Question

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

Answer 1

Step-by-step explanation:

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Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Here we have,

AP - 9 , 17, 25 , .....

Here,

a = 9

d = 8

Also,

S(n) = 636

Using Formula we have,

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

636 = n/2[18 + (n - 1)8]

636 = n/2[9 + (n - 1)4]

636 = n(9 + 4n - 4)

636 = n(5 + 4n)

636 = 5n + 4n²

4n² + 5n - 636 = 0

4n² + 53n - 48n - 636 = 0

4n² - 48n + 53n - 636 = 0

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

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

n = -53/4, 12 (Negative value is not applicable)

So,

n = 12

Therefore,

Twelve terms are required to give sum 636