CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 16

Question 1. Find the general term and 5th term of A.P. \( \frac{1}{p} , \frac{p+1}{p} , \frac{2p+1}{p} , - - - - -. \)
Answer: Given A.P. is \( \frac{1}{p} , \frac{p+1}{p} , \frac{2p+1}{p} , \dots \)
Here, the first term \( a = \frac{1}{p} \)
The common difference \( d = t_2 - t_1 \)
\( d = \frac{p+1}{p} - \frac{1}{p} = \frac{p+1-1}{p} = \frac{p}{p} = 1 \)
The general term (\( t_n \)) of an A.P. is given by:
\( t_n = a + (n-1)d \)
\( t_n = \frac{1}{p} + (n-1)(1) \)
\( t_n = \frac{1}{p} + n - 1 \)
\( t_n = \frac{1 + pn - p}{p} \)
\( t_n = \frac{pn - p + 1}{p} \)
Now, the 5th term (\( t_5 \)) is calculated as:
\( t_5 = \frac{p(5) - p + 1}{p} \)
\( t_5 = \frac{5p - p + 1}{p} \)
\( t_5 = \frac{1 + 4p}{p} \)
Hence, the general term is \( \frac{pn - p + 1}{p} \) and the 5th term is \( \frac{1 + 4p}{p} \).
In simple words: First find the common difference by subtracting the first term from the second. Then, plug the values into the standard A.P. formula to get the expressions for the general term and the fifth term.

Exam Tip: Be careful with fractional terms in an A.P. Simplify the numerator first when finding the common difference to avoid arithmetic errors.

Question 2. The 4th term of an AP. is 14 and its 8th term is 30. Find the first term and common difference.
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
Using the formula \( t_n = a + (n-1)d \):
For the 4th term:
\( t_4 = a + 3d = 14 \) - (Equation 1)
For the 8th term:
\( t_8 = a + 7d = 30 \) - (Equation 2)
Subtracting Equation 1 from Equation 2:
\( (a + 7d) - (a + 3d) = 30 - 14 \)
\( \implies 4d = 16 \)
\( \implies d = 4 \)
Substituting \( d = 4 \) in Equation 1:
\( a + 3(4) = 14 \)
\( \implies a + 12 = 14 \)
\( \implies a = 2 \)
Therefore, the first term \( a = 2 \) and the common difference \( d = 4 \).
In simple words: Write two equations using the given terms. Subtract them to find the difference between steps, and use that to calculate the very first number.

Exam Tip: Setting up and solving linear equations in two variables is a standard method for A.P. problems. Ensure you show the subtraction step clearly to get full marks.

Question 3. The 8th term and 15th terms of an AP are 5 and 33 respectively. Find its 5th term and last term
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
We are given:
\( t_8 = 5 \)
\( \implies a + 7d = 5 \) - (Equation 1)
\( t_{15} = 33 \)
\( \implies a + 14d = 33 \) - (Equation 2)
Subtracting Equation 1 from Equation 2:
\( (a + 14d) - (a + 7d) = 33 - 5 \)
\( \implies 7d = 28 \)
\( \implies d = 4 \)
Substituting \( d = 4 \) in Equation 1:
\( a + 7(4) = 5 \)
\( \implies a + 28 = 5 \)
\( \implies a = -23 \)
Now, the 5th term is:
\( t_5 = a + 4d \)
\( t_5 = -23 + 4(4) \)
\( t_5 = -23 + 16 = -7 \)
The last term (nth term) is:
\( t_n = a + (n-1)d \)
\( t_n = -23 + (n-1)4 \)
\( t_n = -23 + 4n - 4 \)
\( t_n = 4n - 27 \)
Thus, the 5th term is \( -7 \) and the last term is \( 4n - 27 \).
In simple words: Solve the two equations to find the start number and the difference. Use these values to find the 5th term and the formula for the very last term.

Exam Tip: In questions asking for the "last term" without specifying the total number of terms, express your answer in terms of the general variable n.

Question 4. Prove that in an A.P, \( t_{m+n} + t_{m-n} = 2 t_m . \)
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
The general term of an A.P. is defined as:
\( t_k = a + (k-1)d \)
Let us find the terms for \( m+n \) and \( m-n \):
\( t_{m+n} = a + (m+n-1)d \)
\( t_{m-n} = a + (m-n-1)d \)
Adding these two terms together:
\( t_{m+n} + t_{m-n} = [a + (m+n-1)d] + [a + (m-n-1)d] \)
\( \implies t_{m+n} + t_{m-n} = 2a + (m+n-1+m-n-1)d \)
\( \implies t_{m+n} + t_{m-n} = 2a + (2m-2)d \)
Taking out the common factor of 2:
\( \implies t_{m+n} + t_{m-n} = 2[a + (m-1)d] \)
Since \( t_m = a + (m-1)d \), we can substitute it in:
\( \implies t_{m+n} + t_{m-n} = 2 t_m \)
Hence proved.
In simple words: Write the formulas for both terms on the left side and add them up. You will see that the result is exactly double the formula for the middle term.

Exam Tip: For proof-based A.P. questions, start by expressing the terms on the left-hand side using the standard formula, then simplify using basic algebra.

Question 5. Prove that in an AP, \( t_p + t_{p+2q} = 2 t_{p+q} \)
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
Using the term formula \( t_k = a + (k-1)d \):
\( t_p = a + (p-1)d \)
\( t_{p+2q} = a + (p+2q-1)d \)
Adding the two terms on the left-hand side:
\( t_p + t_{p+2q} = [a + (p-1)d] + [a + (p+2q-1)d] \)
\( \implies t_p + t_{p+2q} = 2a + (p - 1 + p + 2q - 1)d \)
\( \implies t_p + t_{p+2q} = 2a + (2p + 2q - 2)d \)
Factoring out 2 from the terms on the right:
\( \implies t_p + t_{p+2q} = 2[a + (p+q-1)d] \)
Since \( t_{p+q} = a + (p+q-1)d \), we get:
\( \implies t_p + t_{p+2q} = 2 t_{p+q} \)
Hence proved.
In simple words: Sum the formulas for the first and the third terms, and you will find that it equals twice the formula of the term right in the middle of them.

Exam Tip: Remember that the sum of two terms equidistant from a middle term in an A.P. is always twice that middle term. This is a key property of arithmetic progressions.

Question 6. In an AP, it is given that \( t_{p+1} = 2 t_{q+1} \) Prove that \( t_{3p+1} = 2 t_{p+q+1} \)
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
We are given:
\( t_{p+1} = 2 t_{q+1} \)
Using \( t_n = a + (n-1)d \):
\( a + (p+1-1)d = 2[a + (q+1-1)d] \)
\( \implies a + pd = 2(a + qd) \)
\( \implies a + pd = 2a + 2qd \)
\( \implies pd - 2qd = 2a - a \)
\( \implies a = (p-2q)d \) - (Equation 1)
Now, we need to prove that:
\( t_{3p+1} = 2 t_{p+q+1} \)
Let's find the left-hand side (LHS):
\( \text{LHS} = t_{3p+1} = a + 3pd \)
Substitute the value of \( a \) from Equation 1 into the LHS:
\( \text{LHS} = (p-2q)d + 3pd \)
\( \implies \text{LHS} = (p - 2q + 3p)d \)
\( \implies \text{LHS} = (4p - 2q)d \)
\( \implies \text{LHS} = 2(2p - q)d \)
Next, let's find the right-hand side (RHS):
\( \text{RHS} = 2 t_{p+q+1} = 2[a + (p+q)d] \)
Substitute the value of \( a \) from Equation 1 into the RHS:
\( \text{RHS} = 2[(p-2q)d + (p+q)d] \)
\( \implies \text{RHS} = 2(p - 2q + p + q)d \)
\( \implies \text{RHS} = 2(2p - q)d \)
Since LHS = RHS, we have:
\( t_{3p+1} = 2 t_{p+q+1} \)
Hence proved.
In simple words: Use the given condition to find a relationship between the first term and the common difference. Then substitute this relation into both sides of the proof equation to show they are equal.

Exam Tip: Expressing one variable in terms of others (like writing a in terms of d, p, and q) helps simplify complex algebraic identities in A.P. proofs.

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Question 7. How many two digit numbers are divisible by 5 ?
Answer: The list of all two-digit numbers divisible by 5 is:
10, 15, 20, 25, ..., 95
This series forms an A.P. where:
First term, \( a = 10 \)
Common difference, \( d = 5 \)
Last term, \( t_n = 95 \)
We know the formula for the nth term of an A.P. is:
\( t_n = a + (n-1)d \)
Substitute the values into the formula:
\( 95 = 10 + (n-1)5 \)
\( \implies 85 = (n-1)5 \)
Dividing by 5 on both sides:
\( \implies n-1 = 17 \)
\( \implies n = 18 \)
Hence, there are 18 two-digit numbers divisible by 5.
In simple words: Start counting from 10 up to 95 by steps of 5. Using the term formula, we calculate that there are exactly 18 such numbers in total.

Exam Tip: Always make sure to include the smallest two-digit number (10) in your list. Skipping 10 is a very common mistake that leads to an incorrect count of 17.

Question 8. How many two digit numbers are there which leaves remainder 1 when divided by 4 ?
Answer: The two-digit numbers that leave a remainder of 1 when divided by 4 are:
13, 17, 21, ..., 97
This sequence forms an A.P. where:
First term, \( a = 13 \)
Common difference, \( d = 4 \)
Last term, \( t_n = 97 \)
Using the formula for the nth term of an A.P.:
\( t_n = a + (n-1)d \)
Substitute the values:
\( 97 = 13 + (n-1)4 \)
\( \implies 84 = (n-1)4 \)
Dividing by 4 on both sides:
\( \implies n-1 = 21 \)
\( \implies n = 22 \)
Thus, there are 22 two-digit numbers that leave a remainder of 1 when divided by 4.
In simple words: Find the smallest two-digit number after a multiple of 4 (which is 13) and the largest one (which is 97). Then, count how many numbers are in this series using the A.P. formula.

Exam Tip: To find the starting number of such series, identify the smallest two-digit multiple of the divisor (12 in this case) and add the required remainder (1) to get the first term (13).

Question 9. If pth term of an AP is q And qth term is p. then show that its nth term is (p+q-n).
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
We are given:
\( t_p = q \)
\( \implies a + (p-1)d = q \) - (Equation 1)
\( t_q = p \)
\( \implies a + (q-1)d = p \) - (Equation 2)
Subtracting Equation 2 from Equation 1:
\( [a + (p-1)d] - [a + (q-1)d] = q - p \)
\( \implies (p - 1 - q + 1)d = q - p \)
\( \implies (p - q)d = -(p - q) \)
Dividing both sides by \( (p-q) \):
\( \implies d = -1 \)
Substitute \( d = -1 \) in Equation 1:
\( a + (p-1)(-1) = q \)
\( \implies a - p + 1 = q \)
\( \implies a = p + q - 1 \)
Now, the nth term of the A.P. is:
\( t_n = a + (n-1)d \)
\( \implies t_n = (p + q - 1) + (n-1)(-1) \)
\( \implies t_n = p + q - 1 - n + 1 \)
\( \implies t_n = p + q - n \)
Hence showed.
In simple words: Solve the equations for the pth and qth terms to get the first term and the common difference. Then, use them to write down the formula for the nth term.

Exam Tip: A common algebraic trick in this standard problem is recognizing that (q - p) is the negative of (p - q), which simplifies the value of d to -1 immediately.

Question 10. Find the 20th term from the end of the sequence 3, 8, 13, - - - - - 253.
Answer: The given sequence is:
3, 8, 13, ..., 253
Here, the first term \( a = 3 \)
Common difference \( d = 8 - 3 = 5 \)
Last term \( l = 253 \)
The formula to find the nth term from the end of an A.P. is:
\( t_{n\text{ from end}} = l - (n-1)d \)
For the 20th term from the end:
\( t_{20\text{ from end}} = 253 - (20-1)5 \)
\( \implies t_{20\text{ from end}} = 253 - (19)5 \)
\( \implies t_{20\text{ from end}} = 253 - 95 \)
\( \implies t_{20\text{ from end}} = 158 \)
Hence, the 20th term from the end of the sequence is 158.
In simple words: To find a term starting from the end of the list, we begin at the last number and step backwards by subtracting the difference each time.

Exam Tip: Alternatively, you can reverse the A.P. to 253, 248, ..., 3. Here, the first term becomes 253 and the common difference becomes -5. Both methods are correct and yield the same result.

Question 11. If 7 times the 7th term of an AP is equal to 11 times the 11th term. Show that the 18th term of an A.P is zero
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
We are given:
7 times the 7th term = 11 times the 11th term
\( 7 \times t_7 = 11 \times t_{11} \)
Using the formula \( t_n = a + (n-1)d \):
\( 7[a + (7-1)d] = 11[a + (11-1)d] \)
\( \implies 7(a + 6d) = 11(a + 10d) \)
\( \implies 7a + 42d = 11a + 110d \)
Rearranging all terms to one side:
\( \implies 11a - 7a + 110d - 42d = 0 \)
\( \implies 4a + 68d = 0 \)
Dividing the equation by 4:
\( \implies a + 17d = 0 \) - (Equation 1)
Now, the 18th term of the A.P. is:
\( t_{18} = a + (18-1)d \)
\( \implies t_{18} = a + 17d \)
From Equation 1, we know that \( a + 17d = 0 \). Therefore:
\( \implies t_{18} = 0 \)
Hence showed.
In simple words: Write out the relation given in the question and simplify it. You will find that it reduces to the exact formula for the 18th term being equal to zero.

Exam Tip: A useful shortcut to remember for competitive exams is that if m times the m-th term of an A.P. is equal to n times the n-th term, then the (m+n)-th term is always zero.

Question 12. Which term is the first negative terms of sequence 18, \( \frac{33}{2} \), 15, - - - - -.
Answer: The given sequence is:
18, \( \frac{33}{2} \), 15, ...
Here, the first term \( a = 18 \)
The common difference \( d = \frac{33}{2} - 18 = \frac{33 - 36}{2} = -\frac{3}{2} \)
Let the nth term be the first negative term, which means \( t_n < 0 \).
Using the formula \( t_n = a + (n-1)d \):
\( a + (n-1)d < 0 \)
Substitute the values of \( a \) and \( d \):
\( 18 + (n-1)\left(-\frac{3}{2}\right) < 0 \)
\( \implies 18 - \frac{3}{2}(n-1) < 0 \)
\( \implies 18 < \frac{3}{2}(n-1) \)
Multiply by 2 on both sides:
\( \implies 36 < 3(n-1) \)
Divide by 3 on both sides:
\( \implies 12 < n-1 \)
\( \implies n > 13 \)
Since \( n \) must be a whole number, the smallest integer value greater than 13 is \( n = 14 \).
Thus, the 14th term is the first negative term of the sequence.
In simple words: We find when the terms of the decreasing sequence cross below zero. Solving the inequality shows that any term after the 13th term is negative, so the 14th term is the first negative one.

Exam Tip: When dealing with inequalities for finding the first negative term, remember that n must always be a positive integer. Always round up to the next integer if you get a fraction or a boundary value.