CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 15

Question 1. For what value of n , the nth terms of the sequences 63, 65, 67, ... and 3, 10, 17, ... are equal ?
Answer: We have two arithmetic progressions. Let us find the general term for each sequence.
For the first progression: \( 63, 65, 67, \dots \)
Here, the first term \( a = 63 \) and the common difference \( d = 65 - 63 = 2 \).
The \( n \)-th term is:
\( a_n = 63 + (n - 1)2 \)

\( \implies a_n = 61 + 2n \)

For the second progression: \( 3, 10, 17, \dots \)
Here, the first term \( a' = 3 \) and the common difference \( d' = 10 - 3 = 7 \).
The \( n \)-th term is:
\( a'_n = 3 + (n - 1)7 \)

\( \implies a'_n = 7n - 4 \)

Since their \( n \)-th terms are given as equal:
\( 61 + 2n = 7n - 4 \)

\( \implies 5n = 65 \)

\( \implies n = 13 \)
Hence, the 13th terms of both sequences are equal.
In simple words: We find the general formula for the \( n \)-th position of both patterns. By setting these formulas equal to each other, we find that at the 13th step, both sequences have the exact same number.

Exam Tip: Always make sure to write down the formula for the \( n \)-th term \( a_n = a + (n-1)d \) clearly before performing algebraic operations to ensure step marks.

Question 2. How many three digit numbers leave the remainder the remainder 2 when divided by 9 ?
Answer: Three-digit numbers lie in the range from 100 to 999.
The smallest three-digit number that leaves a remainder of 2 when divided by 9 is 101.
The largest three-digit number that leaves a remainder of 2 when divided by 9 is 992.
These numbers form an arithmetic progression:
\( 101, 110, 119, \dots, 992 \)
In this progression:
First term, \( a = 101 \)
Common difference, \( d = 9 \)
Last term, \( a_n = 992 \)
Using the general term formula:
\( a_n = a + (n - 1)d \)
\( 992 = 101 + (n - 1)9 \)

\( \implies 891 = (n - 1)9 \)

\( \implies n - 1 = 99 \)

\( \implies n = 100 \)
Thus, there are 100 such three-digit numbers.
In simple words: The first number above 100 that fits our condition is 101, and the last one below 1000 is 992. Counting the numbers in this pattern with a gap of 9 gives a total of 100 numbers.

Exam Tip: To find the first term of such series, divide 100 by the divisor (9), find the remainder (1), and add the necessary difference (8) to 100 to get a remainder of 2, which gives 101.

Question 3. How many three digit numbers are divisible by 7 ?
Answer: Three-digit numbers start at 100 and end at 999.
The smallest three-digit number divisible by 7 is 105.
The largest three-digit number divisible by 7 is 994.
These numbers form the following arithmetic progression:
\( 105, 112, 119, \dots, 994 \)
In this progression:
First term, \( a = 105 \)
Common difference, \( d = 7 \)
Last term, \( a_n = 994 \)
Using the general term formula:
\( a_n = a + (n - 1)d \)
\( 994 = 105 + (n - 1)7 \)

\( \implies 889 = (n - 1)7 \)

\( \implies n - 1 = 127 \)

\( \implies n = 128 \)
Therefore, there are 128 three-digit numbers divisible by 7.
In simple words: The first three-digit multiple of 7 is 105 and the last one is 994. By applying the arithmetic progression formula, we find there are 128 multiples of 7 in this range.

Exam Tip: Be precise when finding the last term of the series by dividing 999 by 7 and subtracting the remainder from 999 to get 994.

Question 4. If the first and the last terms of an AP consisting 60 terms are 7 and 125 respectively. Find its 25th term.
Answer: We are given the following values for the progression:
Total number of terms, \( n = 60 \)
First term, \( a = 7 \)
Last term (\( 60\text{th} \) term), \( a_{60} = 125 \)
Using the formula for the general term:
\( a_n = a + (n - 1)d \)
\( a_{60} = 7 + (60 - 1)d \)
\( 125 = 7 + 59d \)

\( \implies 59d = 118 \)

\( \implies d = 2 \)
Now, we find the 25th term of the progression:
\( a_{25} = a + (25 - 1)d \)
\( a_{25} = 7 + 24(2) \)
\( a_{25} = 7 + 48 \)

\( \implies a_{25} = 55 \)
Thus, the 25th term of the arithmetic progression is 55.
In simple words: Since we know the first and 60th terms, we can find that the numbers increase by 2 at each step. Using this step size, we calculate that the 25th term is 55.

Exam Tip: Always calculate the common difference \( d \) carefully in the first step, as any calculation error there will affect the final term you are asked to find.

Question 5. If the 9th term of an AP is zero, prove that 29th term is double the 19th term.
Answer: Let the first term of the arithmetic progression be \( a \) and its common difference be \( d \).
We are given that the 9th term is zero:
\( a_9 = 0 \)
\( a + (9 - 1)d = 0 \)

\( \implies a + 8d = 0 \)

\( \implies a = -8d \quad \text{- (Equation 1)} \)
Let us write the expression for the 19th term:
\( a_{19} = a + (19 - 1)d = a + 18d \)
Substituting \( a = -8d \) from Equation 1:
\( a_{19} = -8d + 18d \)

\( \implies a_{19} = 10d \quad \text{- (Equation 2)} \)
Next, let us write the expression for the 29th term:
\( a_{29} = a + (29 - 1)d = a + 28d \)
Substituting \( a = -8d \) from Equation 1:
\( a_{29} = -8d + 28d \)

\( \implies a_{29} = 20d \)
We can write this as:
\( a_{29} = 2(10d) \)
Using Equation 2, we substitute \( 10d = a_{19} \):

\( \implies a_{29} = 2a_{19} \)
Hence, it is proved that the 29th term is double the 19th term.
In simple words: Since the 9th term is 0, we can express the first term in terms of the common difference. This helps us show that the 19th term is ten times the difference, and the 29th term is twenty times the difference - which is exactly double.

Exam Tip: In proof-based problems, expressing all higher terms in terms of the common difference \( d \) is a reliable strategy that simplifies the algebra.

Question 6. 4th term of an AP is 3 times the first and the 7th term exceeds twice the 3rd term by 1. Find the first term and the common difference.
Answer: Let the first term of the progression be \( a \) and the common difference be \( d \).
From the first condition, the 4th term is 3 times the first term:
\( a_4 = 3a \)
\( a + 3d = 3a \)

\( \implies 2a = 3d \)

\( \implies a = \frac{3}{2}d \quad \text{- (Equation 1)} \)
From the second condition, the 7th term exceeds twice the 3rd term by 1:
\( a_7 = 2a_3 + 1 \)
\( a + 6d = 2(a + 2d) + 1 \)
\( a + 6d = 2a + 4d + 1 \)

\( \implies 2d - a = 1 \quad \text{- (Equation 2)} \)
Substitute the value of \( a \) from Equation 1 into Equation 2:
\( 2d - \frac{3}{2}d = 1 \)

\( \implies \frac{1}{2}d = 1 \)

\( \implies d = 2 \)
Now, substitute \( d = 2 \) back into Equation 1 to find \( a \):
\( a = \frac{3}{2}(2) \)

\( \implies a = 3 \)
Thus, the first term is 3 and the common difference is 2.
In simple words: We create two algebraic equations using the given conditions. Solving them together reveals that the sequence starts with 3 and increases by 2 each time.

Exam Tip: When dealing with fractions during substitution, multiply the entire equation by the denominator to eliminate fractions and avoid simple calculation errors.

Question 7. Find x, y from the following AP's
(i) 5, x, 11
(ii) 4, x, 14, y
Answer:
(i) Since the terms \( 5, x, 11 \) are in an arithmetic progression, the difference between consecutive terms must be equal:
\( x - 5 = 11 - x \)

\( \implies 2x = 16 \)

\( \implies x = 8 \)

(ii) For the sequence \( 4, x, 14, y \), let the common difference be \( d \).
The third term is related to the first term by:
\( a_3 = a_1 + 2d \)
\( 14 = 4 + 2d \)

\( \implies 2d = 10 \)

\( \implies d = 5 \)
Now, we can find the missing values:
\( x = a_1 + d = 4 + 5 = 9 \)
\( y = a_3 + d = 14 + 5 = 19 \)
In simple words: (i) The middle term of three consecutive numbers in an AP is always the average of the first and third terms, which gives 8. (ii) By comparing the first and third terms, we find the common difference is 5, helping us find that \( x = 9 \) and \( y = 19 \).

Exam Tip: For three consecutive terms in AP, remember that the middle term is the arithmetic mean of the outer terms: \( b = \frac{a+c}{2} \).

Question 8. Find the missing p, q, r from the following AP's
(i) p, 16, q, 8, r
(ii) 3, p, 7, q, 11, r
Answer:
(i) In the sequence \( p, 16, q, 8, r \), let the common difference be \( d \).
The fourth term is related to the second term by:
\( a_4 = a_2 + 2d \)
\( 8 = 16 + 2d \)

\( \implies 2d = -8 \)

\( \implies d = -4 \)
Using this common difference, we calculate the remaining terms:
\( p = a_2 - d = 16 - (-4) = 20 \)
\( q = a_2 + d = 16 + (-4) = 12 \)
\( r = a_4 + d = 8 + (-4) = 4 \)

(ii) In the sequence \( 3, p, 7, q, 11, r \), the third term is related to the first term by:
\( a_3 = a_1 + 2d \)
\( 7 = 3 + 2d \)

\( \implies 2d = 4 \)

\( \implies d = 2 \)
Using this common difference, we calculate the missing terms:
\( p = a_1 + d = 3 + 2 = 5 \)
\( q = a_3 + d = 7 + 2 = 9 \)
\( r = a_5 + d = 11 + 2 = 13 \)
In simple words: (i) The numbers in the first pattern go down by 4 each time, meaning the terms are 20, 12, and 4. (ii) The numbers in the second pattern go up by 2 each time, meaning the terms are 5, 9, and 13.

Exam Tip: Be careful with the sign of the common difference. When terms decrease (like in the first sequence), \( d \) is negative, whereas when terms increase, \( d \) is positive.

Question 9. The 4th term of an AP is equal to 3 times the first term and the 7th term exceeds the third term by 1. Find the 1st term and the AP.
Answer: Let the first term of the arithmetic progression be \( a \) and the common difference be \( d \).
According to the given conditions:
1) The 4th term is 3 times the first term:
\( a_4 = 3a \)
\( a + 3d = 3a \)

\( \implies 2a = 3d \quad \text{- (Equation 1)} \)

2) The 7th term exceeds twice the 3rd term by 1:
\( a_7 = 2a_3 + 1 \)
\( a + 6d = 2(a + 2d) + 1 \)
\( a + 6d = 2a + 4d + 1 \)

\( \implies 2d - a = 1 \quad \text{- (Equation 2)} \)

Multiplying Equation 2 by 2:
\( 4d - 2a = 2 \)
Substitute \( 2a = 3d \) from Equation 1 into this:
\( 4d - 3d = 2 \)

\( \implies d = 2 \)
Now, substitute \( d = 2 \) back into Equation 1 to find \( a \):
\( 2a = 3(2) \)

\( \implies 2a = 6 \)

\( \implies a = 3 \)
The arithmetic progression is:
\( 3, 5, 7, 9, \dots \)
In simple words: By constructing equations from the terms, we find the first term is 3 and the common difference is 2. Thus, our sequence starts at 3 and increases by 2 at each step.

Exam Tip: When the question asks to "find the AP", always write down at least the first four terms of the sequence with dots at the end to represent the infinite series.

Question 10. The tenth term of an AP is (-4) and its 22nd term is (-16). Find its 38th term and the nth term.
Answer: Let the first term be \( a \) and the common difference be \( d \).
We are given:
Tenth term: \( a_{10} = -4 \implies a + 9d = -4 \quad \text{- (Equation 1)} \)
Twenty-second term: \( a_{22} = -16 \implies a + 21d = -16 \quad \text{- (Equation 2)} \)
Subtracting Equation 1 from Equation 2:
\( (a + 21d) - (a + 9d) = -16 - (-4) \)

\( \implies 12d = -12 \)

\( \implies d = -1 \)
Substitute \( d = -1 \) into Equation 1:
\( a + 9(-1) = -4 \)
\( a - 9 = -4 \)

\( \implies a = 5 \)
Now, we calculate the 38th term:
\( a_{38} = a + 37d \)
\( a_{38} = 5 + 37(-1) \)
\( a_{38} = 5 - 37 \)

\( \implies a_{38} = -32 \)
Now, we find the \( n \)-th term expression:
\( a_n = a + (n - 1)d \)
\( a_n = 5 + (n - 1)(-1) \)
\( a_n = 5 - n + 1 \)

\( \implies a_n = 6 - n \)
Therefore, the 38th term is \(-32\) and the \( n \)-th term is \( 6 - n \).
In simple words: We find that the pattern begins at 5 and decreases by 1 with each step. Consequently, the 38th term is \(-32\) and the formula to find any term is \( 6 - n \).

Exam Tip: Be cautious when subtracting negative numbers (like \(-16 - (-4)\)), as sign errors are very common here.