at the sequence - 4 9 16 25 36 -Which expression could be used to find the nth term in this sequence?
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On observing the sequence, we find that the difference between consecutive numbers are odd numbers.
9-4=5, 16-9=7, 25-16=9, 36-25=11.
We cannot solve the question by using the standard formulas for AP.
Here, the common difference themselves are in AP.
The first difference (between 2nd and 1st terms) = 5
The second difference (between 3rd and 2nd terms) = 7
The difference between the nth and (n-1)th terms will be the (n-1)th difference.
To find the nth term of the main sequence,
nth term = first term (ie. 4) + sum of (n-1) differences
Eg: 2nd term = first term + sum of 1 difference = 4+5
3rd term = first term + sum of 2 differences = 4 + (5+7) = 4+12 = 16
The sum of n-1 differences = (n-1)/2*[2(5) + (n-2)2]
{by using the standard formula for summation of n terms in AP}
Thus, the sum of the n-1 differences = [(n-1)(2n+6)]/2
Therefore, the nth term is 4 + [(n-1)(2n+6)]/2
PS: If you find the answer useful, please mark it as brainliest
9-4=5, 16-9=7, 25-16=9, 36-25=11.
We cannot solve the question by using the standard formulas for AP.
Here, the common difference themselves are in AP.
The first difference (between 2nd and 1st terms) = 5
The second difference (between 3rd and 2nd terms) = 7
The difference between the nth and (n-1)th terms will be the (n-1)th difference.
To find the nth term of the main sequence,
nth term = first term (ie. 4) + sum of (n-1) differences
Eg: 2nd term = first term + sum of 1 difference = 4+5
3rd term = first term + sum of 2 differences = 4 + (5+7) = 4+12 = 16
The sum of n-1 differences = (n-1)/2*[2(5) + (n-2)2]
{by using the standard formula for summation of n terms in AP}
Thus, the sum of the n-1 differences = [(n-1)(2n+6)]/2
Therefore, the nth term is 4 + [(n-1)(2n+6)]/2
PS: If you find the answer useful, please mark it as brainliest
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