Question

Find the sum of the following arithmetic series:

5 + (–41) + 9 + (–39) + 13 + (–37) + 17 + ... + (–5) + 81 + (–3)

HINT: Given sum = (5 + 9 + 13 + 17 + ... + 81) + {(–41) + (–39) + ... + (–3)}.

Answer 1

Given: 5 + (–41) + 9 + (–39) + 13 + (–37) + 17 + ... + (–5) + 81 + (–3).

Step-wise calculation:

1. Split into two APs:

Positive terms: 5, 9, 13, ..., 81.

Here a = 5, d = 4, last = 81.

`n_p = ((81 - 5)/4) + 1`

= 19 + 1

= 20

`S_p = n_p/2 xx ("first" + "last")` 

= `20/2 xx (5 + 81)` 

= 10 × 86

= 860

2. Negative terms: –41, –39, –37, ..., –3. 

Here a = –41, d = 2, last = –3. 

`n_n = ((-3 - (-41))/2) + 1` 

= `(38/2) + 1` 

= 19 + 1

= 20

`S_n = n_n/2 xx ("first" + "last")` 

= `20/2 xx (-41 + (-3))` 

= 10 × (–44)

= –440

3. Total sum = S_p + S_n 

= 860 + (–440)

= 420

Check: pairing terms gives pairs –36, –30, –24, ..., 78; 

20 pairs sum = `20/2 xx (-36 + 78)`

= 10 × 42

= 420

The sum of the series is 420.