NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.4
Question 1

Question. 1

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms is 235, find the sum of its first twenty terms.

Answer:

970

Detailed Answer with Explanation:

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.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.4 | Detailed Answers