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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