Let's break down and solve both problems from the image.

Problem 8:

In an Arithmetic Progression (A.P.) the sum of the first three terms is 18, and the sum of the squares of these terms is 126. Find the terms.

Let the three terms be $a - d$, $a$, and $a + d$, where:

• $a$ is the middle term (second term),
• $d$ is the common difference.

1. Sum of the first three terms: $(a - d) + a + (a + d) = 18$ This simplifies to: $3a = 18 \quad \Rightarrow \quad a = 6$

2. Sum of the squares of the first three terms: $(a - d)^2 + a^2 + (a + d)^2 = 126$ Substitute $a = 6$: $(6 - d)^2 + 6^2 + (6 + d)^2 = 126$ This simplifies to: $(36 - 12d + d^2) + 36 + (36 + 12d + d^2) = 126$ Combine like terms: $2d^2 + 108 = 126$ Solve for $d^2$: $2d^2 = 126 - 108 \quad \Rightarrow \quad 2d^2 = 18 \quad \Rightarrow \quad d^2 = 9$ Thus, $d = 3$ (since the common difference is generally taken as positive).

So, the three terms are: $6 - 3 = 3, \quad 6, \quad 6 + 3 = 9$ Thus, the terms are 3, 6, and 9.

Problem 9:

In an A.P. of five terms, the sum of the last four terms is 38, and the product of the second and fifth terms is 70. Find the progression.

Let the five terms be $a - 2d$, $a - d$, $a$, $a + d$, and $a + 2d$, where:

• $a$ is the middle term,
• $d$ is the common difference.

1. Sum of the last four terms: The last four terms are $a - d$, $a$, $a + d$, and $a + 2d$. Their sum is: $(a - d) + a + (a + d) + (a + 2d) = 38$ This simplifies to: $4a + 2d = 38 \quad \Rightarrow \quad 2a + d = 19 \quad \text{(Equation 1)}$

2. Product of the second and fifth terms: The second term is $a - d$ and the fifth term is $a + 2d$. Their product is: $(a - d)(a + 2d) = 70$ Expand this equation: $a^2 + 2ad - ad - 2d^2 = 70 \quad \Rightarrow \quad a^2 + ad - 2d^2 = 70 \quad \text{(Equation 2)}$

Now, solve the system of equations from Equation 1 and Equation 2. I'll solve this next.The solutions to the system of equations are:

1. $a = 8$, $d = 3$
2. $a = 11$, $d = -3$

Now, let's calculate the terms for both possible solutions:

1. For $a = 8$ and $d = 3$:

   • First term: $8 - 2(3) = 2$
   • Second term: $8 - 3 = 5$
   • Third term: $8$
   • Fourth term: $8 + 3 = 11$
   • Fifth term: $8 + 2(3) = 14$

   The progression is 2, 5, 8, 11, 14.

2. For $a = 11$ and $d = -3$:

   • First term: $11 - 2(-3) = 17$
   • Second term: $11 - (-3) = 14$
   • Third term: $11$
   • Fourth term: $11 + (-3) = 8$
   • Fifth term: $11 + 2(-3) = 5$

   The progression is 17, 14, 11, 8, 5.

So, the possible progressions are 2, 5, 8, 11, 14 and 17, 14, 11, 8, 5.

Would you like further details on any step? Here are some related questions for further practice:

1. What is the general formula for the sum of terms in an arithmetic progression?
2. How would the solution change if the sum of squares of the terms was different in problem 8?
3. How can we determine whether $d$ should be positive or negative in an arithmetic progression?
4. What is the relationship between common difference and consecutive terms in an A.P.?
5. Can an arithmetic progression have negative terms throughout?

Tip: When solving systems of equations involving A.P., expressing each term in terms of the common difference helps simplify the relationships.