The interval grows by 2 for each term so this is a geometric sequence involving a n² somewhere.
x0=4
x1=x0+7 x1=4+(5+2(1))
x2=x1+9 x2=4+(5+2(1))+(5+2(2))
x3=4+(5+2(1))+(5+2(2))+(5+2(3))
rearranging
x3=4+(5+5+5)+(2(1)+2(2)+2(3))
The end of this is the triangular sequence, so...
xn=4+5n+2(n2+n)/2
collecting terms and simplifying...
xn=4+5n+(n2+n)
xn=n2+6n+4 ■
+9
10n + 1
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
63
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.
+9
10n + 1
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
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
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