CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 12

Question 1. The \( n^{\text{th}} \) term of a sequence is \( (8-3n) \). Find the sum of 22 terms of the A.P.
Answer: Let the general term of the sequence be \( a_n = 8 - 3n \).
First term, \( a = a_1 = 8 - 3(1) = 5 \).
The \( 22^{\text{th}} \) term is \( a_{22} = 8 - 3(22) = 8 - 66 = -58 \).
The sum of the first \( n \) terms of an arithmetic progression is given by:
\( S_n = \frac{n}{2} (a + a_n) \)
For \( n = 22 \):
\( S_{22} = \frac{22}{2} [a_1 + a_{22}] \)
\( \implies S_{22} = 11 [5 + (-58)] \)
\( \implies S_{22} = 11 \times (-53) = -583 \)
The sum of the first 22 terms of the A.P. is \( -583 \).
In simple words: Find the first term by putting \( n = 1 \) and the 22nd term by putting \( n = 22 \). Then, use the sum formula to get the final total of \( -583 \).

Exam Tip: Always double-check the sign of the common difference or final term. Since the coefficient of \( n \) is negative, the terms are decreasing, resulting in a negative sum.

Question 2. The last term of an A.P is 120. Its 1st term and common difference are 20 and 5 respectively. Find the sum of the A.P.
Answer: Given parameters:
First term \( a = 20 \)
Common difference \( d = 5 \)
Last term \( a_n = 120 \)
First, find the number of terms \( n \):
\( a_n = a + (n-1)d \)
\( 120 = 20 + (n-1)5 \)
\( \implies 100 = 5(n-1) \)
\( \implies n-1 = 20 \)
\( \implies n = 21 \)
Now, calculate the sum of the series:
\( S_n = \frac{n}{2} (a + a_n) \)
\( S_{21} = \frac{21}{2} (20 + 120) \)
\( \implies S_{21} = \frac{21}{2} \times 140 \)
\( \implies S_{21} = 21 \times 70 = 1470 \)
The sum of the arithmetic progression is \( 1470 \).
In simple words: First, use the last term formula to find out how many numbers are in the series, which is 21. Then, add the first and last numbers together, multiply by 21, and divide by 2 to get 1470.

Exam Tip: Clearly show the steps for finding \( n \) before applying the sum formula, as separate marks are usually allocated for both parts.

Question 3. The \( n^{\text{th}} \) term of an A.P. is 18. Its first term is 50 and common difference is \( (-4) \) respectively. Find the sum to n terms of the A.P.
Answer: Given:
First term \( a = 50 \)
Common difference \( d = -4 \)
\( n^{\text{th}} \) term \( a_n = 18 \)
To find the number of terms \( n \):
\( a_n = a + (n-1)d \)
\( 18 = 50 + (n-1)(-4) \)
\( \implies 18 - 50 = -4(n-1) \)
\( \implies -32 = -4(n-1) \)
\( \implies n-1 = 8 \)
\( \implies n = 9 \)
Next, calculate the sum of these \( 9 \) terms:
\( S_n = \frac{n}{2} (a + a_n) \)
\( S_9 = \frac{9}{2} (50 + 18) \)
\( \implies S_9 = \frac{9}{2} \times 68 \)
\( \implies S_9 = 9 \times 34 = 306 \)
The sum of the first \( n \) terms is \( 306 \).
In simple words: Use the given term 18 to find out that there are 9 terms in the sequence. Then, add the first and last terms and calculate the sum for these 9 terms, which gives 306.

Exam Tip: Be careful with the negative common difference when setting up the linear equation for \( n \). Double-negative operations are a common place to make algebraic errors.

Question 4. How many terms of the A.P. 3, 7, 11, ... should be added to get the sum 210?
Answer: Given Arithmetic Progression: 3, 7, 11, ...
First term \( a = 3 \)
Common difference \( d = 7 - 3 = 4 \)
We want the sum \( S_n = 210 \).
Using the sum of \( n \) terms formula:
\( S_n = \frac{n}{2} [2a + (n-1)d] \)
\( 210 = \frac{n}{2} [2(3) + (n-1)4] \)
\( \implies 210 = \frac{n}{2} [6 + 4n - 4] \)
\( \implies 210 = \frac{n}{2} [4n + 2] \)
\( \implies 210 = n(2n + 1) \)
\( \implies 2n^2 + n - 210 = 0 \)
To solve this quadratic equation, split the middle term:
\( 2n^2 + 21n - 20n - 210 = 0 \)
\( \implies n(2n + 21) - 10(2n + 21) = 0 \)
\( \implies (2n + 21)(n - 10) = 0 \)
This gives:
\( n = 10 \text{ or } n = -10.5 \)
Since the number of terms \( n \) must be a positive integer, we discard the negative fraction.
Thus, \( n = 10 \).
Therefore, 10 terms of the progression should be added.
In simple words: Use the sum formula to get a quadratic equation. Solving it gives a positive whole number, 10, which means we must add 10 terms of the sequence to reach 210.

Exam Tip: Discard negative or fractional values of \( n \) because the count of terms must always be a positive integer.

Question 5. Find the number of terms of the A.P. 20, 16, 12, ... which when added gives the sum 56. Explain the reason for double answer.
Answer: For the given A.P. 20, 16, 12, ...
First term \( a = 20 \)
Common difference \( d = 16 - 20 = -4 \)
Sum of terms \( S_n = 56 \)
Using the sum formula:
\( S_n = \frac{n}{2} [2a + (n-1)d] \)
\( 56 = \frac{n}{2} [2(20) + (n-1)(-4)] \)
\( \implies 56 = \frac{n}{2} [40 - 4n + 4] \)
\( \implies 56 = \frac{n}{2} [44 - 4n] \)
\( \implies 56 = n(22 - 2n) \)
\( \implies 2n^2 - 22n + 56 = 0 \)
Divide the entire equation by 2:
\( n^2 - 11n + 28 = 0 \)
\( \implies (n-4)(n-7) = 0 \)
\( \implies n = 4 \text{ or } n = 7 \)

Explanation for the double answer:
The terms of this A.P. are 20, 16, 12, 8, 4, 0, -4, ...
The sum of the first 4 terms is:
\( S_4 = 20 + 16 + 12 + 8 = 56 \)
The sum of the first 7 terms is:
\( S_7 = 20 + 16 + 12 + 8 + 4 + 0 + (-4) = 56 \)
Here, the sum of the \( 5^{\text{th}} \), \( 6^{\text{th}} \), and \( 7^{\text{th}} \) terms is \( 4 + 0 + (-4) = 0 \). Since their sum is zero, adding these extra terms does not alter the overall sum of 56.
In simple words: Solving the sum formula gives two possible answers: 4 or 7. This happens because the terms after the 4th term are 4, 0, and -4, which add up to exactly 0, so they don't change the total sum.

Exam Tip: Whenever you get a quadratic equation for \( n \) with two positive integer solutions, you must calculate and list the extra terms to show why their sum is zero to earn full explanation marks.

Question 6. Find A.P. if \( S_n = 4n^2 - n \).
Answer: Given:
\( S_n = 4n^2 - n \)
First term (\( a_1 \)):
\( a_1 = S_1 = 4(1)^2 - 1 = 3 \)
Second term (\( a_2 \)):
\( S_2 = 4(2)^2 - 2 = 16 - 2 = 14 \)
\( a_2 = S_2 - S_1 = 14 - 3 = 11 \)
Third term (\( a_3 \)):
\( S_3 = 4(3)^2 - 3 = 36 - 3 = 33 \)
\( a_3 = S_3 - S_2 = 33 - 14 = 19 \)
The common difference \( d \) is:
\( d = a_2 - a_1 = 11 - 3 = 8 \)
Thus, the Arithmetic Progression is 3, 11, 19, ...
In simple words: To find the terms, calculate the sum for 1, 2, and 3 terms. Subtracting the first sum from the second sum gives the second term, and doing the same for the next gives the third term, forming the series 3, 11, 19, ...

Exam Tip: Remember that \( a_n = S_n - S_{n-1} \) is a very important formula. You can quickly find any term of the progression using this relationship.

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Question 7. Find the sum of all integers between 50 and 450 which are divisible by 7
Answer: The first integer greater than 50 that is divisible by 7 is 56.
The last integer less than 450 that is divisible by 7 is 448.
The sequence of numbers is:
56, 63, 70, ..., 448.
This is an A.P. with:
First term \( a = 56 \)
Common difference \( d = 7 \)
Last term \( a_n = 448 \)
To find the number of terms \( n \):
\( a_n = a + (n-1)d \)
\( 448 = 56 + (n-1)7 \)
\( \implies 392 = 7(n-1) \)
\( \implies n-1 = 56 \)
\( \implies n = 57 \)
Now, calculate the sum of these 57 terms:
\( S_n = \frac{n}{2} (a + a_n) \)
\( S_{57} = \frac{57}{2} (56 + 448) \)
\( \implies S_{57} = \frac{57}{2} \times 504 \)
\( \implies S_{57} = 57 \times 252 = 14364 \)
The sum of all such integers is 14364.
In simple words: Find the first number after 50 that divides by 7, which is 56, and the last number before 450, which is 448. Find how many such numbers exist (57 of them), and then find their total sum, which is 14364.

Exam Tip: Be careful with the word "between". It means the start and end values are excluded (though 50 and 450 are not divisible by 7 anyway, it's good practice to always identify the correct first and last terms carefully).

Question 8. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the sum of first 30 terms.
Answer: Given:
Third term \( a_3 = 7 \)
Seventh term \( a_7 = 3 a_3 + 2 \)
Substitute \( a_3 = 7 \) into the relation for \( a_7 \):
\( a_7 = 3(7) + 2 = 23 \)
Using the general term formula \( a_n = a + (n-1)d \):
\( a + 2d = 7 \) - (Equation 1)
\( a + 6d = 23 \) - (Equation 2)
Subtracting Equation 1 from Equation 2:
\( (a + 6d) - (a + 2d) = 23 - 7 \)
\( \implies 4d = 16 \)
\( \implies d = 4 \)
Substitute \( d = 4 \) into Equation 1:
\( a + 2(4) = 7 \)
\( \implies a + 8 = 7 \)
\( \implies a = -1 \)
Now, find the sum of the first 30 terms (\( S_{30} \)):
\( S_n = \frac{n}{2} [2a + (n-1)d] \)
\( S_{30} = \frac{30}{2} [2(-1) + (30-1)4] \)
\( \implies S_{30} = 15 [-2 + 29 \times 4] \)
\( \implies S_{30} = 15 [-2 + 116] \)
\( \implies S_{30} = 15 \times 114 = 1710 \)
The sum of the first 30 terms is 1710.
In simple words: Use the clue that the 7th term is 2 more than 3 times the 3rd term to find that the 7th term is 23. Solve the equations to find the first term is -1 and the difference is 4. Finally, calculate the sum of 30 terms to get 1710.

Exam Tip: Be sure to write down the linear equations clearly and state which equation is being subtracted from which to make your working easy for the examiner to follow.

Question 9. How many terms are there in an A.P. Whose first and fifth terms are \( (-14) \) and 2 respectively and the sum of the terms is 40?
Answer: Given:
First term \( a = -14 \)
Fifth term \( a_5 = 2 \)
Sum of terms \( S_n = 40 \)
First, find the common difference \( d \) using \( a_5 \):
\( a_5 = a + 4d \)
\( 2 = -14 + 4d \)
\( \implies 4d = 16 \)
\( \implies d = 4 \)
Now, set up the equation for the sum:
\( S_n = \frac{n}{2} [2a + (n-1)d] = 40 \)
\( \frac{n}{2} [2(-14) + (n-1)4] = 40 \)
\( \implies \frac{n}{2} [-28 + 4n - 4] = 40 \)
\( \implies \frac{n}{2} [4n - 32] = 40 \)
\( \implies n(2n - 16) = 40 \)
\( \implies 2n^2 - 16n - 40 = 0 \)
Divide the quadratic equation by 2:
\( n^2 - 8n - 20 = 0 \)
\( \implies (n-10)(n+2) = 0 \)
\( \implies n = 10 \text{ or } n = -2 \)
Since the number of terms \( n \) cannot be negative, we have:
\( n = 10 \)
There are 10 terms in the A.P.
In simple words: Find the common difference \( d = 4 \) using the fifth term. Then, substitute this into the sum formula to get a quadratic equation. Solving it gives 10 or -2, and since you can't have a negative number of terms, the answer is 10.

Exam Tip: Always state why you are rejecting a negative solution for \( n \) (e.g., "Since the number of terms must be a positive integer...") to ensure you don't lose step marks.

Question 10. If \( S_n = 2n^2 + 3n \) denotes the sum to n terms of a progression. Prove that it is in A.P. Find its \( r^{\text{th}} \) term.
Answer: Given:
\( S_n = 2n^2 + 3n \)
The \( n^{\text{th}} \) term \( a_n \) is defined as:
\( a_n = S_n - S_{n-1} \)
Substitute the expression for \( S_n \):
\( a_n = (2n^2 + 3n) - [2(n-1)^2 + 3(n-1)] \)
\( \implies a_n = (2n^2 + 3n) - [2(n^2 - 2n + 1) + 3n - 3] \)
\( \implies a_n = 2n^2 + 3n - [2n^2 - 4n + 2 + 3n - 3] \)
\( \implies a_n = 2n^2 + 3n - [2n^2 - n - 1] \)
\( \implies a_n = 4n + 1 \)
To prove that this progression is an Arithmetic Progression, find the difference between any two consecutive terms:
\( a_n - a_{n-1} = (4n + 1) - [4(n-1) + 1] \)
\( \implies a_n - a_{n-1} = 4n + 1 - (4n - 3) \)
\( \implies a_n - a_{n-1} = 4 \)
Since the difference \( a_n - a_{n-1} = 4 \), which is a constant, the sequence is an A.P. with a common difference of 4.
The \( r^{\text{th}} \) term is found by substituting \( n = r \) into the formula for \( a_n \):
\( a_r = 4r + 1 \)
In simple words: Find the formula for the \( n^{\text{th}} \) term by subtracting the sum of \( n-1 \) terms from the sum of \( n \) terms, which gives \( 4n + 1 \). Since the difference between any two consecutive terms is always 4, it is an A.P., and replacing \( n \) with \( r \) gives the \( r^{\text{th}} \) term as \( 4r + 1 \).

Exam Tip: To prove a progression is an A.P., you must show that the difference \( a_n - a_{n-1} \) is a constant independent of \( n \). Simply listing the first few terms is often not considered a complete proof by evaluators.