Question

The sum of a certain number of terms of the Arithmetic Progression (A.P.) 20, 17, 14, .... is 65. Find the:

1. number of terms.
2. last term.

Answer 1

Given: A.P.: 20, 17, 14, .... so a = 20, d = 17 – 20 = –3 and S_n = 65.

Use the sum formula `S_n = n/2 [2a + (n - 1)d]`.

Step-wise calculation:

1. Set up equation:

`n/2 [2 xx 20 + (n - 1)(-3)] = 65`

2. Multiply both sides by 2:

n[40 – 3(n – 1)] = 130

3. Simplify inside bracket:

40 – 3n + 3 = 43 – 3n

So, n(43 – 3n) = 130.

4. Rearrange: 43n – 3n² = 130

⇒ 3n² – 43n + 130 = 0

5. Compute discriminant:

Δ = 43² – 4 × 3 × 130 

= 1849 – 1560

= 289

`sqrt(Δ) = 17`

6. Solve:

`n = [43 ± 17]/6`

⇒ `n = 60/6`

= 10

or

⇒ `n = 26/6`

= `13/3` 

Discard non-integer `13/3`, so n = 10.

7. Last term a_n = a + (n – 1)d

= 20 + 9(–3)

= 20 – 27

= –7

8. Quick check: average of first and last

= `(20 + (-7))/2`

= 6.5

Sum = 6.5 × 10 = 65