Find the sum of the first 42 terms of the sequence {12, 16, 20, . . . }.

Question

4,032

3,948

3,717

1,134

Answer 1

3948 !

Remember the small boy Frederick Gauss who grew up to be a great mathematician & scientist. He solved this like this...

The terms are 12+16+20+24+.......+176 If you add the reversed sequence to the original you get

176+172+168+........+ 12 you get

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188+188+188+.........+188 this equals 42*188 = 7896

But you then need to divide it by 2 to obtain 3948

Answer 2

This is an "arithmetic" sequence because you are ADDING 4 to each term to get the new term.

There are formulas for arithmetic sequences.

a₁ is the first term

n is the number of terms

d is the common difference (16-12 = 4, 20 - 16 = 4)

For us

a₁ = 12

n = 42

d = 4

a_n = a₁ + (n-1)*d

The 42nd term is a₄₂=12 + (42-1)(4) = 176

SUM_n = (n/2)(a₁ + a_n) so SUM₄₂=(n/2)(a₁+a₄₂)

The sum of 42 terms is SUM₄₂=(42/2)(12 + 176) = 3948 (Answer is B)