For the first one, you're just adding 14 each time. -62, -48, -34, -20, -6, 8, 22, 36, 50, 64. So the 10th term is 64.
I'm not sure about the second one, but I believe it is neither because an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant and a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.
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For the first one, you're just adding 14 each time. -62, -48, -34, -20, -6, 8, 22, 36, 50, 64. So the 10th term is 64.
I'm not sure about the second one, but I believe it is neither because an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant and a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.
The difference is 14, the first term is -62, the first term number is 1
t(n) = dn + c
-62 = +14(1) + c
c = -76
t(10) = 14(10) -76
t(10) = 64
I use this approach only because it is similar to y = mx + b.
Here diff is d=14
and according to the formula
(n-1)d + a... Where a is first term and n is tle last term
so n=10, a=(-62)
**putting the values**
the answer is 64
a1=-62, d=14
a10 =a1+d(9)
a10 =-62 +126=64
second sequence is neither arithmetic nor geometric