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

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 : -{tex}S_{n}=\\dfrac{n}{2}\\{2a+(n-1)d\\}, where S_{n} {/tex}is 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,\xa0⇒ {tex}636 = \\dfrac{n}{2} \\{2(9) + (n - 1)8 \\} \\\\ \\\\ \\\\ = > 1272 = n(18 + 8n - 8) \\\\ \\\\ = > 1272 = 10n + 8n {}^{2} \\\\ \\\\ = > 636 = 5n + 4n {}^{2}{/tex}\xa0⇒ 4n² + 5n - 636 = 0\xa0⇒ 4n² + ( 53 - 48 )n - 636 = 0\xa0⇒ 4n² + 53n - 48n - 636 = 0\xa0⇒ 4n² - 48n + 53n - 636 = 0\xa0⇒ 4n( n - 12 ) + 53( n - 12 ) = 0\xa0⇒ ( n - 12 )( 4n + 53 ) = 0By Zero Product Rule,\xa0⇒ n - 12 = 0\xa0⇒ n = 12Hence,Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.\xa0

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