t(n) = 6*n - 1 where n = 1, 2, 3, ...
11, 14, 19, 26 and 35.
The simplest and most obvious sequence is given by the formula: X[n]=136 - n*7 This gives X[11]=59.
Let the numbers by 'm' & 'n' Hence m + n = 12 mn = 35 Hence m = 35/n Substitute 35/n + n = 12 Multiply through by 'n' 35 + n^2 = 12n n^2 - 12n + 35 = 0 It is now in Quadratic Form ; Factor ( n - 7)(n - 5) = 0 Hence n= 7 or n= 5 It follows that m = 5 , or m = 7
47 and 53. This sequence can be 6n+5. For n= 2,3,4,5,6,7 etc.
n + 6 * * * * * I suggest you try t(n) = 6n + 5 instead.
Let the smaller be n, then the larger is n+1; and: n + 4(n+1) = 59 → n + 4n + 4 = 59 → 5n = 55 → n = 11 → the two consecutive integers are 11 and 12.
23-8 = 15 so 50-N = 15 N = 35
35 * * * * * That is the next term. The question, however, is about the nth term. And that is 6*n - 1
35° 9′ 47″ N, 33° 18′ 59″ E35.163056, 33.316389
35°37′59″n 88°49′15″w
49° 59′ 37″ N, 15° 43′ 23″ E49.993611, 15.723056
31°42′11″N 35°11′44″E
53°45′N 071°59′W
35° 41′ 46″ N, 51° 25′ 23″ E
Approximate latitude: 17°35′N to 18°35′N Approximate longitude: 76°'W to 78°30′W The coordinates of its capital, Kingston: Latitude: 17°59′N Longitude: 76°48′W
a (sub n) = 35 - (n - 1) x d