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An arithmetic series has first term a and common difference d, where d is a prime number. The sum of the first n terms of the series is Sn and Sm=39 , S2m=320 . Find the value of d and the value of m. Show clear algebraic working.

Please give me the answer and working. Ive been stuck for days and deadline is on monday

An arithmetic series has first term a and common difference d, where d is a prime-example-1
User Ylgwhyh
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Final answer:

To solve for the common difference (d) and the value of m in the arithmetic series, we use the formulas for the sum of an arithmetic series. After setting up two equations with the given conditions, we find that the common difference (d) is 11 and the value of m is also 11.

Step-by-step explanation:

To find the value of d and m for an arithmetic series with first term a and common difference d (which is a prime number), we utilize the formulas for the sum of the first n terms of an arithmetic series, Sn, and apply the given conditions Sm=39 and S2m=320.

The sum of the first n terms of an arithmetic series is given by:

Sn = n/2 (2a + (n-1)d)

Using the conditions, we set up two equations:

  • Sm = m/2 (2a + (m-1)d) = 39
  • S2m = 2m/2 (2a + (2m-1)d) = 320

We then simplify and solve these equations for a, d, and m.

First, simplify the second equation by dividing by m to get:

(2a + (2m-1)d) = 160

Subtract the first equation from the second equation to eliminate a:

(2m-1)d - (m-1)d = 160 - 39

md = 121

Since d is known to be a prime number and it must divide 121, which is 11 squared, we deduce that d = 11.

Plugging in d = 11 into the equation md = 121, we get:

m= 121/11 = 11

User Nodots
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