Question

How many terms of the A.P. 45, 39, 33... must be taken so that their sum is 180? Explain the double answer.

Answer 1

Given:

AP: 45, 39, 33, ... (first term a = 45, common difference d = –6).

Required: find n such that the sum of n terms S_n = 180.

Step-wise calculation:

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

2. Substitute a = 45 and d = –6:

`S_n = n/2 [2 xx 45 + (n - 1)(-6)]`

= `n/2 [90 - 6(n - 1)]`

3. Simplify inside bracket:

90 – 6(n – 1) = 90 – 6n + 6 

= 96 – 6n

So `S_n = (n(96 - 6n))/2`

= n(48 – 3n)

4. Set S_n = 180 and solve for n:

n(48 – 3n) = 180

⇒ 48n – 3n² – 180 = 0 

⇒ 3n² – 48n + 180 = 0 

⇒ n² – 16n + 60 = 0

5. Solve the quadratic:

Discriminant Δ = 16² – 4 × 1 × 60 

= 256 – 240

= 16

`n = (16 ± sqrt(16))/2` 

= `(16 ± 4)/2`

⇒ `n = (16 + 4)/2 = 10` or `n = (16 - 4)/2 = 6`

6. Verify by direct summation:

For n = 6: 45 + 39 + 33 + 27 + 21 + 15 = 180.

For n = 10: first 10 terms are 45, 39, 33, 27, 21, 15, 9, 3, –3, –9; their sum = 180.

n = 6 or n = 10.

Because the common difference is negative, the partial-sum S_n = n(48 – 3n) is a concave-down quadratic in n. Therefore, the equation S_n = 180 is a quadratic and can have two positive integer roots; both n = 6 and n = 10 are valid, which is why there are two answers.