Question

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?​

Answer 1

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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Answer 2

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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