Question

36. How many terms of the AP: 9, 17, 25, ................. must be taken to give a sum of 636
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Answer 1

$\bf\huge\underline{Question}$

How many terms of the AP: 9, 17, 25, ...... must be taken to give a sum of 636?

$\bf\huge\underline{Answer}$

Here, a = 9, d = 17 - 9 = 8, ${S_n}$ = 636

${S_n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

$\dfrac{n}{2}$ [(2 × 9) + (n - 1) × 8] = 636

=> n[18 + (n + 1) × 8] = 1272

=> 18n + 8n² - 8n = 1272

=> 8n² + 10n = 1272

=> 4n² + 5n - 636 = 0

=> 4n² - 48n + 53n - 636 = 0

=> 4n (n - 12) + 53(n - 12) = 0

=> (n - 12)(4n + 53) = 0 => n = 12, $\dfrac{-53}{4}$

As n = $\dfrac{-53}{4}$ is neglected.

Required number of terms is 12.

Answer 2

Answer:

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