1,4,6,5,11,6 .......find the sum of first 100 term.
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Answers
Answer:
1+6+11+......4+5+6....
50terms + 50 terms = 100 terms
hence,
(n/2) (2a+(n-1)d)
25(2+50×5)+25(8+50×1) = 6300 +1450
=> 7750 Answer !!
Concept
Arithmetic progression is the sequence in which all the terms have common difference. It is also known as arithmetic sequence.
Given
A sequence 1,4,6,11,6....
To find
The sum of first 100 terms.
Explanation
Difference between first term and third term=6-1=5
Difference between third term and fifth term=11-6=5
Difference between second term and fourth term=5-4=1
Difference between fourth term and sixth term=6-5=1
Because the difference between first term and sixth term is equal to the difference between third and fifth term so we will collect these terms.
The difference between second term and fourth term is equal to the difference between fourth and sixth term so will collect these terms also.
Sum of the given sequence will be =(1+6+11+.......)+(4+5+6+........)
Because we have to find sum of 100 terms so there are 50 terms each for each group)
The groups which we have formed look like arithmetic progression so will use the formula of sum of terms of arithmetic progression.
Sum of terms of an arithmetic progression=n/2[2a+(n-1)d]
Sum of first group=50/2[2*1+(50-1)*5]
=25(2+49*5)
=25(2+245)
=25*247
=6175
Sum of second group=50/2[2*4+(50-1)*1]
=25(8+49)
=25*57
=1425
Total sum=sum of first group+sum of second group
=6175+1425
=7600
Hence the sum of first 100 terms of the given sequence is 7600.
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