How many term of sequence 57,54,51... show be taken to get their sum zero
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5
Answer:
39
Step-by-step explanation:
a= 57
d= -3
Sn = 0
Sn = n/2 {2a+(n-1)d}
n/2 {2a+(n-1)d} = 0
n/2 {2×57+(n-1)×-3} = 0
n/2 {114-3n+3} = 0
n/2 {117-3n} = 0
117n/2 - 3n²/2 = 0
Multiply Eq by 2
117n - 3n² = 0
- 3n² = -117n
n² = 117n/3
n = 39
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