The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
To find the nth term of the sequence -4, -1, 4, 11, 20, 31, we first identify the pattern in the differences between the terms: 3, 5, 7, 9, 11, which increases by 2 each time. This suggests a quadratic relationship. The nth term can be expressed as ( a_n = n^2 + n - 4 ). Thus, the nth term of the sequence is ( a_n = n^2 + n - 4 ).
+9
Difference is 5,7,9,11,13 Second difference is 2 (2x)^2 gives 4,9,16,25 Difference between 2x^2 and sequence is -5 Thus, the nth term will be (2n)^2-5
2n^2-1
31 - n
10n + 1
The nth term in the arithmetic progression 10, 17, 25, 31, 38... will be equal to 7n + 3.
The nth term is 6n+1 and so the next term will be 31
the nth term is = 31 + (n x -9) where n = 1,2,3,4,5 ......... so the 1st term is 31+ (1x -9) = 31 - 9 =22 so the 6th tern is 31 + (6 x -9) = -23 Hope this helps
63
11