Series And Sequence Solved Examples
1.Find the 13th term of G.P. with first term -1 and common ratio 3.
2. Find the common ratio and 8th term of the G.P. 1,-3,9, ...
Solution:
3. Find the common ratio of a G.P. whose 5th term is 2 and first term is 1/8.
4. Which term of the G.P. is 64/27 whose first term is 27 and the second term is 18?
5. The third term of a geometric series is 27 times the 6th term and the 4th term is 9. Find the series.
6. Find the sum of the geometric series 2+4+8+ ... where there are 8 terms in the series.
7. The sum of first 8 terms of a G.P. with common ratio 2 is 1530. Find the first term.
8. How many terms are there in the G.P. whose first terms is 2, common ratio 3 and the sum 728?
9.The sum of three consecutive terms in G.P. is 62 and their product is 1000, find the terms.
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.