FOR FREE YEAR SOLVED
Series 14
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Here the program is written to print the series
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# For loop is used to print the series.
# For the even terms the value will be: 2 × (previous term)
# For the odd terms the value will be: (3/2) × (previous term)
# The value of the first term is 1.
If the given number of terms is 4,
1st term:1
2nd term (even term): 2 × 1 = 2
3rd term (odd term): (3/2) × 2 = 3
4th term (even term): 2 × 3 = 6
Therefore, the output will be: 1 2 3 6
Algorithm
INPUT: Number of terms
OUTPUT: The aforesaid series upto n terms
PROCESS:
Step 1: [Taking the input]
Read n [Number of terms]
Step 2: [Printing the series]
Set t<-1
[Printing the series]
Print "The series is: t "
For i=2 to n repeat
[If the term is even]
If i mod 2=0 then
Set t<-2×t
[If the term is odd]
Else
Set t<-(3×t)/2
[End of ‘if-else’]
[Printing the result]
Print t
[End of ‘for’ loop]
Step 3: Stop.
Code
TIME COMPLEXITY:
for(i=2;i<=n;i++)------------------------------------ O(n)
{
//if the term is even
if(i%2==0)------------------------------- O(1)
t=2*t;
//if the term is odd
else--------------------------------------- O(1)
t=(3*t)/2;
//printing the result
printf("%d ",t);
}
The time complexity of this program is O(n) where ‘n’ is the number of terms of the series.
SPACE COMPLEXITY:
The space complexity of this program is O(1) as it requires a constant number of memory spaces for any given input.
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