The given sequence is an arithmetic sequence with a common difference of 8. To find the nth term, we use the formula for the nth term of an arithmetic sequence: ( a_n = a_1 + (n-1)d ), where ( a_n ) is the nth term, ( a_1 ) is the first term, ( n ) is the position of the term, and ( d ) is the common difference. In this case, the first term ( a_1 = 15 ) and the common difference ( d = 8 ). Therefore, the nth term for this sequence is ( a_n = 15 + (n-1)8 = 15 + 8n - 8 = 8n + 7 ).
2n^2-1
+9
a (sub n) = 35 - (n - 1) x d
Type your answer here... The next numbers in the sequence are 55, 70, 87, 106, 127, etc.
The nth term is 7n-4 and so the next number in the sequence is 31
Assuming this is a linear or arithmetic sequence, the nth term is Un = 31 - 8n. But, there are infinitely many polynomials of order 5 or higher, and many other functions that will fit the above 5 numbers.
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
2n^2-1
31 - n
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 pattern is: +11, +15, +19, +23, +27 (4n+7) So, the next number would be: (4*6 + 7) = 24 + 7 = +31 Therefore, the answer is: 105 + 31 = 136
+9
(5/31) / (15/23) = (5/31)*(23/15) = (5*23)/(31*15) = (1*23)/(31/3) = 23/93
a (sub n) = 35 - (n - 1) x d
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.
5 to 7 is 27 to 17 is 1017 to 19 is 219 to 29 is 1029 to 31 is 2there fore following the pattern the nth term is 4131 to 41 is 10