Question

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Answer 1

First term that is a = 5.
Last term that is = tn = 45
Sum is 400 that is sn= 400
So sn = n/2(t1 + tn)
400=n/2(5+45)
400=n/2(50)
400= 25n
N= 16

Common difference is 8/3

Answer 2

There is a formula in Arithmetic progressions which relates sum, first term and last term with the number of terms. It is;

Sum = ⁿ/₂ [a₁ + a₀]

where a₁ is the first term

a₀ is the last term

n is the number of terms

400 = ⁿ/₂ [5 + 45]

800 = n [50]

n = 80/5 = 16

Now, the last term is a₀ = 45

that is a + (n-1)d = 45

5 + (16-1)d = 45

15d = 40

d = 40/15 = 8/3

Therefore the common difference, d = 8/3 while the number of terms are 16