NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3

Question 26

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

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

where:

• \(a\) = first term
• \(d\) = common difference
• \(n\) = number of terms

Step 2: Find \(S_{12}\).

\[ S_{12} = \dfrac{12}{2} \big( 2a + (12-1)d \big) \]

\[ S_{12} = 6(2a + 11d) \]

\[ S_{12} = 12a + 66d \]

Step 3: Find \(S_8\).

\[ S_8 = \dfrac{8}{2} \big( 2a + (8-1)d \big) \]

\[ S_8 = 4(2a + 7d) \]

Step 4: Find \(S_4\).

\[ S_4 = \dfrac{4}{2} \big( 2a + (4-1)d \big) \]

\[ S_4 = 2(2a + 3d) \]

Step 5: Calculate \(S_8 - S_4\).

\[ S_8 - S_4 = 4(2a + 7d) - 2(2a + 3d) \]

\[ = (8a + 28d) - (4a + 6d) \]

\[ = 4a + 22d \]

Step 6: Multiply the result by 3.

\[ 3(S_8 - S_4) = 3(4a + 22d) \]

\[ = 12a + 66d \]

Step 7: Compare with \(S_{12}\).

We found that:

• \(S_{12} = 12a + 66d\)
• \(3(S_8 - S_4) = 12a + 66d\)

Since both are equal, the given relation is proved.