The sum of the first k terms of the arithmetic progression -10, -7, -4, ...., is 104. What is the value of \(\frac{k+9}{k-5}\) ?

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AAI Junior Assistant (Fire Service) Official Paper (Held On: 15 Nov, 2022 Shift 2)
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  1. \(\frac{10}{3}\)
  2. \(\frac{12}{3}\)
  3. \(\frac{11}{4}\)
  4. \(\frac{3}{7}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{11}{4}\)
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Detailed Solution

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Given:

Arithmetic progression: -10, -7, -4, ...

Sum of first k terms (Sk) = 104

Formula used:

Sk = (k/2)[2a + (k - 1)d]

where a is the first term and d is the common difference.

Calculation:

a = -10

d = -7 - (-10) = 3

Now, substitute the values into the sum formula:

⇒ 104 = (k/2)[2(-10) + (k - 1)3]

⇒ 104 = (k/2)[-20 + 3k - 3]

⇒ 104 = (k/2)[3k - 23]

⇒ 208 = k(3k - 23)

⇒ 3k2 - 23k - 208 = 0

⇒ 3k2 - 39k + 16k - 208 = 0

 3k(k - 13) + 16(k - 13) = 0

⇒ (3k + 16)(k - 13)

k = 13, or -16/3(Negative not possible)

The value of (k + 9) / (k - 5) is

⇒ (13 + 9) / (13 - 5) =  22/8 = 11/4.

Hence the correct answer is 11/4.

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