a,b,c,d are in GP are in ascending order such a+d=112 and b+c=48 that If the GP is continued with a as the first term, then the sum of the first six term is :
Answers
Given : a,b,c,d are in GP are in ascending order such a+d=112 and b+c=48 that If the GP is continued with a as the first term
To find : sum of the first six term is
Solution:
Let say GP is
a , ar , ar² , ar³ .......................
a = a
b = ar
c = ar²
d = ar³
a + d = 112
=> a + ar³ = 112
=> a (1 + r³ ) = 112
b + c = 48
=> ar + ar² = 48
=> ar(1 + r) = 48
a (1 + r³ ) / ar(1 + r) = 112/48
using x³ + y³ = (x + y)(x² + y² - xy)
=> ( 1 + r)(1 + r² - r) / r( 1 + r) = 7/3
=> 3 + 3r² - 3r = 7r
=> 3r² - 10r + 3 = 0
=> 3r² - 9r - r + 3 = 0
=> (3r - 1)(r - 3) = 0
=> r = 1/3 , 3
GP is in ascending order
hence r = 3
a (1 + r³ ) = 112
=> a( 1+ 3³) = 112
=> a( 28) = 112
=> a = 4
GP is
4 , 12 , 36 , 108 , .......
Sum of GP = a(rⁿ - 1)/(r - 1)
for n = 6
= 4 ( 3⁶ - 1)/(3 - 1)
= 2 ( 729 - 1)
= 1456
sum of the first six term is 1456
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