Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.5

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers And Sequences Ex 2.5

 

Question 1. Check whether the following sequences are in A.P.?
(i) a – 3, a – 5, a – 7,...
(ii) \( \frac { 1 }{ 2 }, \frac { 1 }{ 3 }, \frac { 1 }{ 4 }, \frac { 1 }{ 5 }, .......... \)
(iii) 9, 13, 17, 21, 25,..
(iv) \( \frac {-1}{ 3 }, 0, \frac { 1 }{ 3 }, \frac { 2 }{ 3 } \)
(v) 1,-1, 1,-1, 1, -1, ...
Answer:
(i) Given sequence: \( a - 3, a - 5, a - 7, ... \)
To check if it's an A.P., we find the common difference between consecutive terms.
First, find the difference between the second and first terms:
\( t_2 - t_1 = (a - 5) - (a - 3) = a - 5 - a + 3 = -2 \)
Next, find the difference between the third and second terms:
\( t_3 - t_2 = (a - 7) - (a - 5) = a - 7 - a + 5 = -2 \)
Since \( t_2 - t_1 = t_3 - t_2 \), the common difference is the same (which is -2). This consistency is the hallmark of an Arithmetic Progression.
Therefore, the given sequence is an A.P.
(ii) Given sequence: \( \frac { 1 }{ 2 }, \frac { 1 }{ 3 }, \frac { 1 }{ 4 }, \frac { 1 }{ 5 }, .......... \)
First, find the difference between the second and first terms:
\( t_2 - t_1 = \frac { 1 }{ 3 } - \frac { 1 }{ 2 } = \frac { 2 - 3 }{ 6 } = \frac { -1 }{ 6 } \)
Next, find the difference between the third and second terms:
\( t_3 - t_2 = \frac { 1 }{ 4 } - \frac { 1 }{ 3 } = \frac { 3 - 4 }{ 12 } = \frac { -1 }{ 12 } \)
Since \( t_2 - t_1 \neq t_3 - t_2 \) (because \( \frac{-1}{6} \neq \frac{-1}{12} \)), the common difference is not the same. This means the sequence does not follow a constant addition pattern.
Therefore, the sequence is not in A.P.
(iii) Given sequence: 9, 13, 17, 21, 25,..
First, find the difference between the second and first terms:
\( t_2 - t_1 = 13 - 9 = 4 \)
Next, find the difference between the third and second terms:
\( t_3 - t_2 = 17 - 13 = 4 \)
Then, the difference between the fourth and third terms:
\( t_4 - t_3 = 21 - 17 = 4 \)
And the difference between the fifth and fourth terms:
\( t_5 - t_4 = 25 - 21 = 4 \)
Since the common difference is equal throughout the sequence (which is 4), it means a constant value is added to each term to get the next one.
Therefore, the sequence is in A.P.
(iv) Given sequence: \( \frac {-1}{ 3 }, 0, \frac { 1 }{ 3 }, \frac { 2 }{ 3 } \)
First, find the difference between the second and first terms:
\( t_2 - t_1 = 0 - \left( -\frac { 1 }{ 3 } \right) = 0 + \frac { 1 }{ 3 } = \frac { 1 }{ 3 } \)
Next, find the difference between the third and second terms:
\( t_3 - t_2 = \frac { 1 }{ 3 } - 0 = \frac { 1 }{ 3 } \)
Since \( t_2 - t_1 = t_3 - t_2 \), the common difference is the same (which is \( \frac{1}{3} \)). This steady increase confirms it's an Arithmetic Progression.
Therefore, the sequence is in A.P.
(v) Given sequence: 1, -1, 1, -1, 1, -1, ...
First, find the difference between the second and first terms:
\( t_2 - t_1 = -1 - 1 = -2 \)
Next, find the difference between the third and second terms:
\( t_3 - t_2 = 1 - (-1) = 1 + 1 = 2 \)
Since \( t_2 - t_1 \neq t_3 - t_2 \) (because \( -2 \neq 2 \)), the common difference is not the same. This alternating pattern means it's not an A.P.
Therefore, the sequence is not an A.P.
In simple words: To check if a sequence is an A.P., subtract each term from the one after it. If all these differences are the same, it's an A.P. If they are different, it is not.

🎯 Exam Tip: Always calculate at least two differences between consecutive terms to confirm a common difference. If the first two differences are not equal, you can stop there and conclude it's not an A.P.

 

Question 2. First term a and common difference d are given below. Find the corresponding A.P.?
(i) a = 5, d = 6
(ii) a = 7, d = -5
(iii) a = \( \frac { 3 }{ 4 }, d = \frac { 1 }{ 2 } \)
Answer:
The general form of an Arithmetic Progression (A.P.) is given by \( a, a + d, a + 2d, a + 3d, ... \), where \( a \) is the first term and \( d \) is the common difference. This formula helps us construct any A.P. if we know these two values.
(i) Given: \( a = 5, d = 6 \)
Using the general form:
First term: \( a = 5 \)
Second term: \( a + d = 5 + 6 = 11 \)
Third term: \( a + 2d = 5 + 2(6) = 5 + 12 = 17 \)
Fourth term: \( a + 3d = 5 + 3(6) = 5 + 18 = 23 \)
So, the A.P. is 5, 11, 17, 23, ...
(ii) Given: \( a = 7, d = -5 \)
Using the general form:
First term: \( a = 7 \)
Second term: \( a + d = 7 + (-5) = 7 - 5 = 2 \)
Third term: \( a + 2d = 7 + 2(-5) = 7 - 10 = -3 \)
Fourth term: \( a + 3d = 7 + 3(-5) = 7 - 15 = -8 \)
So, the A.P. is 7, 2, -3, -8, ...
(iii) Given: \( a = \frac { 3 }{ 4 }, d = \frac { 1 }{ 2 } \)
Using the general form:
First term: \( a = \frac { 3 }{ 4 } \)
Second term: \( a + d = \frac { 3 }{ 4 } + \frac { 1 }{ 2 } = \frac { 3 }{ 4 } + \frac { 2 }{ 4 } = \frac { 5 }{ 4 } \)
Third term: \( a + 2d = \frac { 3 }{ 4 } + 2\left(\frac { 1 }{ 2 }\right) = \frac { 3 }{ 4 } + 1 = \frac { 3 }{ 4 } + \frac { 4 }{ 4 } = \frac { 7 }{ 4 } \)
Fourth term: \( a + 3d = \frac { 3 }{ 4 } + 3\left(\frac { 1 }{ 2 }\right) = \frac { 3 }{ 4 } + \frac { 3 }{ 2 } = \frac { 3 }{ 4 } + \frac { 6 }{ 4 } = \frac { 9 }{ 4 } \)
So, the A.P. is \( \frac { 3 }{ 4 }, \frac { 5 }{ 4 }, \frac { 7 }{ 4 }, \frac { 9 }{ 4 }, ... \)
In simple words: To find the terms of an A.P., start with the first number given (a). Then, keep adding the common difference (d) to get the next number in the list.

🎯 Exam Tip: Remember that adding a negative common difference means the terms will decrease. Pay careful attention to signs when calculating terms.

 

Question 3. Find the first term and common difference of the Arithmetic Progressions whose nth terms are given below
(i) \( t_n = -3 + 2n \)
(ii) \( t_n = 4 – 7n \)
Answer:
(i) Given: \( t_n = -3 + 2n \)
To find the first term (\( a \)), substitute \( n = 1 \) into the formula:
\( a = t_1 = -3 + 2(1) = -3 + 2 = -1 \)
To find the common difference (\( d \)), we first need the second term. Substitute \( n = 2 \):
\( t_2 = -3 + 2(2) = -3 + 4 = 1 \)
The common difference is the difference between any term and its preceding term:
\( d = t_2 - t_1 = 1 - (-1) = 1 + 1 = 2 \)
So, the first term is -1 and the common difference is 2.
(ii) Given: \( t_n = 4 – 7n \)
To find the first term (\( a \)), substitute \( n = 1 \) into the formula:
\( a = t_1 = 4 - 7(1) = 4 - 7 = -3 \)
To find the common difference (\( d \)), we first need the second term. Substitute \( n = 2 \):
\( t_2 = 4 - 7(2) = 4 - 14 = -10 \)
The common difference is the difference between any term and its preceding term:
\( d = t_2 - t_1 = -10 - (-3) = -10 + 3 = -7 \)
So, the first term is -3 and the common difference is -7.
In simple words: To get the first term, put \( n=1 \) into the given formula. To get the common difference, find the second term by putting \( n=2 \), then subtract the first term from the second.

🎯 Exam Tip: Remember that \( t_n \) represents the nth term of the sequence. If the expression for \( t_n \) is linear in \( n \), the sequence is an A.P. and the coefficient of \( n \) is the common difference.

 

Question 4. Find the 19th term of an A.P. -11, -15, -19,...
Answer:
Given A.P.: -11, -15, -19,...
First, identify the first term (\( a \)) and the common difference (\( d \)).
The first term \( (a) = -11 \)
The common difference \( (d) = t_2 - t_1 = -15 - (-11) = -15 + 11 = -4 \)
The formula for the \( n \)th term of an A.P. is \( t_n = a + (n - 1)d \). This formula allows us to find any term in an A.P. without listing all the terms.
We need to find the 19th term, so \( n = 19 \).
Substitute the values into the formula:
\( t_{19} = -11 + (19 - 1)(-4) \)
\( t_{19} = -11 + (18)(-4) \)
\( t_{19} = -11 - 72 \)
\( t_{19} = -83 \)
Therefore, the 19th term of the A.P. is -83.
In simple words: To find a specific term in an A.P., start with the first term, then add the common difference "n-1" times.

🎯 Exam Tip: Be very careful with negative signs, especially when calculating the common difference or when multiplying it by \((n-1)\).

 

Question 5. Which term of an A.P. 16, 11, 6,1, ........... is -54?
Answer:
Given A.P.: 16, 11, 6, 1, ...
We are given that the \( n \)th term (\( t_n \)) is -54.
First, identify the first term (\( a \)) and the common difference (\( d \)):
\( a = 16 \)
\( d = t_2 - t_1 = 11 - 16 = -5 \)
The formula for the \( n \)th term of an A.P. is \( t_n = a + (n - 1)d \). We use this formula to find the value of \( n \).
Substitute the known values into the formula:
\( -54 = 16 + (n - 1)(-5) \)
Now, solve for \( n \):
\( -54 = 16 - 5n + 5 \)
\( -54 = 21 - 5n \)
Subtract 21 from both sides:
\( -54 - 21 = -5n \)
\( -75 = -5n \)
Divide both sides by -5:
\( n = \frac { -75 }{ -5 } \)
\( n = 15 \)
So, the 15th term of the A.P. is -54. This means that if you continue the sequence, the 15th number will be -54.
In simple words: We are looking for the position of -54 in the list. We use the A.P. formula, put in the first term, the common difference, and -54 as the nth term, then solve to find "n", which is the position.

🎯 Exam Tip: When solving for 'n', isolate the term containing 'n' first, then perform division. Double-check your arithmetic, especially with negative numbers.

 

Question 6. Find the middle term(s) of an A.P. 9, 15, 21, 27, ..., 183.
Answer:
Given A.P.: 9, 15, 21, 27, ..., 183
First, identify the first term (\( a \)), last term (\( l \)), and common difference (\( d \)).
\( a = 9 \)
\( l = 183 \)
\( d = t_2 - t_1 = 15 - 9 = 6 \)
To find the middle term(s), we first need to know the total number of terms (\( n \)) in the A.P. The formula for the last term is \( l = a + (n - 1)d \).
Substitute the known values:
\( 183 = 9 + (n - 1)6 \)
\( 183 - 9 = (n - 1)6 \)
\( 174 = 6(n - 1) \)
Divide by 6:
\( \frac { 174 }{ 6 } = n - 1 \)
\( 29 = n - 1 \)
Add 1 to both sides:
\( n = 29 + 1 \)
\( n = 30 \)
Since the number of terms \( n \) is 30 (an even number), there will be two middle terms. These are the \( \frac{n}{2} \)th term and the \( \left(\frac{n}{2} + 1\right) \)th term. For \( n=30 \), the middle terms are the 15th and 16th terms.
Now, calculate these terms using \( t_n = a + (n - 1)d \):
For the 15th term (\( t_{15} \)):
\( t_{15} = 9 + (15 - 1)6 = 9 + (14)6 = 9 + 84 = 93 \)
For the 16th term (\( t_{16} \)):
\( t_{16} = 9 + (16 - 1)6 = 9 + (15)6 = 9 + 90 = 99 \)
So, the middle terms of the A.P. are 93 and 99. Finding 'n' first helps us understand the structure of the A.P. before pinpointing the middle values.
In simple words: First, find how many total numbers are in the list. If it's an odd number, there's one middle number. If it's an even number, there are two middle numbers. Then, use the A.P. formula to find those specific middle numbers.

🎯 Exam Tip: Remember that if 'n' is even, there are two middle terms (\( \frac{n}{2} \) and \( \frac{n}{2} + 1 \)); if 'n' is odd, there is one middle term (\( \frac{n+1}{2} \)).

 

Question 7. If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.
Answer:
Let the first term of the A.P. be \( a \) and the common difference be \( d \).
The formula for the \( n \)th term of an A.P. is \( t_n = a + (n - 1)d \).
Given condition: Nine times the ninth term is equal to fifteen times the fifteenth term.
\( 9 \times t_9 = 15 \times t_{15} \)
Substitute the formula for \( t_n \):
\( 9(a + (9 - 1)d) = 15(a + (15 - 1)d) \)
\( 9(a + 8d) = 15(a + 14d) \)
Expand both sides:
\( 9a + 72d = 15a + 210d \)
Move all terms to one side to form an equation equal to zero:
\( 0 = 15a - 9a + 210d - 72d \)
\( 0 = 6a + 138d \)
We can factor out 6 from the expression:
\( 0 = 6(a + 23d) \)
Now, we need to show that six times the twenty-fourth term (\( 6 \times t_{24} \)) is zero.
The twenty-fourth term \( t_{24} \) is given by \( a + (24 - 1)d = a + 23d \).
From our derived equation, we have \( 6(a + 23d) = 0 \).

\( \implies \) \( 6t_{24} = 0 \)
Thus, it is proved that six times the twenty-fourth term is zero. This type of relationship often emerges in A.P.s due to the linear nature of term progression.
In simple words: We are given a rule about two terms in an A.P. We use the A.P. formula to write this rule as an equation. Then we simplify that equation to show that six times the 24th term must be zero.

🎯 Exam Tip: When proving a relationship, always start with the given condition and manipulate it algebraically to arrive at the desired result. Clearly state the \( t_n \) formula and each step of substitution and simplification.

 

Question 8. If \( 3 + k, 18 – k, 5k + 1 \) are in A.P. then find k?
Answer:
Given that the terms \( 3 + k, 18 - k, 5k + 1 \) are in an Arithmetic Progression (A.P.).
In an A.P., the common difference between consecutive terms is constant. This means the difference between the second and first term must be equal to the difference between the third and second term.
\( t_2 - t_1 = t_3 - t_2 \)
Substitute the given terms into this equation:
\( (18 - k) - (3 + k) = (5k + 1) - (18 - k) \)
Now, simplify and solve for \( k \):
\( 18 - k - 3 - k = 5k + 1 - 18 + k \)
Combine like terms on both sides:
\( 15 - 2k = 6k - 17 \)
Move all terms involving \( k \) to one side and constants to the other side. Add \( 2k \) to both sides:
\( 15 = 6k + 2k - 17 \)
\( 15 = 8k - 17 \)
Add 17 to both sides:
\( 15 + 17 = 8k \)
\( 32 = 8k \)
Divide by 8 to find \( k \):
\( k = \frac { 32 }{ 8 } \)
\( k = 4 \)
The value of \( k \) is 4. This ensures the terms form an A.P., maintaining a constant difference between them.
In simple words: For three numbers to be in an A.P., the middle number should be exactly halfway between the first and third. So, the gap between the first two numbers must be the same as the gap between the last two. We set up an equation using this idea and solve for \( k \).

🎯 Exam Tip: For any three terms \( A, B, C \) to be in A.P., the condition \( 2B = A + C \) (where B is the arithmetic mean) is often quicker to use than \( B - A = C - B \). Both methods yield the same result.

 

Question 9. Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
Answer:
Given that the numbers \( x, 10, y, 24, z \) are in an Arithmetic Progression (A.P.).
In an A.P., the common difference \( d \) between consecutive terms is constant. We can express this using the terms given:
\( t_2 - t_1 = 10 - x \) .......... (1)
\( t_3 - t_2 = y - 10 \) .......... (2)
\( t_4 - t_3 = 24 - y \) .......... (3)
\( t_5 - t_4 = z - 24 \) .......... (4)
Since the common difference is the same throughout, we can equate these differences.
Equating (2) and (3):
\( y - 10 = 24 - y \)
Add \( y \) to both sides and add 10 to both sides:
\( y + y = 24 + 10 \)
\( 2y = 34 \)
Divide by 2:
\( y = \frac { 34 }{ 2 } \)
\( y = 17 \)
Now we have the value of \( y \). Let's use it to find \( x \) and \( z \).
Equating (1) and (2), and substituting \( y = 17 \):
\( 10 - x = y - 10 \)
\( 10 - x = 17 - 10 \)
\( 10 - x = 7 \)
Subtract 10 from both sides:
\( -x = 7 - 10 \)
\( -x = -3 \)

\( \implies \) \( x = 3 \)
Equating (3) and (4), and substituting \( y = 17 \):
\( 24 - y = z - 24 \)
\( 24 - 17 = z - 24 \)
\( 7 = z - 24 \)
Add 24 to both sides:
\( z = 7 + 24 \)
\( z = 31 \)
So, the solutions are \( x = 3, y = 17, \) and \( z = 31 \). By using the constant common difference property, we can solve for any missing terms in an A.P.
In simple words: Since all numbers are in an A.P., the gap between any two numbers next to each other must be the same. We use this idea to make equations for \( x, y, \) and \( z \), and then solve them one by one.

🎯 Exam Tip: When you have multiple unknowns in an A.P., write out all expressions for the common difference. Equating pairs of these expressions will give you a system of equations that you can solve sequentially.

 

Question 10. In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
Answer:
This problem describes an Arithmetic Progression because the number of seats increases by a constant amount (2 seats) in each successive row.
Let \( t_n \) be the number of seats in the \( n \)th row.
The number of seats in the first row (\( a \)) = 20.
So, \( t_1 = 20 \).
Each successive row has two additional seats, so the common difference (\( d \)) = 2.
The total number of rows (\( n \)) = 30.
We need to find the number of seats in the last row, which is the 30th term (\( t_{30} \)).
The formula for the \( n \)th term of an A.P. is \( t_n = a + (n - 1)d \). This formula is key to finding the number of seats in any given row.
Substitute the values into the formula:
\( t_{30} = 20 + (30 - 1)2 \)
\( t_{30} = 20 + (29)2 \)
\( t_{30} = 20 + 58 \)
\( t_{30} = 78 \)
Therefore, there are 78 seats in the last row (the 30th row).
In simple words: The number of seats starts at 20 and goes up by 2 each row. We want to find the number of seats in the 30th row. We use the A.P. formula for the nth term to quickly calculate this.

🎯 Exam Tip: When a problem describes a quantity increasing or decreasing by a constant amount, it's a strong indicator of an Arithmetic Progression. Clearly identify 'a', 'd', and 'n' before applying the formula.

 

Question 11. The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Answer:
Let the three consecutive terms in A.P. be \( a - d, a, \) and \( a + d \). Choosing terms in this form simplifies calculations, especially for sums.
Given that the sum of these three terms is 27:
\( (a - d) + a + (a + d) = 27 \)
\( 3a = 27 \)
Divide by 3:
\( a = \frac { 27 }{ 3 } \)
\( a = 9 \)
Now, given that the product of these three terms is 288:
\( (a - d)(a)(a + d) = 288 \)
Substitute the value of \( a = 9 \):
\( (9 - d)(9)(9 + d) = 288 \)
Divide both sides by 9:
\( (9 - d)(9 + d) = \frac { 288 }{ 9 } \)
\( (9 - d)(9 + d) = 32 \)
Use the algebraic identity \( (x - y)(x + y) = x^2 - y^2 \):
\( 9^2 - d^2 = 32 \)
\( 81 - d^2 = 32 \)
Rearrange to solve for \( d^2 \):
\( 81 - 32 = d^2 \)
\( 49 = d^2 \)
Take the square root of both sides:
\( d = \pm\sqrt{49} \)
\( d = \pm 7 \)
We have two possible values for \( d \), so we find two sets of terms:
Case 1: When \( a = 9 \) and \( d = 7 \)
The terms are:
\( a - d = 9 - 7 = 2 \)
\( a = 9 \)
\( a + d = 9 + 7 = 16 \)
So, the A.P. is 2, 9, 16.
Case 2: When \( a = 9 \) and \( d = -7 \)
The terms are:
\( a - d = 9 - (-7) = 9 + 7 = 16 \)
\( a = 9 \)
\( a + d = 9 + (-7) = 9 - 7 = 2 \)
So, the A.P. is 16, 9, 2.
Both sets of terms are valid as they represent the same sequence in reverse order. The problem asks for the terms, and both (2, 9, 16) and (16, 9, 2) satisfy the conditions.
In simple words: To solve, we first assumed the three numbers are \( a-d, a, a+d \). We used their sum to find \( a \), then used their product to find \( d \). Since \( d \) can be positive or negative, we got two possible sets of numbers.

🎯 Exam Tip: When dealing with an odd number of consecutive terms in an A.P., representing them as \( a-d, a, a+d \) (for 3 terms) or \( a-2d, a-d, a, a+d, a+2d \) (for 5 terms) often simplifies the sum equation nicely, as the 'd' terms cancel out.

 

Question 12. The ratio of 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term.
Answer:
Let the first term of the A.P. be \( a \) and the common difference be \( d \).
The formula for the \( n \)th term of an A.P. is \( t_n = a + (n - 1)d \).
Given that the ratio of the 6th term to the 8th term is 7:9.
\( \frac { t_6 }{ t_8 } = \frac { 7 }{ 9 } \)
Substitute the formula for \( t_n \):
\( \frac { a + (6 - 1)d }{ a + (8 - 1)d } = \frac { 7 }{ 9 } \)
\( \frac { a + 5d }{ a + 7d } = \frac { 7 }{ 9 } \)
Cross-multiply to solve for \( a \) in terms of \( d \):
\( 9(a + 5d) = 7(a + 7d) \)
\( 9a + 45d = 7a + 49d \)
Rearrange the terms to group \( a \) and \( d \):
\( 9a - 7a = 49d - 45d \)
\( 2a = 4d \)
Divide both sides by 2:
\( a = 2d \)
Now, we need to find the ratio of the 9th term to the 13th term (\( t_9 : t_{13} \)).
\( t_9 = a + (9 - 1)d = a + 8d \)
\( t_{13} = a + (13 - 1)d = a + 12d \)
Substitute \( a = 2d \) into the expressions for \( t_9 \) and \( t_{13} \):
\( t_9 = 2d + 8d = 10d \)
\( t_{13} = 2d + 12d = 14d \)
Now, find the ratio:
\( \frac { t_9 }{ t_{13} } = \frac { 10d }{ 14d } \)
Cancel out \( d \) and simplify the fraction:
\( \frac { t_9 }{ t_{13} } = \frac { 10 }{ 14 } = \frac { 5 }{ 7 } \)
So, the ratio of the 9th term to the 13th term is 5:7. This demonstrates how relationships between terms in an A.P. can be used to find other ratios within the same sequence.
In simple words: We are given a relationship between the 6th and 8th terms. We use the A.P. formula to turn this into an equation with \( a \) and \( d \), which helps us find a link between \( a \) and \( d \). Then we use this link to find the ratio of the 9th and 13th terms.

🎯 Exam Tip: Always express \( a \) in terms of \( d \) (or vice versa) from the initial given ratio. This substitution simplifies the expressions for other terms dramatically.

 

Question 13. In a winter season, the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.
Answer:
Let the five temperatures in A.P. for Monday, Tuesday, Wednesday, Thursday, and Friday be \( a - 2d, a - d, a, a + d, \) and \( a + 2d \) respectively. This representation is convenient for sums as the \( d \) terms often cancel out.
Given: The sum of temperatures from Monday to Wednesday is 0°C.
\( (a - 2d) + (a - d) + a = 0 \)
\( 3a - 3d = 0 \)
Divide by 3:
\( a - d = 0 \) .......... (1)
Given: The sum of temperatures from Wednesday to Friday is 18°C.
\( a + (a + d) + (a + 2d) = 18 \)
\( 3a + 3d = 18 \)
Divide by 3:
\( a + d = 6 \) .......... (2)
Now we have a system of two linear equations with two variables:
1) \( a - d = 0 \)
2) \( a + d = 6 \)
Add equation (1) and equation (2):
\( (a - d) + (a + d) = 0 + 6 \)
\( 2a = 6 \)
Divide by 2:
\( a = 3 \)
Substitute the value of \( a = 3 \) into equation (1):
\( 3 - d = 0 \)
\( d = 3 \)
So, the first term \( a = 3 \) and the common difference \( d = 3 \). These values define the entire sequence of temperatures.
Now, find the temperature for each of the five days:
Monday: \( a - 2d = 3 - 2(3) = 3 - 6 = -3^\circ\text{C} \)
Tuesday: \( a - d = 3 - 3 = 0^\circ\text{C} \)
Wednesday: \( a = 3^\circ\text{C} \)
Thursday: \( a + d = 3 + 3 = 6^\circ\text{C} \)
Friday: \( a + 2d = 3 + 2(3) = 3 + 6 = 9^\circ\text{C} \)
The temperatures for the five days are -3°C, 0°C, 3°C, 6°C, and 9°C.
In simple words: We set up the temperatures for the five days using 'a' and 'd'. Then we used the given sums to create two simple equations. Solving these equations gave us 'a' and 'd', which we used to find each day's temperature.

🎯 Exam Tip: For problems involving consecutive terms, always try to represent the terms using \( a \) as the middle term (if odd number of terms) or halfway between the middle terms (if even number of terms), as this often simplifies the algebraic equations significantly.

 

Question 14. Priya earned Rs 15,000 in the first month. Thereafter her salary increases by Rs 1500 per year. Her expenses are Rs 13,000 during the first year and the expenses increases by Rs 900 per year. How long will it take her to save Rs 20,000 per month.
Answer:
First, let's tabulate Priya's monthly salary, expenses, and savings for the initial years to see the pattern.
Monthly salary for the first year (a_s) = Rs 15,000.
Increase in salary per year (d_s) = Rs 1,500.
Monthly expenses for the first year (a_e) = Rs 13,000.
Increase in expenses per year (d_e) = Rs 900.
Monthly saving = Monthly salary - Monthly expenses.

From the table, the monthly savings form an Arithmetic Progression:
2000, 2600, 3200, ...
First term of savings A.P. (\( a \)) = Rs 2,000.
Common difference of savings A.P. (\( d \)) = \( 2600 - 2000 = 600 \).
We want to find the number of years (\( n \)) when her monthly savings (\( t_n \)) reach Rs 20,000.
The formula for the \( n \)th term of an A.P. is \( t_n = a + (n - 1)d \). This formula will help us predict her savings in any given year.
Substitute the known values into the formula:
\( 20000 = 2000 + (n - 1)600 \)
Subtract 2000 from both sides:
\( 20000 - 2000 = (n - 1)600 \)
\( 18000 = (n - 1)600 \)
Divide both sides by 600:
\( \frac { 18000 }{ 600 } = n - 1 \)
\( 30 = n - 1 \)
Add 1 to both sides:
\( n = 30 + 1 \)
\( n = 31 \)
Priya will take 31 years to save Rs 20,000 per month. This highlights the power of consistent small increases over time.
In simple words: First, we find her savings for the first few years by subtracting expenses from salary. We see that her savings increase by a fixed amount each year, forming an A.P. Then, we use the A.P. formula to find out in which year her savings will reach Rs 20,000.

🎯 Exam Tip: For problems involving financial growth (salary, savings, etc.) that increase by a constant amount, treat them as an A.P. Breaking down the problem into smaller steps (salary A.P., expenses A.P., then savings A.P.) makes it easier to solve.