Understanding the Arithmetic Progression Problem

This problem involves an arithmetic progression (A.P.), which is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, often denoted by '$d$'. We are given information about the sums of specific groups of terms and asked to find the sum of the first 10 terms.

Key Information Given:

• The sequence is an arithmetic progression.
• The sum of the first three terms ($a_1, a_2, a_3$) is $24$.
• The sum of the next three terms ($a_4, a_5, a_6$) is $51$.
• We need to find the sum of the first $10$ terms ($S_{10}$).

Solving for the First Term and Common Difference

Let the first term of the A.P. be '$a$' and the common difference be '$d$'. The terms of the A.P. can be represented as: $a, a+d, a+2d, a+3d, a+4d, a+5d, \dots$

Sum of the First Three Terms

The sum of the first three terms is given as $24$. We can write this using the terms of the A.P.: $a_1 + a_2 + a_3 = a + (a+d) + (a+2d)$ $$a + (a+d) + (a+2d) = 24$$ Combining like terms, we get: $$3a + 3d = 24$$ Dividing the entire equation by $3$, we get a simpler equation: $$a + d = 8 \quad (*)$$ This tells us that the second term ($a_2 = a+d$) is $8$. From this, we can express '$a$' in terms of '$d$': $$a = 8 - d$$

Sum of the Next Three Terms

The sum of the next three terms (the $4^{th}$, $5^{th}$, and $6^{th}$ terms) is given as $51$. These terms are $a+3d$, $a+4d$, and $a+5d$. $a_4 + a_5 + a_6 = (a+3d) + (a+4d) + (a+5d)$ $$(a+3d) + (a+4d) + (a+5d) = 51$$ Combining like terms: $$3a + 12d = 51$$ Dividing this equation by $3$: $$a + 4d = 17 \quad (**)$$

Finding the Values of 'a' and 'd'

Now we have a system of two linear equations with two variables ('$a$' and '$d$'): 1. $a + d = 8$ 2. $a + 4d = 17$ We can solve this system. Substitute the expression for '$a$' from equation ($*$) into equation ($**$): $$(8 - d) + 4d = 17$$ $$8 + 3d = 17$$ Subtract $8$ from both sides: $$3d = 17 - 8$$ $$3d = 9$$ Divide by $3$: $$d = \frac{9}{3}$$ $$d = 3$$ Now that we have the common difference '$d$', we can find the first term '$a$' using equation ($*$): $$a = 8 - d$$ $$a = 8 - 3$$ $$a = 5$$ So, the first term of the A.P. is $5$ and the common difference is $3$. The sequence starts $5, 8, 11, 14, 17, 20, \dots$

Calculating the Sum of the First 10 Terms

We need to find the sum of the first $10$ terms ($S_{10}$). The formula for the sum of the first '$n$' terms of an A.P. is: $$S_n = \frac{n}{2} [2a + (n-1)d]$$ In this case, $n=10$, $a=5$, and $d=3$. Plugging these values into the formula: $$S_{10} = \frac{10}{2} [2(5) + (10-1)(3)]$$ $$S_{10} = 5 [10 + (9)(3)]$$ $$S_{10} = 5 [10 + 27]$$ $$S_{10} = 5 [37]$$ $$S_{10} = 185$$

Therefore, the sum of the first $10$ terms of this arithmetic progression is $185$.