How many terms of ap 9 17 25

Asked by admin @ 25/08/2021 in Math viewed by 278 People

How many terms of the AP: 9,17,25,.......... must be taken so that their sum is 636

Answered by admin @ 25/08/2021

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 = 9

Second term = a + d = 17

Common difference = d = a + d - a = 17 - 9 = 8

From the indentities of arithmetic progressions, we know : -
$S_{n}=\dfrac{n}{2}\{2a+(n-1)d\}$ , where $S_{n}$ 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,
$= > 636 = \dfrac{n}{2} \{2(9) + (n - 1)8 \} \\ \\ \\ = > 1272 = n(18 + 8n - 8) \\ \\ = > 1272 = 10n + 8n {}^{2} \\ \\ = > 636 = 5n + 4n {}^{2}$

= > 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 ) = 0

By Zero Product Rule,

= > n - 12 = 0

= > n = 12

Hence,
Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.

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