Consider the series 212+199+186+173+... Find the number of terms, n, if Sn = –4488.

Question

Any Help???

Answer 1

First, PLZ note that S is a series.  That is, the SUM of the terms thus far in a sequence.

 

So, the sequence is:

  212, 199, 186, 173, ... a_n

 

and the sum of the first n terms is:

  S_n = a₁ + a₂ + a₃ + ... + a_n                [note: begin numbering terms with 1]

 

Find the number of terms, n, if Sn = –4488.

 

This is an Arithmetic Sequence with a common difference of (-13).  That is, each term in the sequence is 13 less than the preceding term.

 

Now, the formula for the Series (the sum of the terms thus far in an Arithmetic Sequence) is:

     S_n = (n/2)(2a₁ + (n-1)d)

 

     -4488 =(n/2)(2(212) + (n-1)(-13))                [a₁=212;  d = -13]

     -4488 = (n/2)(424 -13n + 13)                        [distribute]

     -4488 = (n/2)(-13n + 437)

     -8976 = -13n² + 437n                              [multiply both sides by 2; distribute]

    13n² - 437n - 8976 = 0                       [get 0 on right side]

     (13n+187)(n-48)                             [either factor or use quadratic formula]

   Either n=-187/13 or else n=48  or both           [Multiplicative Property of Zero]

    n, the number of terms, must be positive, so

          n = 48

 

Check:

 

S_n = (n/2)(2a₁ + (n-1)d)

S₄₈ = (24)(2(212) + 47(-13))

      =  (24)(424 - 611)

      =  (24)(-187)

      =  -4488

  Check !

Answer 2

The common difference is -13

 

212-13 = 199

199-13 = 186

and so forth

 

To find a term:  S_n = 212 - 13(n-1)  given that S₁ is 212

 

212 - 13(n-1) = -4488

212 - 13n + 13 = -4488

-13n + 225 = -4488

-13n = -4713

n = 362.5

 

Are you certain of the final term as -4488??