i) The first term of an AP is 5, the last
temui 45 and the sum is 400. find
the number
of
terms and the common
difference?
Answers
Answered by
0
solution=
first term a=5, last term l=45 and S
n
= 400
Since, S
n
=
2
n
(a+l)
⇒400=
2
n
(5+45)
⇒400=
2
n
(50)
⇒400=n(25)
⇒n=16
Since, l=a+(n-1)d
⇒45=5+(16−1)d
⇒45−5=(15)d
⇒40=15d
⇒d= 40/15
⇒d=8/3
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Answered by
0
Answer:
n = 16
Step-by-step explanation:
a = 5, l/an = 45
Sn =400 To find: n
l/an = a + (n-1) d
45 = 5 + (n-1) d
(n-1) d = 45 -5 = 40------->[1]
Sn = n/2(2a + (n-1) d)
400 = n/2(2a + (n-1) d)
By [1]
400×2= n(2×5 + 40)
800 = n (50)
n = 800/50
n = 16
Putting this in [1]
(n-1) d = 45 -5 = 40
40 = (16-1) d
40 = 15d
d = 40/ 15
d = 8/3 OR 2.67
n =16 , d = 8/3
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