Which term of the arithmetic progression5,15,25.....will be 130 more than its 31th term?
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a = 5
d = 10
ATQ
a + 30d + 130 = a +(n-1)d
30*19 +130 = (n-1)*10
700/10 = n-1
n = 70 +1 = 71 ans
d = 10
ATQ
a + 30d + 130 = a +(n-1)d
30*19 +130 = (n-1)*10
700/10 = n-1
n = 70 +1 = 71 ans
neosingh:
sorry for the mistake,30*19 should be 30*10. so ans should be 430/10 + 1 = 44th teem
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