What is the sum of the arithmetic sequence 3, 9, 15, if there are 34 terms?

7 Answers

• ♥♦♣♠

  first term, a = 3

  difference = 9 – 3 = 6

  ……………..= 15 – 9 = 6

  n = 34

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

  S_34 = ( 34 / 2 ) [ 2 ( 3 ) + ( 34 – 1 ) ( 6 ) ]

  ………= 17 [ 6 + ( 33 ) ( 6 ) ]

  ………= 17 ( 6 + 198 )

  ………= 17 ( 204 )

  ………= 3,468

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• 3, 9, 15, 21, . . . .

  a = 3, d = 9-3 = 6, n = 34

  Sn = (34/2)[2*3 + (34-1)*(6)]

  Sn = 17*[6 +198] = 3468 >==============< ANSWER

• 3+9+15+21+….. 3+(33*6). =n/2*[2a+(n-1)d]=34/2*[2*3+33*6]

  =17 * [6+198]

  =3468

  where a=3 first term

  d=6 difference between two consequtive terms

  n= no of terms

• Use Sn formula with n = 34, a = 3 and d = 6. Simple!

• It depends what the rest of the terms are. Probably it averages out to 40 or 50, so I would say the answer is 45 or 45 and a half.

• 3468

  Source(s): Used Excel

• 3,103