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For a G.P. the ratio of the 7^{th} and the third terms is 16. The sum of 9 terms is 2555. What is the first term?
ar^{6}/ar^{2} = 16
r^{4} = 16
r^{4} = (2)^{4}
r = 2
S_{9} = a(r^{n}  1)
2555 = a((2)^{9}  1)
2555 = a(5121)
2555/511 = a
a = 5
The geometric mean of two numbers, say x, and y is the square root of their product x * y.
Mean = [x * y]^{1/2}
= [5 * 8]^{1/2}
= [40]^{1/2}
= 2(10)^{1/2}
The 10^{th} term of an G.P. is and the 5th term is.What is the Common Ratio?
ar^{9 }= 1/(2)^{8} (1)
ar^{3} = 1/(2)^{2} (2)
Comparing eq (1) & (2)
1/(2)^{8} r^{9} = 1/(2)^2 r^{4}
1/r^{5} = (2)^{5}
r^{5} = 1/(2)^{5}
r = 1/2
A sequence a_{1}, a_{2}, a_{3},…, a_{n} is called ______ progression, if each term is nonzero and = constant for
The expression depicts the ratio of two numbers, if the ratio between the numbers is constant, then it will definately form a GP.
Which of the following sequeces in GP will have common ratio 3,where n is an Integer?
g_{n} = 6( 3^{n1}) it is a geometric expression with coefficient of constant as 3^{n1}.So it is GP with common ratio 3.
The third term of a G.P. is 3, the product of first five terms of this progression is:
8(1) + 8(11) + 8(111) + 8(1111) +........up to n times
= 8{ 1 + 11 + 111 + 1111 + ....... up to n times}
= 8/9{9 + 99 + 999 + 9999 + ..... up to n times}
= 8/9 {(101 )+ (1001) + (10001) + (100001) + ........up to n times}
= 8/9 {(10 + 100 + 1000 + 10000 +...... n times)  (n×1)}
= 8/9{( (10×(10^{n} 1))/(101)) n}
= 80/81(10^{n}  1)  n
= 8/81(10^{n+1}  10 9n)
Which term of the following sequence is 64 ?
2 , 2√2, 4, .....
Given sequence : 2, 2√2, 4….
First term a_{1} = a = 2 and 2nd term a_{2} = 2√2, then
Common ratio r = a_{2}/a = (2√2)/2
Let a_{n} = 64
∴ ar^{(n1)} = 64
⇒ 2.(√2)^{(n1)} = 32
⇒ (2)^{(n1)/2 }= 32
∴ (2)^{(n1)/2} = (2)^{5}
⇒ (n − 1)/2 = 5,
⇒ n = 11
How many terms of the G.P. 4 + 16 + 64 + … will make the sum 5460?
Sum (S_{n}) = a x (r^{n} 1)/(r1)
5460 = 4 x (4^{n} 1)/3
16380 = 4^{n+1}  4
16384 = 4^{n+1}
4^{7 }= 4^{n+1}
7 = n + 1
n = 6
What is the 50^{th} term of the sequence √3, 3, 3√3, 9, ......
an = ar^{(n1)}
Given, a = √3, r = 3/√3
r = √3
a_{50} = ar^{(n1)}
= (√3)(√3)^{(501)}
= (√3)(√3)^{49}
= (√3)^{50}
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