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Question

# How many terms of the AP: 9, 17, 25, ... must be added to get a sum of 636?___

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Solution

## First term = a = 9 Common difference = d = 17 -9 = 8 Sn=636 Applying formula, Sn=n2(2a+ (n−1) d) to find sum of n terms of AP, we get 636=n2(18+ (n−1) (8)) ⇒1272=n (18+8n−8) ⇒1272=18n+8n2−8n ⇒ 8n2+10n−1272=0 ⇒4n2+5n−636=0 It is a quadratic equation. We can solve it using different methods. Let’s solve it using quadratic formula. Comparing equation 4n2+5n−636=0 with general form an2+bn+c=0, we get a=4, b=5 and c=−636 Applying quadratic formula, n = (−b±√b2−4ac)2 and putting values of a, b and c, we get n= (−5± √25−4 (4) (−636)8 ⇒ n= (−5 ± √25−4 )(4) (−636)8 = (−5± √25 +10176)8 = −5±1018 ⇒n= (−5+101)8, (−5−101)8 ⇒ n=968, −1068 =12, −1068 We discard negative value of n here because n cannot be in negative, n can only be a positive integer. Therefore, n=12. Therefore, 12 terms of the given sequence make sum equal to 636.

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