How many terms of the AP:9,17,25,... must be taken to give a sum of 636
Answers
hope this helps u
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 identities of arithmetic progressions, we know : -
sn = n/2(2a+(n-1) * d, where is the sum of first term to nth term of the AP, a is the first term, d 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
⇒4n= -53
⇒n = -53/4 no.of terms cannot be negative
= > 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.
Step-by-step explanation:
AP : 9,17,25,....
first term, a = 9
common difference, d = 17 – 9 = 8
Sn = 636, n = ?
Sn = n / 2 [ 2a + (n–1) d ]
636 = n / 2 [ 2×9 + (n– 1)8 ]
636 × 2 = n [ 18 + 8n – 8 ]
636 × 2= n [ 10 + 8n ]
636 × 2 = 2n [ 5 + 2n ]
636 = 5n + 4n²
4n² + 5n – 636 = 0
4n² + 53n – 48n - 636 =0
n ( 4n + 53 ) – 12 ( 4n + 53 ) = 0
( n – 12 ) ( 4n + 53 ) = 0
n = 12
n= – 53 /4
No of terms cannot be negative hence neglecting negative
therefore no of terms required to make a sum of 636 = 12