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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±b24ac)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, (5101)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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