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Step-by-Step Solution
Step 1: Represent the five terms in A.P.
Let the common difference of the arithmetic progression (A.P.) be $d$, and let the middle term be $a$. The five consecutive terms can be written as:
$a - 2d,\; a - d,\; a,\; a + d,\; a + 2d.$
Step 2: Use the sum condition
We are given that the sum of these five terms is 25. Therefore,
$(a - 2d) + (a - d) + a + (a + d) + (a + 2d) = 25.
$
Simplify the left-hand side:
$5a = 25
\quad \Longrightarrow \quad
a = 5.
$
Step 3: Apply the product condition
The product of the five terms is given as 2520. Thus,
$(a - 2d)\cdot(a - d)\cdot a \cdot(a + d)\cdot(a + 2d) = 2520.
$
Substitute $a = 5$ into this expression:
$(5 - 2d)(5 - d)(5)(5 + d)(5 + 2d) = 2520.
Notice that we can group the factors as
$(5 - 2d)(5 + 2d)$ and $(5 - d)(5 + d)$, which simplify to $(25 - 4d^2)$ and $(25 - d^2)$ respectively. Hence, the product becomes:
$(25 - 4d^2)(25 - d^2)\cdot 5 = 2520.
$
Step 4: Form and solve the equation for $d$
Let us set
$
(25 - 4d^2)(25 - d^2) = k
$
and note that $k \cdot 5 = 2520,$ so $k = \frac{2520}{5} = 504.$
Hence,
$(25 - 4d^2)(25 - d^2) = 504.
$
Upon expansion or direct factorization (and comparing with the known factorization steps), we get the equivalent condition that leads to the possible values of $d$:
$d^2 = 1
\quad \text{or} \quad
\; d^2 = \frac{121}{4}.
$
Thus,
$
d = \pm 1
\quad \text{or} \quad
d = \pm \frac{11}{2}.
$
Step 5: Use the given fraction term to eliminate invalid $d$ values
The problem states that one of the five numbers is $-\frac{1}{2}.$ If $d = \pm 1,$ our terms become integers (since $a=5$ is integer and $d=1$ is integer), and we would not get a fractional term like $-\frac{1}{2}.$ Thus, $d = \pm 1$ is not suitable.
The valid solutions occur for $d = \pm \frac{11}{2}.$
Step 6: Identify the greatest term
We want the largest term. If we take $d = \frac{11}{2},$ then the terms become:
$a - 2d = 5 - 11 = -6, \\
a - d = 5 - \frac{11}{2} = \frac{10 - 11}{2} = -\frac{1}{2}, \\
a = 5, \\
a + d = 5 + \frac{11}{2} = \frac{21}{2}, \\
a + 2d = 5 + 11 = 16.
$
The greatest term here is $16.$ (One can check similarly for $d = -\frac{11}{2},$ but it will not produce a larger maximum term than 16.)
Final Answer
The greatest number among the five terms of the A.P. is 16.
