Solving Arithmetic Progressions: Finding First and Last Terms and Sum

Arithmetic progressions (AP) are sequences of numbers where each term after the first is obtained by adding a constant, called the common difference, to the previous term. This article will guide you through finding the first term, last term, and the sum of an AP with 15 terms and a common difference of -3.

Understanding the Given Information

We are given an arithmetic progression (AP) with:

Number of terms, ( n 15 )Common difference, ( d -3 )

The goal is to find the first term (( a )), the last term (( t_{15} )), and the sum of all terms (( S_{15} )) of this AP.

Step-by-Step Solution

To find the first term (( a )):

Step 1: Use the Sum Formula

The sum of the first ( n ) terms of an AP is given by:

[ S_n frac{n}{2} [2a (n-1)d] ]

Given ( n 15 ), ( d -3 ), and ( S_{15} 120 ), substitute these values into the sum formula:

[ 120 frac{15}{2} [2a (15-1)(-3)] ]

Calculate the expression inside the brackets:

[ 120 frac{15}{2} [2a - 42] ]

Multiply both sides by 2 to clear the fraction:

[ 240 15 (2a - 42) ]

Divide both sides by 15:

[ 16 2a - 42 ]

Add 42 to both sides:

[ 58 2a ]

Divide by 2:

[ a 29 ]

So, the first term ( a 29 ).

Step 2: Find the Last Term (( t_{15} ))

The last term ( t_{15} ) of an AP can be calculated using the formula:

[ t_n a (n-1)d ]

Substitute ( n 15 ), ( a 29 ), and ( d -3 ) into the formula:

[ t_{15} 29 (15-1)(-3) ]

Simplify the expression:

[ t_{15} 29 14(-3) ][ t_{15} 29 - 42 ][ t_{15} -13 ]

So, the last term ( t_{15} -13 ).

Step 3: Calculate the Sum of Terms

The sum of the first ( n ) terms of an AP can also be found using the formula:

[ S_n frac{n}{2} [a l] ]

Where ( l ) is the last term. Substitute ( n 15 ), ( a 29 ), and ( l -13 ) into the formula:

[ S_{15} frac{15}{2} [29 (-13)] ]

Simplify the expression inside the brackets:

[ S_{15} frac{15}{2} [16] ][ S_{15} 15 times 8 ][ S_{15} 120 ]

This confirms our earlier sum value.

General Term of the AP

The general term rule of this arithmetic sequence when ( n 1, 2, 3, ldots ) is:

[ t_n a (n-1)d ]

Substitute ( a 29 ) and ( d -3 ) into the formula:

[ t_n 29 (n-1)(-3) ][ t_n 29 - 3(n-1) ][ t_n 29 - 3n 3 ][ t_n 32 - 3n ]

This formula represents the ( n )-th term of the given AP.

Conclusion

To summarize, for an arithmetic progression with 15 terms and a common difference of -3:

The first term ( a 29 )The last term ( t_{15} -13 )The sum of the terms ( S_{15} 120 )

The sequence is:29, 26, 23, 20, 17, 14, 11, 8, 5, 2, -1, -4, -7, -10, -13.