How many
terms of the ap is 9,17,25.....must be taken to give a sum of 636
Answers
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 = 9
Second term = a + d = 17
Common difference = d = a + d - a = 17 - 9 = 8
From the indentities of arithmetic progressions, we know : -
, where 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,
= > 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.
hope it helps u
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