NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 5: Arithematic Progressions

Exercise 5.1

We are asked to find the first term \(a\) of the Arithmetic Progression (AP).

Step 1: Recall the formula for the \(n^{th}\) term of an AP:

\(a_n = a + (n - 1) d\)

Step 2: Substitute the given values: \(a_n = 4\), \(n = 7\), \(d = -4\).

So, \(4 = a + (7 - 1)(-4)\).

Step 3: Simplify inside the brackets:

\(7 - 1 = 6\). So,

\(4 = a + (6)(-4)\).

Step 4: Multiply:

\(4 = a - 24\).

Step 5: Add 24 on both sides to find \(a\):

\(a = 4 + 24\).

Step 6: Calculate:

\(a = 28\).

So, the first term \(a\) is 28.

Step 1: Recall the formula for the n-th term of an arithmetic progression (AP):

\(a_n = a + (n-1) \times d\)

Step 2: Here, \(a = 3.5\), \(d = 0\), and \(n = 101\).

Step 3: Put these values in the formula:

\(a_{101} = 3.5 + (101 - 1) \times 0\)

Step 4: Simplify inside the brackets:

\(a_{101} = 3.5 + 100 \times 0\)

Step 5: Multiply \(100 \times 0 = 0\).

So, \(a_{101} = 3.5 + 0 = 3.5\).

Step 6: This means all terms of the AP are the same (because \(d=0\)), so every term is equal to \(3.5\).

Therefore, the answer is option B (3.5).

Step 1: Write the numbers clearly: \(-10, -6, -2, 2, \ldots\)

Step 2: In an Arithmetic Progression (AP), the difference between one number and the next must be the same each time. This difference is called the common difference (\(d\)).

Step 3: Find the difference between the second and the first number: \(-6 - (-10) = -6 + 10 = 4\).

Step 4: Find the difference between the third and the second number: \(-2 - (-6) = -2 + 6 = 4\).

Step 5: Find the difference between the fourth and the third number: \(2 - (-2) = 2 + 2 = 4\).

Step 6: Since the difference is the same (\(4\)) every time, the numbers form an AP with common difference \(d = 4\).

Final Answer: The list is an AP with \(d = 4\). (Option B)

Step 1: Recall the formula for the \(n^{th}\) term of an AP:

\(a_n = a + (n-1)\,d\)

Here, \(a\) = first term, \(d\) = common difference, and \(n\) = term number.

Step 2: From the given AP \(-5, -\tfrac{5}{2}, 0, \tfrac{5}{2}, \ldots\)

First term, \(a = -5\).

Common difference, \(d = -\tfrac{5}{2} - (-5) = -\tfrac{5}{2} + 5 = \tfrac{5}{2}.\)

Step 3: We want the 11th term, so \(n = 11\).

Using the formula: \(a_{11} = a + (11-1)\,d\)

\(a_{11} = -5 + 10 \times \tfrac{5}{2}\)

Step 4: Simplify:

\(10 \times \tfrac{5}{2} = 25\)

So, \(a_{11} = -5 + 25 = 20\).

Final Answer: The 11th term is 20. Option B.

We are given:

• First term (\(a\)) = −2
• Common difference (\(d\)) = −2

An Arithmetic Progression (AP) means we get the next term by adding the common difference (\(d\)) to the previous term.

Step 1: First term = −2

Step 2: To find the second term, add \(d\) to the first term:
\(-2 + (-2) = -4\)

Step 3: To find the third term, add \(d\) to the second term:
\(-4 + (-2) = -6\)

Step 4: To find the fourth term, add \(d\) to the third term:
\(-6 + (-2) = -8\)

So, the first four terms are: −2, −4, −6, −8.

Correct option: C

Step 1: In an Arithmetic Progression (AP), the first term is called \(a\). Here, the first term is \(-3\). So, \(a = -3\).

Step 2: The common difference \(d\) is found by subtracting the first term from the second term.
\(d = 4 - (-3) = 4 + 3 = 7\).

Step 3: The formula for the \(n\)-th term of an AP is:
\(a_n = a + (n-1)\,d\).

Step 4: We need the 21st term. So, put \(n = 21\).
\(a_{21} = a + (21 - 1) d\).

Step 5: Substitute the values:
\(a_{21} = -3 + 20 \times 7\).

Step 6: Multiply: \(20 \times 7 = 140\).

Step 7: Add: \(-3 + 140 = 137\).

Final Answer: The 21st term is 137.

Step 1: Recall the formula for the n^th term of an AP: \(a_n = a + (n-1) d\), where \(a\) is the first term, and \(d\) is the common difference.

Step 2: The 2nd term is given as 13. Using the formula: \(a_2 = a + (2-1)d = a + d\). So, \(a + d = 13\). (Equation 1)

Step 3: The 5th term is given as 25. Using the formula: \(a_5 = a + (5-1)d = a + 4d\). So, \(a + 4d = 25\). (Equation 2)

Step 4: Subtract Equation (1) from Equation (2): \((a + 4d) - (a + d) = 25 - 13\) \(3d = 12\) \(d = 4\).

Step 5: Put the value of \(d\) into Equation (1): \(a + 4 = 13\) \(a = 9\).

Step 6: Now find the 7th term: \(a_7 = a + (7-1)d = 9 + 6 \times 4 = 9 + 24 = 33\).

Final Answer: The 7th term is 33.

Step 1: In an Arithmetic Progression (AP), the first term is called a and the common difference is called d.

Here, the AP is \(21, 42, 63, 84, \ldots\).

So, \(a = 21\) and \(d = 42 - 21 = 21\).

Step 2: The formula for the \(n^{\text{th}}\) term of an AP is:

\(a_n = a + (n-1)\,d\)

Step 3: We are asked: which term is \(210\)?

This means we set \(a_n = 210\).

So, \(210 = 21 + (n-1)\times 21\).

Step 4: Simplify the equation:

\(210 = 21 + 21(n-1)\)

\(210 = 21 + 21n - 21\)

\(210 = 21n\)

Step 5: Divide both sides by 21:

\(n = \dfrac{210}{21} = 10\).

Final Answer: The \(210\) is the 10th term of the AP.

We are asked to find \(a_{18} - a_{13}\), which means the difference between the 18th term and the 13th term of an Arithmetic Progression (AP).

Step 1: Recall the property of AP. The difference between the \(m\)-th term and the \(n\)-th term is given by:

\(a_m - a_n = (m - n) \times d\)

Here, \(d\) is the common difference.

Step 2: In this question, \(m = 18\), \(n = 13\), and \(d = 5\).

Step 3: Substitute the values:

\(a_{18} - a_{13} = (18 - 13) \times 5\)

Step 4: Simplify the bracket:

\(18 - 13 = 5\)

Step 5: Multiply:

\(5 \times 5 = 25\)

So, \(a_{18} - a_{13} = 25\).

Correct option: C

We are given: \(a_{18} - a_{14} = 32\).

Step 1: Recall the formula for the \(n^{th}\) term of an AP: \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Step 2: The difference between two terms in an AP depends only on the difference of their positions: \(a_m - a_n = (m - n)d\).

Step 3: Here, \(m = 18\) and \(n = 14\). So, \(a_{18} - a_{14} = (18 - 14)d = 4d\).

Step 4: We are told this difference is 32. So, \(4d = 32\).

Step 5: Divide both sides by 4: \(d = \dfrac{32}{4} = 8\).

Therefore, the common difference is 8.

Step 1: The formula for the \(n^{th}\) term of an AP is:

\(a_n = a + (n-1) \times d\)

Step 2: For the first AP, the first term \(a = -1\) and the common difference is \(d\).

So, the 4th term = \(-1 + (4-1) \times d = -1 + 3d\).

Step 3: For the second AP, the first term \(a = 8\) and the common difference is the same \(d\).

So, the 4th term = \(8 + (4-1) \times d = 8 + 3d\).

Step 4: Now, find the difference between the 4th terms:

\((-1 + 3d) - (8 + 3d) = -1 + 3d - 8 - 3d = -9\).

Step 5: Therefore, the difference between their 4th terms is \(-9\).

Step 1: Recall the formula for the \(n\)-th term of an AP.

\(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Step 2: Write the 7th term.

\(a_7 = a + (7-1)d = a + 6d\).

Step 3: Write the 11th term.

\(a_{11} = a + (11-1)d = a + 10d\).

Step 4: Use the condition given in the question.

7 times the 7th term = 11 times the 11th term.

So, \(7(a + 6d) = 11(a + 10d)\).

Step 5: Expand both sides.

\(7a + 42d = 11a + 110d\).

Step 6: Bring like terms together.

\(7a - 11a = 110d - 42d\).

\(-4a = 68d\).

Step 7: Simplify.

\(a = -17d\).

Step 8: Find the 18th term.

\(a_{18} = a + (18-1)d = a + 17d\).

Step 9: Substitute \(a = -17d\).

\(a_{18} = -17d + 17d = 0\).

Final Answer: The 18th term is 0 (Option D).

Step 1: Identify the first term and common difference.

The first term \(a = -11\).

The numbers are increasing by 3 each time, so the common difference \(d = 3\).

Step 2: Find how many terms are there in the AP.

The last term is given as \(l = 49\).

The formula to find the number of terms in an AP is:

\(n = \dfrac{l - a}{d} + 1\).

Substitute the values: \(n = \dfrac{49 - (-11)}{3} + 1 = \dfrac{60}{3} + 1 = 20 + 1 = 21\).

So, there are 21 terms in this AP.

Step 3: Understand what "4th term from the end" means.

If there are 21 terms in total, the last term is the 21st term. The 4th from the end means:

21st → last term

20th → 2nd from the end

19th → 3rd from the end

18th → 4th from the end

So we need to find the 18th term.

Step 4: Find the 18th term.

The formula for the \(n\)th term is: \(a_n = a + (n-1)d\).

Put \(n = 18\):

\(a_{18} = -11 + (18 - 1) \times 3\).

\(a_{18} = -11 + 17 \times 3\).

\(a_{18} = -11 + 51 = 40\).

Final Answer: The 4th term from the end is 40.

When Gauss was a small schoolboy, his teacher asked the class to add all the numbers from 1 to 100.

Instead of adding one by one, Gauss thought of a smart trick:

He paired the first and last numbers:

\(1 + 100 = 101\)

Then the second and the second-last numbers:

\(2 + 99 = 101\)

Similarly, \(3 + 98 = 101\), and so on.

He noticed that every pair adds up to 101.

There are 50 such pairs (because 100 numbers make 50 pairs).

So the total sum is:

\(50 \times 101 = 5050\)

This clever method is why Gauss is remembered for this problem.

We are asked to find the sum of the first 6 terms of an Arithmetic Progression (AP).

Step 1: Recall the formula for the sum of the first \(n\) terms of an AP:

\[ S_n = \frac{n}{2} \big(2a + (n-1)d \big) \]

where \(a\) = first term, \(d\) = common difference, \(n\) = number of terms.

Step 2: Here, \(a = -5\), \(d = 2\), \(n = 6\).

Step 3: Substitute these values into the formula:

\[ S_6 = \frac{6}{2} \big(2(-5) + (6-1)(2) \big) \]

Step 4: Simplify step by step:

\(\frac{6}{2} = 3\)

Inside the bracket: \(2(-5) + 5 \cdot 2 = -10 + 10 = 0\)

Step 5: Multiply: \(3 \times 0 = 0\).

Final Answer: The sum of the first 6 terms is 0.

Step 1: Identify the first term (\(a\)) and the common difference (\(d\)).

The first term is \(a = 10\).

The numbers are decreasing by 4 each time (\(6 - 10 = -4\)).

So, \(d = -4\).

Step 2: Recall the formula for the sum of first \(n\) terms of an AP:

\[ S_n = \dfrac{n}{2} [2a + (n-1)d] \]

Step 3: Put the values into the formula:

Here, \(n = 16\), \(a = 10\), and \(d = -4\).

\[ S_{16} = \dfrac{16}{2} [2(10) + (16-1)(-4)] \]

Step 4: Simplify step by step:

\( \dfrac{16}{2} = 8 \)

Inside the brackets: \( 2(10) = 20 \)

\( (16-1)(-4) = 15 \times -4 = -60 \)

So, inside the brackets: \(20 - 60 = -40\)

Step 5: Multiply:

\( S_{16} = 8 \times (-40) = -320 \)

Final Answer: \(-320\), which is Option A.

We are given:

• First term, \(a = 1\)
• Last term, \(a_n = 20\)
• Sum of \(n\) terms, \(S_n = 399\)

Step 1: Recall the formula for the sum of \(n\) terms of an AP:

\[ S_n = \frac{n}{2} (a + a_n) \]

Step 2: Substitute the given values:

\[ 399 = \frac{n}{2} (1 + 20) \]

Step 3: Simplify inside the bracket:

\[ 399 = \frac{n}{2} (21) \]

Step 4: Multiply:

\[ 399 = \frac{21n}{2} \]

Step 5: Remove the fraction by multiplying both sides by 2:

\[ 798 = 21n \]

Step 6: Divide both sides by 21:

\[ n = \frac{798}{21} = 38 \]

Therefore, \(n = 38\).

Step 1: A multiple of 3 means a number that comes in the table of 3.

Step 2: The first five multiples of 3 are: \(3, 6, 9, 12, 15\).

Step 3: Now, add them one by one:

\(3 + 6 = 9\)

\(9 + 9 = 18\)

\(18 + 12 = 30\)

\(30 + 15 = 45\)

Step 4: Therefore, the sum of the first five multiples of 3 is \(45\).