How many terms of the AP : 9, 17, 25, must be taken to give a sum of 6

1.	How many terms of the AP : 9, 17, 25, must be taken to give a sum of 636?
Answer» Let the number of terms required to make the sum of 636 be n and common difference be d.Given Arithmetic Progression : 9 , 17 , 25 ....First term = a = 9Second term = a + d = 17Common difference = d = a + d - a = 17 - 9 = 8From the indentities of arithmetic progressions, we know : -, where\xa0\xa0is the sum of first term to nth term of the AP, a is the first term, d is the common difference and n is the number of terms of AP.In the given Question, sum of APs is 636.Therefore,= > 4n² + 5n - 636 = 0= > 4n² + ( 53 - 48 )n - 636 = 0= > 4n² + 53n - 48n - 636 = 0= > 4n² - 48n + 53n - 636 = 0= > 4n( n - 12 ) + 53( n - 12 ) = 0= > ( n - 12 )( 4n + 53 ) = 0By Zero Product Rule,= > n - 12 = 0= > n = 12Hence,Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.\xa0

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