If the sum of the first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, then find the sum of the first 10 terms.

Steps to Solve

1. Identify the given values

We know that the sum of the first 6 terms (S₆) is 36 and the sum of the first 16 terms (S₁₆) is 256.

2. Write the formula for the sum of an A.P.

The formula for the sum of the first $n$ terms of an arithmetic progression (A.P.) is given by:

$$ S_n = \frac{n}{2} (2a + (n-1)d) $$

where $a$ is the first term and $d$ is the common difference.

3. Write equations for the given sums

Substituting into the formulas:

For $S₆$: $$ S_6 = \frac{6}{2} (2a + (6-1)d) = 36 $$

This simplifies to: $$ 3(2a + 5d) = 36 $$ $$ 2a + 5d = 12 \quad \text{(Equation 1)} $$

For $S₁₆$: $$ S_{16} = \frac{16}{2} (2a + (16-1)d) = 256 $$

This simplifies to: $$ 8(2a + 15d) = 256 $$ $$ 2a + 15d = 32 \quad \text{(Equation 2)} $$

4. Set up a system of equations

We have two equations now:

1. $2a + 5d = 12 \quad \text{(Equation 1)}$

2. $2a + 15d = 32 \quad \text{(Equation 2)}$

3. Subtract Equation 1 from Equation 2

Subtracting gives: $$ (2a + 15d) - (2a + 5d) = 32 - 12 $$ $$ 10d = 20 $$ $$ d = 2 $$

6. Substitute back to find (a)

Now substitute $d = 2$ into Equation 1: $$ 2a + 5(2) = 12 $$ $$ 2a + 10 = 12 $$ $$ 2a = 2 $$ $$ a = 1 $$

7. Find the sum of the first 10 terms (S_{10})

Now we can find $S_{10}$ using the formula again: $$ S_{10} = \frac{10}{2} (2a + (10-1)d) $$ Substituting $a = 1$ and $d = 2$: $$ S_{10} = 5(2(1) + 9(2)) $$ $$ S_{10} = 5(2 + 18) $$ $$ S_{10} = 5 \times 20 = 100 $$