the sum of first 100 terms of an arithmetic sequence is 5,9,13... is how much more than the sum of first 100 terms of an arithmetic sequence 4,7,10...
Answers
Answer:
The sum is the average of the first and last terms multiplied by the number of terms.
The first term is 7. The general term is 7 + 3*(n - 1) = 3n + 4
The hundredth term is 304. We are adding 100 terms.
Sum = (7+304)/2 * 100 = 311*50 = 15,550
Step-by-step explanation:
Given :-
The AP's are 5,9,13... and 4,7,10,...
To find :-
The sum of first 100 terms of an arithmetic sequence is 5,9,13... is how much more than the sum of first 100 terms of an arithmetic sequence 4,7,10... ?
Solution :-
The first Arithmetic Progression is 5,9,13,...
First term (a) = 5
Common difference (d) = 9-5 = 4
We know that
The sum of first n terms of an AP is
(Sn) = (n/2)[2a+(n-1)d]
The sum of first 100 terms of the AP is
S100 = (100/2)[2(5)+(100-1)(4)]
=> S100 = (50)[10+99(4)]
=> S100 = (50)(10+396)
=> S100 = 50×406
=> S100 =20300 ------------------(1)
The second Arithmetic Progression is 4,7,10,...
First term (a) = 4
Common difference (d) = 7-4 = 3
We know that
The sum of first n terms of an AP is
(Sn) = (n/2)[2a+(n-1)d]
The sum of first 100 terms of the AP is
S100 = (100/2)[2(4)+(100-1)(3)]
=> S100 = (50)[8+99(3)]
=> S100 = (50)(8+297)
=> S100 = 50×305
=> S100 =15250 ------------------(2)
On Subtracting (2) from (1)
=> (1)-(2)
=>20300-15250
=> 550
The difference between the two AP's is 550
Answer:-
The sum of the first 100 terms of the AP :5,9,13,... is 550 more than the sum of the first 100 terms of the AP : 4,7,10,..
Used formulae:-
★The sum of first n terms of an AP
(Sn) = (n/2)[2a+(n-1)d]
Where , a = First term
d = Common difference
n = Number of terms