Problem Analysis: We need to find the sum of the first 14 terms ($S_{14}$) of an Arithmetic Progression (A.P.). We are given the 10th term ($a_{10} = 29.25$) and the 6th term ($a_6 = 25.25$).

Finding the Common Difference (d)

The formula for the nth term of an A.P. is $a_n = a + (n-1)d$, where '$a$' is the first term and '$d$' is the common difference.

• $a_{10} = a + (10-1)d = a + 9d = 29.25$
• $a_6 = a + (6-1)d = a + 5d = 25.25$

Subtracting the equation for $a_6$ from the equation for $a_{10}$: $(a + 9d) - (a + 5d) = 29.25 - 25.25$ $4d = 4$ $d = \frac{4}{4}$ $d = 1$

Finding the First Term (a)

Substitute the value of $d=1$ into the equation for $a_6$: $a + 5d = 25.25$ $a + 5(1) = 25.25$ $a + 5 = 25.25$ $a = 25.25 - 5$ $a = 20.25$

Calculating the Sum of the First 14 Terms (S₁₄)

The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Using $n=14$, $a=20.25$, and $d=1$: $S_{14} = \frac{14}{2}[2(20.25) + (14-1)(1)]$ $S_{14} = 7[40.5 + (13)(1)]$ $S_{14} = 7[40.5 + 13]$ $S_{14} = 7[53.5]$ $S_{14} = 374.5$

Therefore, the sum of the first 14 terms is 374.5.