សូមគណនាលេខ៖ 2 + 4 + 6 + ... + 146 និង 193 - 189 - 185 - ... - 21 - 17 ។

Steps to Solve

1. Calculating the first series: 2 + 4 + 6 + ... + 146

This is an arithmetic series where the first term ( a = 2 ), the common difference ( d = 2 ), and the last term ( l = 146 ).

First, let's find the number of terms ( n ) in the series using the formula for the last term of an arithmetic sequence: $$ l = a + (n-1)d $$

Substituting in the values: $$ 146 = 2 + (n-1) \cdot 2 $$

Solving for ( n ): $$ 146 - 2 = (n-1) \cdot 2 $$ $$ 144 = (n-1) \cdot 2 $$ $$ n-1 = 72 $$ $$ n = 73 $$

2. Finding the sum of the first series

The sum ( S_n ) of an arithmetic series is given by: $$ S_n = \frac{n}{2} \cdot (a + l) $$

Substituting ( n = 73 ), ( a = 2 ), and ( l = 146 ): $$ S_{73} = \frac{73}{2} \cdot (2 + 146) $$

Calculating: $$ S_{73} = \frac{73}{2} \cdot 148 $$ $$ S_{73} = 73 \cdot 74 $$ $$ S_{73} = 5402 $$

3. Calculating the second series: 193 - 189 - 185 - ... - 21 - 17

This is also an arithmetic series with ( a = 193 ), ( d = -4 ), and ( l = 17 ).

Finding the number of terms ( n ): $$ 17 = 193 + (n-1)(-4) $$

Solving for ( n ): $$ 17 - 193 = (n-1)(-4) $$ $$ -176 = (n-1)(-4) $$ $$ n-1 = 44 $$ $$ n = 45 $$

4. Finding the sum of the second series

Using the sum formula for an arithmetic series: $$ S_n = \frac{n}{2} \cdot (a + l) $$

Substituting ( n = 45 ), ( a = 193 ), and ( l = 17 ): $$ S_{45} = \frac{45}{2} \cdot (193 + 17) $$

Calculating: $$ S_{45} = \frac{45}{2} \cdot 210 $$ $$ S_{45} = 45 \cdot 105 $$ $$ S_{45} = 4725 $$