Understanding AP and GP Concepts

This problem involves two types of sequences: Arithmetic Progression (AP) and Geometric Progression (GP).

• An Arithmetic Progression (AP) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by '\(d\)'.
• A Geometric Progression (GP) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio, denoted by '\(r\)'.

Setting up the Equations

Let the 5 consecutive terms of the AP be represented symmetrically around a central term '\(a\)' with a common difference '\(d\)'. The terms are:

\( a-2d, \quad a-d, \quad a, \quad a+d, \quad a+2d \)

We are given that the product of these 5 terms is 229635:

\( (a-2d)(a-d)(a)(a+d)(a+2d) = 229635 \)

Using the difference of squares formula, \((x-y)(x+y) = x^2 - y^2\), we can simplify the product:

\( a \times [(a-d)(a+d)] \times [(a-2d)(a+2d)] = 229635 \)

\( a (a^2 - d^2) (a^2 - 4d^2) = 229635 \quad (Equation 1) \)

We are also given that the first term (\(t_1 = a-2d\)), second term (\(t_2 = a-d\)), and fifth term (\(t_5 = a+2d\)) are in GP. For terms to be in GP, the square of the middle term is equal to the product of the other two terms. In this case, \((t_2)^2 = t_1 \times t_5\).

\( (a-d)^2 = (a-2d)(a+2d) \)

Expanding both sides:

\( a^2 - 2ad + d^2 = a^2 - 4d^2 \)

Now, let's solve for the relationship between '\(a\)' and '\(d\)':

\( -2ad + d^2 = -4d^2 \)

\( 5d^2 = 2ad \)

Since the terms are consecutive terms of an AP and their product is non-zero, the common difference '\(d\)' cannot be zero. Therefore, we can divide both sides by '\(d\)':

\( 5d = 2a \)

This gives us a relationship between '\(a\)' and '\(d\)':

\( a = \frac{5}{2}d \quad (Equation 2) \)

Calculating the Common Difference

Substitute the expression for '\(a\)' from Equation 2 into Equation 1:

\( \left(\frac{5}{2}d\right) \left( \left(\frac{5}{2}d\right)^2 - d^2 \right) \left( \left(\frac{5}{2}d\right)^2 - 4d^2 \right) = 229635 \)

Calculate the terms inside the parentheses:

\( \left(\frac{5}{2}d\right)^2 = \frac{25}{4}d^2 \)

Substitute this back:

\( \left(\frac{5}{2}d\right) \left( \frac{25}{4}d^2 - d^2 \right) \left( \frac{25}{4}d^2 - 4d^2 \right) = 229635 \)

Simplify the terms in the parentheses:

\( \frac{25}{4}d^2 - d^2 = \frac{25d^2 - 4d^2}{4} = \frac{21}{4}d^2 \)

\( \frac{25}{4}d^2 - 4d^2 = \frac{25d^2 - 16d^2}{4} = \frac{9}{4}d^2 \)

Now substitute these simplified terms back into the equation:

\( \left(\frac{5}{2}d\right) \left(\frac{21}{4}d^2\right) \left(\frac{9}{4}d^2\right) = 229635 \)

Multiply the coefficients and the powers of '\(d\)':

\( \frac{5 \times 21 \times 9}{2 \times 4 \times 4} d^{(1+2+2)} = 229635 \)

\( \frac{945}{32} d^5 = 229635 \)

Now, solve for '\(d^5\)':

\( d^5 = 229635 \times \frac{32}{945} \)

Perform the division:

\( \frac{229635}{945} = 243 \)

So the equation becomes:

\( d^5 = 243 \times 32 \)

Recognize the values as powers of 3 and 2:

\( 243 = 3^5 \)

\( 32 = 2^5 \)

Substitute these powers back:

\( d^5 = 3^5 \times 2^5 \)

\( d^5 = (3 \times 2)^5 \)

\( d^5 = 6^5 \)

Taking the fifth root of both sides:

\( d = 6 \)

Verifying the Solution

If the common difference '\(d\)' is 6, we can find the value of '\(a\)' using Equation 2:

\( a = \frac{5}{2}d = \frac{5}{2}(6) = 15 \)

The 5 consecutive terms of the AP are:

\( a-2d = 15 - 2(6) = 15 - 12 = 3 \)

\( a-d = 15 - 6 = 9 \)

\( a = 15 \)

\( a+d = 15 + 6 = 21 \)

\( a+2d = 15 + 2(6) = 15 + 12 = 27 \)

The terms are 3, 9, 15, 21, 27.

Check the product:

\( 3 \times 9 \times 15 \times 21 \times 27 = 27 \times 15 \times 21 \times 27 = 405 \times 567 = 229635 \)

This matches the given product.

Check the GP condition: The first (3), second (9), and fifth (27) terms must be in GP.

Ratio between the second and first term: \( \frac{9}{3} = 3 \)

Ratio between the fifth and second term: \( \frac{27}{9} = 3 \)

Since the ratios are the same, the terms 3, 9, and 27 are indeed in GP.

Thus, the calculated common difference '\(d=6\)' is correct.