Geometric progression?
In a geometric progression, the sum of the fifth and seventh terms is 10 and the second term is 16 times the sixth term. Find:
i. The common rate
ii. The first term
iii. The sum of the first 4 terms.
In a geometric progression, the sum of the fifth and seventh terms is 10 and the second term is 16 times the sixth term. Find:
i. The common rate
ii. The first term
iii. The sum of the first 4 terms.
G-boy S.S.
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a r^4 + a r^6 = 10 ................... equation (1)
a r = 16 a r ^5
1 = 16 r^4
r^4 = 1 / 16
common rate (ratio), r = ± ½
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plug in the computed value of r into eq (1);
a r^4 + a r^6 = 10
..................... .................... let's just use +½ because r is raised to even numbers
a ( ½ )^4 + a ( ½ )^6 = 10
a ( 1 / 16 ) + a ( 1 / 64 ) = 10
4a + a = 640
5a = 640
first term, a = 128
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for r = + ½ :
sum of the first 4 terms is :
n
Σ a r^k = a ( 1 - r^(n+1)) / ( 1 - r )
k=0
3
Σ 128 ( ½ )^k = 128 ( 1 - ( ½ )^(3+1)) / ( 1 - ½ )
k=0
...................... = 128 ( 1 - ( ½ )^4 ) / ½
...................... = 128 ( 1 - ( 1 / 16 ) / ½
...................... = 128 ( 15 / 16 ) / ½
...................... = 8 ( 15 ) / ½
...................... = 120 / ½
...................... = 240
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for r = – ½ :
sum of the first 4 terms is :
n
Σ a r^k = a ( 1 - r^(n+1)) / ( 1 - r )
k=0
3
Σ 128 ( –½ )^k = 128 ( 1 - ( –½ )^(3+1)) / ( 1 - (–½ ) )
k=0
...................... = 128 ( 1 - ( –½ )^4 ) / 1½
...................... = 128 ( 1 - ( 1 / 16 ) / 1½
...................... = 128 ( 15 / 16 ) / 1½
...................... = 8 ( 15 ) / 1½
...................... = 120 / 1½
...................... = 80
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?
ar=16ar^5 , 1/16=r^4 , r=+-1/2 , ar^4+ar^6=10 , a=128 , terms 128 , 64 , 32 , 16 sum=240 or terms 128 , -64 , 32 , -16 sum=80
?
i 1/2
ii 128
iii 128+64+32+16
Rajiv Nair
1.) Common rate = (16)^(1/(6-2) = 4th root of 16 = 2
I cant help you with the rest I am afraid.