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kt22
can't we use the formula: S(n) = n/2 [2a + (n-1)d]. we know d =-1, and that S = 19
st 1 gives us a. 1 variable, 1 equation.
n= 37/19.

st2 is obviously not suff.

where am i wrong? would appreciate your pov chetansharma.
TIA.

kt22 Good question! The problem does not specify the relation between the 1st and 2nd term, however, so we cannot use that formula as that's for strictly arithmetic progressions.
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kt22
can't we use the formula: S(n) = n/2 [2a + (n-1)d]. we know d =-1, and that S = 19
st 1 gives us a. 1 variable, 1 equation.
n= 37/19.

st2 is obviously not suff.

where am i wrong? would appreciate your pov.
TIA.

You are not wrong. You CAN use this formula. Let's try that.

S(n) = n/2 [2a + (n-1)d]
For statement 1,
Given: S(n) = 19 (in question)
a = 10
n=?

Putting values --> n/2 [2*10 + (n-1)(-1)] = 19
n[20-n+1]=19*2
n(21-n)=38
n^2 - 21n +38 = 0
n(n-19) - 2(n-19) = 0
n = either 19 or 2

Not sufficient.
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