Geometric Progression,Mean & Sum of Geometric Series

Series And Sequence Solved                           Examples

1.Find the 13th term of G.P. with first term -1 and common ratio 3.
 Series and sequence
2. Find the common ratio and 8th term of the G.P. 1,-3,9, ...
Solution:
 Series and sequence

3. Find the common ratio of a G.P. whose 5th term is 2 and first term is 1/8.

 Series and sequence
4. Which term of the G.P. is 64/27 whose first term is 27 and the second term is 18?
 Series and sequence


5. The third term of a geometric series is 27 times the 6th term and the 4th term is 9. Find the series.
 Series and sequence


6. Find the sum of the geometric series 2+4+8+ ... where there are 8 terms in the series.
 Series and sequence
7. The sum of first 8 terms of a G.P. with common ratio 2 is 1530. Find the first term.
 Series and sequence
8. How many terms are there in the G.P. whose first terms is 2, common ratio 3 and the sum 728?
 Series and sequence
9.The sum of three consecutive terms in G.P. is 62 and their product is 1000, find the terms.
 Series and Sequence
10. The sum of first four terms is a G.P. is 30 and that of the last four terms is 960.If its first term is 2, find the common ratio.
Solution:
Let r be the common ratio then,
2+2r+2r²+2r³ = 30
or, 1+r+r²+r³ = 15
or r³+r²+r+14 = 0
or r³-2r²+3r²-6r+7r -14 = 0
or r²(r-2)+3r(r-2)+7(r-2) = 0
or (r-2) (r²+3r+7)
∴ r=2

11. Some geometrical means are inserted between 5 and 160. If the third mean between them is 40, find the number of means.
Solution:
Here, the third mean =40
we also know, third mean=4th term
So, ar³=40
or, 5.r³ = 40
or, r³ =8 =2³
∴r = 2
Now,
tₙ=arⁿ⁻¹
or, 160 = 5.2ⁿ⁻¹
or, 32= 2ⁿ⁻¹
or, 2⁵ = 2ⁿ⁻¹
or, n-1 = 5
∴n=6

So, the number of mean =n-2 =4

12. If A.M. and G.M. between two numbers are 10 and 8 respectively, find the two numbers.
Solution: 
Let a and b be the two numbers whose A.M. is 10 and G.M. is 8.
Then, (a+b)/2 = 10
or, a+b = 20 ...(i)
And √ab =8
or, ab =64
or, b = 64/a ...(ii)

Putting the value of b in equation (i) we get
a+64/a =20
or, a²+64 =20a
or, a²-20a+64 =0
or, a²-16a-4a+64 =0
or, a(a-16) -4(a-16) =0
or, (a-16)(a-4) =0
∴a = 16 or 4

when a=16, then b= 64/16 =4
when a = 4, then b =64/4 =16
Hence, required two numbers are 16 and 4 or 4 and 16.

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