The sum of the first 12 terms of an arithmetic series is 186, and the 20th term is 83. What is the sum of the first 40 terms?

Question

arithmetic series question

Answer 1

S₁₂=186    =  6(a₁+a₁₂)=6(2a₁+11d)

a₂₀= 83     =  a₁+19d

S₄₀= ?

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2a₁+11d = 31

  a₁+19d = 83  

 

27d= 135

 

d=5,  a₁=-12

 

 

Now

S40=20*(a₁+a₄₀)=20(2a₁+39d)=20(-24+195)=20*171=3420.

Answer 2

Sum of an arithmetic series is given by:  Sum = (n/2)(2a₁ + (n-1)d)

 

To find term a_n:  a_n = a₁ + (n-1)d

 

Since the 20th term is 83 we have:

a₁ + 19d = 83

a₁ = 83 - 19d

 

Sum of 1st 12 terms is 186 so:

186 = (12/2)[2(83-19d) + (12-1)d]

186 = 6(166 - 38d + 11d)

31 = 166 - 27d

31 - 166 = -27d

-135 = -27d

d = 5

 

a₁ = 83-19(5) = -12

 

a₄₀ = -12 + (40-1)(5)

a₄₀ = 183

 

Sum₄₀ = (40/2)[2(-12) + (40-1)(5)]

= 20(-24 + 195)

= 20(171)

= 3420

 

The sum of the first 40 terms is 3420