write down the arithmetic sequence with first term 40 and the sum of the first five term is 500
Answers
Example:
Consider the arithmetic sequence 6,10,14,.....
(a) What is the sum of first n consequence terms of the above sequence?
(b) How many consecutive terms from the beginning should be added to get the sum 240?
(c) Is the sum of first-few consecutive terms becomes 250? Why?
Solution:
(a) 6,10,14,....
Since t
2
−t
1
=t
3
−t
2
Common difference, d=10−6=4
First term a=6
S
n
=
2
n
2a+(n−1)d=
2
n
2×6+(n−1)4=
2
n
12+4n−4=
2
n
4n+8
=n(2n+4)
2n
2
+4n
(b) For the sum to be 240 from the beginning
240=n
2
+4n
⇒240=n(2n+4)
⇒240=n(2n+4)
⇒120=n(n+2)
⇒n
2
+2n−120=0
⇒n
2
+12n−10n−120=0
⇒n(n+12)−10(n+12)=0
⇒(n+12)(n−10)=0
∴ n=−12,10
10 terms should be added to make the sum to be 240
(c) To make the sum to be 240, 'n' should be natural number.
250=n(2n+4)
⇒125=n(n+2)
⇒n
2
+2n−125=0
⇒n
2
+n−125=0
n=
2
−1
1−4×125
Since n is not a natural number, therefore the term 250 cannot exist in the series.
Answer:
40, 70, 100, 130, 160...
Step-by-step explanation:
a1 = 40(a)
sum of first five terms = 500
here five numbers are there(odd number)
a+b+c+d+e =500
(To get the middle number divide 500/5)
5c= 500
c=500/5
c= 100
let a2 = b
a3 = c
a4 = d
a5 = e
a+b+c+d+e= 500
40+b+100+d+e= 500
Common difference = a3-a1/3-1
= 100-40/3-1
=30
a2= 40+30 = 70
a4=100+30= 130
a5= 130+30= 160
therefore the arithmetic sequence are:
40, 70, 100, 130, 160