Find the first three terms of the arithmetic series in which a1 = 6, an = 201, and Sn = 4140.

a1 = a = 6

a(n) = 201

S(n) = 4140

first three terms of the arithmetic series

We know that,

nth term of an AP – a(n) = a + (n - 1)d

Hence,

→ 201 = 6 + (n - 1)d

→ 201 - 6 = (n - 1)d

→ 195/(n - 1) = d -- equation(1).

Now,

Sum of first n terms of an AP = n/2 * [ a + a(n) ]

→ 4140 = n/2 * [ 6 + 201 ]

→ 4140 * (1/207) = n/2

→ 4140 * (1/207) * 2 = n

→ 40 = n

Putting n = 40 in equation (1) we get,

→ 195 / (40 - 1) = d

→ 195/39 = d

→ 5 = d

Now,

The first three terms of an AP are a , a + d , a + 2d.

Hence,

I hope it will help you.

Regards.

→ 4140 = n/2 * [ 6 + 201 ]

→ 4140 * (1/207) = n/2

→ 4140 * (1/207) * 2 = n

→ 40 = n

Putting n = 40 in equation (1) we get,

→ 195 / (40 - 1) = d

→ 195/39 = d

→ 5 = d

Now,

The first three terms of an AP are a , a + d , a + 2d.

Hence,

1st term = a1 = a = 6

2nd term = a + d = 6 + 5 = 11

3rd term = a + 2d = 6 + 2 * 5 = 6 + 10 = 16

Step-by-step explanation:

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times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3