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QUESTION

# 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?

S_40 = 3220

Use the sum formula :

S_n = n/2 [ 2a + (n-1)d]

and the n-th term

T_n = A +(n-1) d

to solve for a and d.

This will get you

S_12 = 186 and T_20 = 83.

The equations are

"i) " 12/2 [ 2a +11d ] = 186

and

"ii) " a + 19d = 83

Solve like a linear system :

{(12a + 66d = 186), (a + 19d = 83) :}

a = 83 - 19d

sub in "i)"

12 (83 - 19d ) + 66d = 186

996 - 228d + 66d = 186

162 d = - 810

d = -5

This means that

a + 19 ( -5 ) = 83

a -95 = 83

a = 178

Now

S_40 = 20 [ 2(178) + 39 ( -5) ]

 S_40 = 20 [ 356 - 195 ]

S_40 = 3220