The first term of an AP is 3 and the sum of all 6 terms is 48. If the sum of all the terms is 168, calculate: a. the common difference, d b. the last term.

Part (a): Sum of all terms

Step 1: Sum formula for an AP
$S_n = \frac{n}{2} \left[2a + (n-1)d\right]$

Step 2: Substitute $n=2025$, $a=4$, $d=3$
$S_{2025} = \frac{2025}{2} \left[2(4) + (2025-1)(3)\right]$
$S_{2025} = \frac{2025}{2} \left[8 + 2024 \times 3\right]$

Step 3: Simplify inside brackets
$2024 \times 3 = 6072$
$8 + 6072 = 6080$
$S_{2025} = \frac{2025}{2} \times 6080$

Step 4: Simplify
$\frac{6080}{2} = 3040$
$S_{2025} = 2025 \times 3040$

Step 5: Compute $2025 \times 3040$
$2000 \times 3040 = 6\,080\,000$
$25 \times 3040 = 76\,000$
$6\,080\,000 + 76\,000 = 6\,156\,000$
$6156000$

Part (b): Last term

Step 1: Last term formula
$l = a + (n-1)d$

Step 2: Substitute $n=2025$, $a=4$, $d=3$
$l = 4 + (2025-1) \times 3$
$l = 4 + 2024 \times 3$

Step 3: Simplify
$2024 \times 3 = 6072$
$4 + 6072 = 6076$
$6076$