Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is the square of an integer. The least possible value of the number of digits of $c$ is

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day,$4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

Let $S = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a+b+c=21, a \leq b \leq c\}$ and $T = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a, b, c \text{ are in } AP\}$, where $\mathbb{N}$ is the set of all natural numbers. Then, the number of elements in the set $S \cap T$ is:

If $a, b, c, d, e$ are in an arithmetic progression,then what is the value of $a - 4b + 6c - 4d + e$?

Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression $3, 7, 11, \ldots$. If $40 < \left(\frac{6}{n(n+1)} \sum_{k=1}^{n} S_{k}\right) < 42$,then $n$ equals

If $x, y, z$ are in an arithmetic progression,and $a$ is the arithmetic mean of $x$ and $y$,and $b$ is the arithmetic mean of $y$ and $z$,then what is the arithmetic mean of $a$ and $b$?