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

Exercise 5.4

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

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

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

Step 2: Write the given conditions using this formula.

• For the first 5 terms: \( S_5 = \dfrac{5}{2}[2a + 4d] = 5(a + 2d) \)
• For the first 7 terms: \( S_7 = \dfrac{7}{2}[2a + 6d] = 7(a + 3d) \)

It is given that: \( S_5 + S_7 = 167 \)

So, \( 5(a + 2d) + 7(a + 3d) = 167 \)

Simplify: \( 5a + 10d + 7a + 21d = 167 \)

\( 12a + 31d = 167 \) … (Equation 1)

Step 3: Write the condition for 10 terms.

\( S_{10} = \dfrac{10}{2}[2a + 9d] = 5(2a + 9d) = 10a + 45d \)

It is given: \( S_{10} = 235 \)

So, \( 10a + 45d = 235 \) … (Equation 2)

Step 4: Solve the two equations to find \(a\) and \(d\).

Equation (1): \( 12a + 31d = 167 \)

Equation (2): \( 10a + 45d = 235 \)

Multiply Equation (2) by 6: \( 60a + 270d = 1410 \)

Multiply Equation (1) by 5: \( 60a + 155d = 835 \)

Subtract the two: \( (60a + 270d) - (60a + 155d) = 1410 - 835 \)

\( 115d = 575 \)

\( d = 5 \)

Now put \( d = 5 \) in Equation (2): \( 10a + 45(5) = 235 \)

\( 10a + 225 = 235 \)

\( 10a = 10 \)

\( a = 1 \)

Step 5: Find the sum of the first 20 terms.

Formula: \( S_{20} = \dfrac{20}{2}[2a + 19d] \)

\( S_{20} = 10[2(1) + 19(5)] \)

\( S_{20} = 10[2 + 95] \)

\( S_{20} = 10 \times 97 = 970 \)

Final Answer: The sum of the first 20 terms is 970.

(i) and (ii)

To be a multiple of both 2 and 5, a number must be a multiple of 10 (because 10 is the least common multiple of 2 and 5).

So the numbers we are adding are: 10, 20, 30, ..., 500.

This is an Arithmetic Progression (AP) with:

• First term \(a = 10\)
• Common difference \(d = 10\)
• Last term \(l = 500\)

Number of terms: \(n = \frac{l - a}{d} + 1 = \frac{500 - 10}{10} + 1 = 50\).

Formula for sum of AP: \(S = \frac{n}{2}(a + l)\).

So, \(S = \frac{50}{2}(10 + 500) = 25 \times 510 = 12750\).

Thus, answer for (i) and (ii) = 12750.

(iii)

Now we need the sum of numbers that are multiples of 2 or 5.

This means we add all multiples of 2 and all multiples of 5, but be careful: numbers that are multiples of 10 will be counted twice (once in 2’s table, once in 5’s table). So we must subtract them once. This method is called the Inclusion–Exclusion Principle.

Step 1: Sum of multiples of 2

Multiples of 2 up to 500 are: 2, 4, 6, ..., 500.

There are \(250\) such numbers (since \(500/2 = 250\)).

Sum = \(2(1 + 2 + 3 + ... + 250)\).

We know formula: \(1 + 2 + ... + n = \frac{n(n+1)}{2}\).

So, sum = \(2 \times \frac{250 \times 251}{2} = 250 \times 251 = 62750\).

Step 2: Sum of multiples of 5

Multiples of 5 up to 500 are: 5, 10, 15, ..., 500.

There are \(100\) such numbers (since \(500/5 = 100\)).

Sum = \(5(1 + 2 + ... + 100) = 5 \times \frac{100 \times 101}{2}\).

So, sum = \(5 \times 5050 = 25250\).

Step 3: Subtract sum of multiples of 10

Multiples of 10 are already included in both the above sums, so they are counted twice.

We already calculated this in (i): sum = 12750.

Step 4: Apply Inclusion–Exclusion

Total sum = (sum of multiples of 2) + (sum of multiples of 5) − (sum of multiples of 10).

Total = \(62750 + 25250 − 12750 = 75250\).

So, answer for (iii) = 75250.

Step 1: Write the general formula of an AP

The (n^{th}) term of an AP is: \(a_n = a + (n-1)d\), where \(a\) = first term, \(d\) = common difference.

Step 2: Translate the first condition

The 8th term is half of the 2nd term.

8th term = \(a + 7d\)

2nd term = \(a + d\)

So, \(a + 7d = \tfrac{1}{2}(a + d)\).

Step 3: Simplify this equation

Multiply both sides by 2: \(2a + 14d = a + d\)

Bring terms together: \(2a - a + 14d - d = 0\)

\(a + 13d = 0\)

So, \(a = -13d\). … (Equation 1)

Step 4: Translate the second condition

The 11th term is 1 more than one-third of the 4th term.

11th term = \(a + 10d\)

4th term = \(a + 3d\)

Condition: \(a + 10d = \tfrac{1}{3}(a + 3d) + 1\)

Step 5: Simplify this equation

Multiply everything by 3 to remove the fraction: \(3a + 30d = a + 3d + 3\)

Bring terms together: \(3a - a + 30d - 3d = 3\)

\(2a + 27d = 3\) … (Equation 2)

Step 6: Substitute value of a

From (Equation 1): \(a = -13d\)

Substitute in (Equation 2): \(2(-13d) + 27d = 3\)

\(-26d + 27d = 3\)

\(d = 3\)

Now, \(a = -13d = -13 × 3 = -39\)

Step 7: Find the 15th term

15th term = \(a + 14d\)

= \(-39 + 14 × 3\)

= \(-39 + 42\)

= 3

Final Answer: The 15th term is 3.

Step 1: Identify the middle terms.

The total number of terms is 37. The middle term is the \(19^{\text{th}}\) term. So, the three middle-most terms are the \(18^{\text{th}}, 19^{\text{th}}, 20^{\text{th}}\).

Step 2: Use the given sum of middle terms.

Sum of these three terms = 225.

Notice that the middle three terms are consecutive. Their sum is the same as \(3 \times a_{19}\) (because in an AP, the middle term is the average of three consecutive terms).

So, \(3a_{19} = 225 \Rightarrow a_{19} = 75\).

Step 3: Identify the last three terms.

The last three terms are the \(35^{\text{th}}, 36^{\text{th}}, 37^{\text{th}}\).

Their sum is given as 429.

Similarly, this sum is equal to \(3a_{36}\) (since \(a_{36}\) is the middle term of these three).

So, \(3a_{36} = 429 \Rightarrow a_{36} = 143\).

Step 4: Relating two terms of the AP.

We know:

• \(a_{19} = a + 18d = 75\)
• \(a_{36} = a + 35d = 143\)

Step 5: Eliminate \(a\) to find \(d\).

Subtract the two equations:

\((a + 35d) - (a + 18d) = 143 - 75\)

\(17d = 68 \Rightarrow d = 4\).

Step 6: Find the first term \(a\).

Substitute \(d = 4\) in \(a + 18d = 75\):

\(a + 18(4) = 75\)

\(a + 72 = 75 \Rightarrow a = 3\).

Final Answer:

First term \(a = 3\), common difference \(d = 4\). So the AP is: 3, 7, 11, 15, …

Step (i): Integers divisible by 9

1. First, write down the numbers between 100 and 200 which are divisible by 9.

The smallest number greater than 100 divisible by 9 is 108, and the largest below 200 is 198.

So the numbers are: 108, 117, 126, …, 198.

2. This is an arithmetic progression (AP) with first term \(a = 108\), last term \(l = 198\), and common difference \(d = 9\).

3. To find how many terms there are, use the AP formula: \(n = \frac{l - a}{d} + 1\).

\(n = \frac{198 - 108}{9} + 1 = \frac{90}{9} + 1 = 10 + 1 = 11\).

4. The sum of an AP is given by: \(S = \frac{n}{2}(a + l)\).

\(S = \frac{11}{2}(108 + 198) = \frac{11}{2}(306) = 11 \times 153 = 1683\).

So, the sum of numbers divisible by 9 is 1683.

Step (ii): Integers not divisible by 9

1. Integers strictly between 100 and 200 are: 101, 102, …, 199.

This is another AP with first term \(a = 101\), last term \(l = 199\), and number of terms \(n = 99\).

2. The sum of these 99 numbers is:

\(S = \frac{n}{2}(a + l) = \frac{99}{2}(101 + 199) = \frac{99}{2}(300) = 99 \times 150 = 14850\).

3. Out of this total sum, the part divisible by 9 (already found in step (i)) is 1683.

4. Therefore, the sum of integers not divisible by 9 is:

\(14850 - 1683 = 13167\).

So, the sum of numbers not divisible by 9 is 13167.

Step 1: General term of an AP

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

Step 2: Write the 11th and 18th terms

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

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

Step 3: Use the given ratio

We are told that \(a_{11}:a_{18} = 2:3\).

This means: \(\dfrac{a+10d}{a+17d} = \dfrac{2}{3}\).

Step 4: Solve for relation between (a) and (d)

Cross multiply: \(3(a + 10d) = 2(a + 17d)\)

\(3a + 30d = 2a + 34d\)

\(3a - 2a = 34d - 30d\)

\(a = 4d\)

Step 5: Find the 5th and 21st terms

\(a_5 = a + 4d = 4d + 4d = 8d\)

\(a_{21} = a + 20d = 4d + 20d = 24d\)

So, ratio \(a_5:a_{21} = 8d:24d = 1:3\).

Step 6: Formula for sum of (n) terms of an AP

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

Step 7: Find sum of first 5 terms

\(S_5 = \dfrac{5}{2}[2a + 4d]\)

Substitute (a = 4d): \(S_5 = \dfrac{5}{2}[2(4d) + 4d]\)

\(= \dfrac{5}{2}(8d + 4d) = \dfrac{5}{2}(12d) = 30d\)

Step 8: Find sum of first 21 terms

\(S_{21} = \dfrac{21}{2}[2a + 20d]\)

Substitute (a = 4d): \(S_{21} = \dfrac{21}{2}[2(4d) + 20d]\)

\(= \dfrac{21}{2}(8d + 20d) = \dfrac{21}{2}(28d) = 294d\)

Step 9: Ratio of sums

\(S_5:S_{21} = 30d:294d = 30:294 = 5:49\)

Final Answer: \(a_5:a_{21} = 1:3\), \(S_5:S_{21} = 5:49\)

Step 1: Recall formula for common difference.

The common difference of an AP is given by:

\(d = \text{second term} - \text{first term} = b - a.\)

Step 2: Express last term in terms of \(n\).

In an AP, the \(n^{\text{th}}\) term is: \(a_n = a + (n-1)d.\)

Here, last term \(c = a + (n-1)d.\)

So, solving for \(n\):

\(n-1 = \dfrac{c-a}{d} \;\Rightarrow\; n = \dfrac{c-a}{d} + 1.\)

Step 3: Formula for sum of \(n\) terms.

We know: \(S = \dfrac{n}{2}(\text{first term} + \text{last term}).\)

So, \(S = \dfrac{n}{2}(a+c).\)

Step 4: Substitute value of \(n\).

\(S = \dfrac{a+c}{2}\Big(\dfrac{c-a}{d} + 1\Big).\)

Step 5: Simplify.

\(S = \dfrac{a+c}{2}\cdot\dfrac{c-a+d}{d}.\)

Step 6: Replace \(d = b-a.\)

Then numerator: \(c-a+d = c-a+(b-a) = b+c-2a.\)

Final Result:

\(S = \dfrac{(a+c)(b+c-2a)}{2(b-a)}.\)

Hence proved.

Step 1: Identify the sequence.

The numbers are: \(-4, -1, 2, ... , x\).

This is an arithmetic progression (AP), because the difference between terms is constant.

Step 2: Find the first term (a) and common difference (d).

First term: \(a = -4\).

Second term: \(-1\).

Common difference: \(d = -1 - (-4) = 3\).

Step 3: General formula for the n^th term.

In an AP, the n^th term is:

\(T_n = a + (n-1) \cdot d\).

Here, the last term is \(x\). So:

\(x = -4 + (n-1) \cdot 3\).

Step 4: Rearrange to find n.

\(x + 4 = (n-1) \cdot 3\).

\(n - 1 = \dfrac{x+4}{3}\).

\(n = \dfrac{x+7}{3}\).

Step 5: Use the sum formula for an AP.

The sum of n terms is:

\(S_n = \dfrac{n}{2}(a + \text{last term})\).

Here, \(S_n = 437, a = -4, \text{last term} = x\).

So: \(437 = \dfrac{n}{2}(-4 + x)\).

Step 6: Substitute n.

\(437 = \dfrac{1}{2} \cdot \dfrac{x+7}{3} (x-4)\).

Simplify:

\(437 = \dfrac{(x+7)(x-4)}{6}\).

Step 7: Eliminate the fraction.

Multiply both sides by 6:

\(437 \times 6 = (x+7)(x-4)\).

\(2622 = x^2 + 3x - 28\).

\(x^2 + 3x - 2650 = 0\).

Step 8: Solve the quadratic equation.

Equation: \(x^2 + 3x - 2650 = 0\).

Using factorization (or quadratic formula), we get:

\(x = 50\) or \(x = -53\).

Step 9: Choose the valid answer.

The sequence starts from \(-4\) and increases by 3 each time. So a negative value like \(-53\) cannot appear in this sequence.

Therefore, the valid solution is:

\(x = 50\)

Step 1: Identify the pattern of payments

Jaspal Singh starts by paying Rs 1000 in the first month. Each month he pays Rs 100 more than the previous month. So, the instalments are:

1000, 1100, 1200, 1300, …

This is an arithmetic progression (AP) where:

• First term, \(a = 1000\) (in Rs)
• Common difference, \(d = 100\) (in Rs)

Step 2: Find the 30th instalment

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

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

For the 30th instalment (\(n = 30\)):

\(a_{30} = 1000 + (30-1)\times100 = 1000 + 2900 = 3900\)

So, the 30th instalment is Rs 3900.

Step 3: Find the total paid in 30 months

The formula for the sum of first \(n\) terms of an AP is:

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

For \(n = 30\):

\(S_{30} = \dfrac{30}{2}[2(1000) + (30-1)(100)]\)

\(= 15[2000 + 2900]\)

\(= 15 \times 4900\)

\(= 73500\)

So, the total paid in 30 months = Rs 73,500.

Step 4: Find the unpaid amount

Total loan = Rs 118000

Amount paid in 30 months = Rs 73500

Unpaid = Rs 118000 – Rs 73500 = Rs 44,500

Final Answer:

30th instalment = Rs 3900

Unpaid after 30 instalments = Rs 44,500

Step 1: Total number of flags = 27.

Step 2: The middle-most flag is number 14 (because 13 flags on the left, 1 in the middle, 13 on the right). The store is exactly at this 14th flag’s position.

Step 3: The middle flag does not need any movement, because it is already at the store position.

Step 4: On each side (left and right), there are 13 flags. These flags are placed at distances of 2 m, 4 m, 6 m, …, up to 26 m away from the store.

Step 5: Ruchi carries one flag at a time. For a flag kept at distance \(d\) metres, she walks from the store to that place and then comes back. So the distance covered = \(2d\) metres (a round trip).

Step 6: Total distance for one side:

\[ 2(2 + 4 + 6 + \cdots + 26) \]

This is 2 times the sum of the first 13 even numbers.

Sum of first 13 even numbers = \(2 + 4 + 6 + \cdots + 26 = 2(1+2+3+\cdots+13)\).

\[ 1+2+3+\cdots+13 = \frac{13 \times 14}{2} = 91. \]

So, sum = \(2 \times 91 = 182\).

Therefore, distance for one side = \(2 \times 182 = 364\,\text{m}.\)

Step 7: The other side is exactly the same, so total distance both sides = \(364\,\text{m}\).

Step 8: The farthest flag is at 26 m. This means the maximum distance she carries a flag away from the store is \(26\,\text{m}\).

Final Answer: Total distance = 364 m, Maximum carrying distance = 26 m.