3,5,7,9 find sum of first 100 terms
Answers
What is the sum of 1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+…+95+97+99?
The above series forms an Arithemetic Progression (A.P).
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant.
For example,
the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1.
Second example: the sequence 3, 5, 7, 9, 11,... is an arithmetic progression
with common difference 2.
Now the above series
1+3+5+………+97+99
We can solve this in two ways
Solution:
Using Progression Concept
The Common Difference is 3–1 = 2
Total no of terms can be found using below formula
T1 = a+(n-1)d
Tn=a+(n−1)d
Tn is the nth Term
a i
Method :1
1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+…+95+97+99
The series is in AP
a=1,d=2and n=50
and
last term l=99
Formula....
Sum of series = n/2∗(a+l)
= 50/2∗(1+99)
=25∗100
=2500 Answer
Method second (short trick)
1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+…+95+97+99
If we see the series it is addition of odd terms from 1 to 99
So total number of terms available =50
And addition of n odd consecutive terms is n²
So 50²
=2500 answer
3 , 5 , 7 , 9
This is in A . P
Sum of first 100 terms is 10200.
