If a,b,c,d,e,f are sequentially the terms of an A.P. belong to set `{1, 2 , 3, ……, 9}` where all the terms a,b,c,… are in increasing order, then the last digit of `a^b xx c^d xx e^f` is :

Options

A. 5
B. 2
C. 7
D. either of (a) and (b)

Answer

To solve the problem, we need to find the last digit of the expression \( a^b \times c^d \times e^f \), where \( a, b, c, d, e, f \) are sequential terms of an arithmetic progression (A.P.) and belong to the set \{1, 2, 3, ..., 9\}.

### Step-by-step Solution:

1. **Identify the Terms of the A.P.:**
Since \( a, b, c, d, e, f \) are sequential terms of an A.P., we can express them as:
\[ a, a+1, a+2, a+3, a+4, a+5 \]
The last term \( f \) must be less than or equal to 9. Therefore, the maximum value for \( a \) is 4 (since \( 4 + 5 = 9 \)).

2. **Consider Possible Values of \( a \):**
The possible values for \( a \) are 1, 2, 3, and 4. We will calculate the last digit of \( a^b \times c^d \times e^f \) for each case.

3. **Calculate for Each Case:**
- **Case 1:** \( a = 1 \)
\[ a = 1, b = 2, c = 3, d = 4, e = 5, f = 6 \]
\[ 1^2 \times 3^4 \times 5^6 = 1 \times 81 \times 15625 \]
Last digit = \( 1 \times 1 \times 5 = 5 \).

- **Case 2:** \( a = 2 \)
\[ a = 2, b = 3, c = 4, d = 5, e = 6, f = 7 \]
\[ 2^3 \times 4^5 \times 6^7 = 8 \times 1024 \times 279936 \]
Last digit = \( 8 \times 4 \times 6 = 8 \times 4 = 32 \) (last digit 2).

- **Case 3:** \( a = 3 \)
\[ a = 3, b = 4, c = 5, d = 6, e = 7, f = 8 \]
\[ 3^4 \times 5^6 \times 7^8 = 81 \times 15625 \times 5764801 \]
Last digit = \( 1 \times 5 \times 1 = 5 \).

- **Case 4:** \( a = 4 \)
\[ a = 4, b = 5, c = 6, d = 7, e = 8, f = 9 \]
\[ 4^5 \times 6^6 \times 8^7 = 1024 \times 46656 \times 2097152 \]
Last digit = \( 4 \times 6 \times 2 = 48 \) (last digit 8).

4. **Conclusion:**
The last digits we found are:
- For \( a = 1 \): last digit = 5
- For \( a = 2 \): last digit = 2
- For \( a = 3 \): last digit = 5
- For \( a = 4 \): last digit = 8

The possible last digits are 5, 2, and 8. The last digit of \( a^b \times c^d \times e^f \) can be either 2 or 5 depending on the starting term of the A.P.

### Final Answer:
The last digit of \( a^b \times c^d \times e^f \) can be either **2 or 5**.