How many terms of an A.P:9,6,3,0,-3.. will be needed to give sum (-216)
Answers
Answered by
14
Just use the formula for S to n terms
S=n/2[2*first term + (n-1)*common difference]
-432=n/2[2(9)+ (n-1)(-3)]
=> n^2-7n -144=0
=>n=16
Hope it helps :)
S=n/2[2*first term + (n-1)*common difference]
-432=n/2[2(9)+ (n-1)(-3)]
=> n^2-7n -144=0
=>n=16
Hope it helps :)
sweetyalora:
How you got 16 ???
From the quadratic formula
[-(b)+-sqrt(b^2 - 4ac)]/2a
But i got b2-4ac as root -527
You must have taken the wrong signs
Answered by
15
AP : 9 , 6 , 3 , 0 , -3 ,...... N
d = common difference = -3
a = 9.
Final term N = a + (n-1) d
N = 9 + (n - 1)(-3) = 12 - 3 n
Sum desired: S = - 216
- 216 = sum = [ a + N ] * n/2 = [9 + 12 - 3n ] * n/2
- 216 = 21 n/2 - 3 n² /2
n² - 7 n -144 = 0
-144 = -16 * 9 ,,,,, -16+9 = -7
n² - 16 n + 9 n - 144 = 0
(n - 16) (n + 9) = 0 , n cannot be negative.
So n = 16. Answer.
d = common difference = -3
a = 9.
Final term N = a + (n-1) d
N = 9 + (n - 1)(-3) = 12 - 3 n
Sum desired: S = - 216
- 216 = sum = [ a + N ] * n/2 = [9 + 12 - 3n ] * n/2
- 216 = 21 n/2 - 3 n² /2
n² - 7 n -144 = 0
-144 = -16 * 9 ,,,,, -16+9 = -7
n² - 16 n + 9 n - 144 = 0
(n - 16) (n + 9) = 0 , n cannot be negative.
So n = 16. Answer.
:-)
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