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find the first term of arithmetic sequence 
with a common difference of 11 and whose 27th 
term is 263?
The formula is 
Tn = T1 + (n-1)d
although some teachers and books use the letter "a"
for the terms where I use the letter "T".  
Substitute 27 for n, and 11 for d:
T27 = T1 + (27-1)(11)
Now substitute 263 for T27:
263 = T1 + (26)(11)
263 = T1 + 286
263 - 286 = T1
      -23 = T1
The sequence is:
T1 = -23;  T2 = -12;  T3 = -1;  T4 = 10;  T5 = 21;
T6 = 32;  T7 = 43;  T8 = 54;  T9 = 65;  T10 = 76;
T11 = 87;  T12 = 98;  T13 = 109;  T14 = 120;  T15 = 131;
T16 = 142;  T17 = 153;  T18 = 164;  T19 = 175;  T20 = 186;
T21 = 197;  T22 = 208;  T23 = 219;  T24 = 230;  T25 = 241;
T26 = 252;  T27 = 263 
Edwin