum of three terms = 39/10
Product of three terms = 1
(a/r) + a + a r = 39/10 ………..(1)
(a/r) x a x a r = 1
a³ = 1³
a = 1
On Putting a = 1 in equation 1
(a/r) + a + a r = 39/10
(1/r) + 1 + 1 × r = 39/10
1/r) + 1 + r = 39/10
(1 + r + r²)/r = 39/10
[Taking LCM r ]
10(1 + r + r²) = 39 r
10 + 10r + 10r² = 39r
10 r² + 10 r - 39 r + 10 = 0
10 r² - 29 r + 10 = 0
10 r² -25r -4r +10 = 0 [by middle term splitting]
5r (2r -5) - 2(2r - 5) =0
(2r - 5) (5r - 2) = 0
2 r - 5 = 0 or 5r - 2 = 0
2 r = 5 or 5 r = 2
r = 5/2 or r = 2/5
a/r = 1/(5/2) [r = 5/2]
a/r= 2/5
a/r = 1/(2/5) [r = ⅖]
a/r = 5/2
ar = 1× (5/2) [ a = 1 , r = 5/2 ]
ar = = 5/2
ar = 1×(2/5) [ a = 1 ,r= 2/5 ]
ar = 2/5
Hence, the three terms are (2/5),1,(5/2) or (5/2),1,(2/5)
HOPE THIS WILL HELP YOU….
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:
Sum of three terms = 39/10
Product of three terms = 1
(a/r) + a + a r = 39/10 ………..(1)
(a/r) x a x a r = 1
a³ = 1³
a = 1
On Putting a = 1 in equation 1
(a/r) + a + a r = 39/10
(1/r) + 1 + 1 × r = 39/10
1/r) + 1 + r = 39/10
(1 + r + r²)/r = 39/10
[Taking LCM r ]
10(1 + r + r²) = 39 r
10 + 10r + 10r² = 39r
10 r² + 10 r - 39 r + 10 = 0
10 r² - 29 r + 10 = 0
10 r² -25r -4r +10 = 0 [by middle term splitting]
5r (2r -5) - 2(2r - 5) =0
(2r - 5) (5r - 2) = 0
2 r - 5 = 0 or 5r - 2 = 0
2 r = 5 or 5 r = 2
r = 5/2 or r = 2/5
a/r = 1/(5/2) [r = 5/2]
a/r= 2/5
a/r = 1/(2/5) [r = ⅖]
a/r = 5/2
ar = 1× (5/2) [ a = 1 , r = 5/2 ]
ar = = 5/2
ar = 1×(2/5) [ a = 1 ,r= 2/5 ]
ar = 2/5
Hence, the three terms are (2/5),1,(5/2) or (5/2),1,(2/5)
HOPE THIS WILL HELP YOU….
Answered by
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Solution :
Let a/r , a , ar are three terms in
G.P .
i ) Their product = 1
=> ( a/r ) × a × ar = 1
=> a³ = 1
a = 1 ----( 1 )
ii ) Sum of the three terms = 39/10
=> a/r + a + ar = 39/10
=> a( 1/r + 1 + r ) = 39/10
=> ( 1 + r + r² )/r = 39r/10 [ from ( 1 ) ]
=> 10( 1 + r + r² ) = 39r
=> 10 + 10r + 10r² - 39r = 0
=> 10r² - 29r + 10 = 0
=> 10r² - 4r - 25r + 10 = 0
=> 2r( 5r - 2 ) - 5( 5r - 2 ) = 0
=> ( 5r - 2 )( 2r - 5 ) = 0
5r - 2 = 0 or 2r - 5 = 0
r = 2/5 or r = 5/2
r = 2/5 or r = 5/2
Now ,
iii ) If a = 1 , r = 2/5 ,
Required three terms are
1/(2/5) , 1 , 1 × ( 2/5 )
5/2 , 1 , 2/5
iii ) If a = 1 , r = 5/2
Then required three terms are
2/5 , 1 , 5/2
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