Please help me to solve this
How many terms of the AP must be taken so that the sum will be zero
18, 16, 14, 12,......
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sum of n terms = n/2[2a+(n-1)d]. 0=n/2[2*18+(n-1)*-2]. 0=32n-2n^2+2n. n^2=17n. =>n=17
Answered by
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Hi !
we have to find "n"
Sn = 0
a = 18
d = -2
Sn = n/2 [ 2a + (n-1)d]
0 = n/2 [ 36 -2n + 2 ]
0 = n/2 [ 38 - 2n ]
0 = n/2*2 [ 19 - n ]
n² - 19n = 0
n² - 19n - 0 = 0
[ a = 1 , b = -19 ,c = 0 ]
n = -b ± √b²- 4ac / 2a
= 19 ±√ 361/2
= 19 ± 19/2
if n = 19 + 19/2
n = 38/2 = 19
if n = 19 - 19/2
n = 0/2 = 0
As the no: of terms (n) cannot be zero , n = 19
No of terms = 19
we have to find "n"
Sn = 0
a = 18
d = -2
Sn = n/2 [ 2a + (n-1)d]
0 = n/2 [ 36 -2n + 2 ]
0 = n/2 [ 38 - 2n ]
0 = n/2*2 [ 19 - n ]
n² - 19n = 0
n² - 19n - 0 = 0
[ a = 1 , b = -19 ,c = 0 ]
n = -b ± √b²- 4ac / 2a
= 19 ±√ 361/2
= 19 ± 19/2
if n = 19 + 19/2
n = 38/2 = 19
if n = 19 - 19/2
n = 0/2 = 0
As the no: of terms (n) cannot be zero , n = 19
No of terms = 19
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