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A.s Problem : ____________ Find A.s if the sum of all terms starting from the second term is (-36) and the sum of all terms except the last term is Zero and the difference between the sixth term and the tenth term is equal to 16. _________________________

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No LaTex, not interested.
@experimentX :If u have time :)
let, \( n\) be the number of terms!! given, \( S_n - a = -36 \implies \frac{n (a+l)}{2} - a = -36\) or \( \frac{n (2a+(n-1)d)}{n} - a = -36\) -------- 1 \( S_n - l = 0 \implies \frac{n (a+l)}{2} - l = 0\) or, \frac{n(2a+(n-1)d)}{n} - a - (n-1)d = 0\) -------- 2 <-- also it implies that AP must be And \( a + (5-1)d - (a + (10-1)d = 16 \) ---- 3 3 unknown variables and three equations ... it should be solveable!!

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Other answers:

Ah ... that n in 1 down there is supposed to be 2
aha ,I got it :) ,TYSM for your time :) @experimentX
yw ..

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