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How many terms of the A.P. 3, 5, 7, 9, ... must be added to get the sum 80? - Mathematics

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Question

How many terms of the A.P. 3, 5, 7, 9, ... must be added to get the sum 80?

Sum
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Solution

Given: A.P. 3, 5, 7, 9, ... ; a = 3, d = 2; required sum Sn = 80.

Step-wise calculation:

1. Use the sum formula:

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

2. Substitute a = 3 and d = 2:

`S_n = n/2 [2 xx 3 + (n − 1) xx 2]`

= `n/2 [6 + 2n - 2]`

= `n/2 [2n + 4]`

= n(n + 2)

3. Set n(n + 2) = 80

⇒ n2 + 2n – 80 = 0

4. Solve the quadratic:

Discriminant = 22 – 4(1)(–80) 

= 4 + 320

= 324 

`sqrt(324) = 18 xx n` 

= `(-2 ± 18)/2` 

⇒ n = 8 or n = –10   ...(Discard negative)

The first 8 terms must be added to get the sum 80.

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2025-2026 (March) Basic - 430/5/2
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