Give the sequence 7,16,21,......84 . Find the number of term

An AP has a constant difference between consecutive terms.
Calculate the first differences:
$d_1 = 16 - 7 = 9$
$d_2 = 21 - 16 = 5$
Since $d_1 \neq d_2$, the sequence is not an Arithmetic Progression.

A GP has a constant ratio between consecutive terms.
Calculate the ratios:
$r_1 = \frac{16}{7}$
$r_2 = \frac{21}{16}$
Since $r_1 \neq r_2$, the sequence is not a Geometric Progression.

A quadratic sequence has a constant second difference.
First differences: $9, 5$
Second difference: $5 - 9 = -4$
Since the second difference is constant, the sequence is a quadratic sequence of the form $a_n = An^2 + Bn + C$.
Using the second difference, $2A = -4 \implies A = -2$.
Using the first differences:
$3A + B = 9 \implies 3(-2) + B = 9 \implies -6 + B = 9 \implies B = 15$.
Using the first term:
$A + B + C = 7 \implies -2 + 15 + C = 7 \implies 13 + C = 7 \implies C = -6$.
So, the general term for this quadratic sequence is $a_n = -2n^2 + 15n - 6$.

Set $a_n = 84$:
$84 = -2n^2 + 15n - 6$
Rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$:
$2n^2 - 15n + 6 + 84 = 0$
$2n^2 - 15n + 90 = 0$

The discriminant is $\Delta = b^2 - 4ac$.
$\Delta = (-15)^2 - 4(2)(90)$
$\Delta = 225 - 720$
$\Delta = -495$

Since the discriminant $\Delta < 0$, there are no real solutions for $n$. This means that 84 is not a term in the sequence defined by the pattern of the first three terms ($a_n = -2n^2 + 15n - 6$).

Conclusion: Based on the given terms, the sequence is neither an Arithmetic Progression nor a Geometric Progression. Furthermore, if we assume a quadratic pattern, the number 84 does not fit this pattern as a term in the sequence. Therefore, it is not possible to find the number of terms ending at 84 with the given sequence.