Math  /  Algebra

QuestionThe 15 th term of the arithmetic sequence is 33 and the 50 th term is 103 . What is the 79th partial sum of the arithmetic sequence?

Studdy Solution

STEP 1

1. The sequence is arithmetic, meaning it has a common difference.
2. We are given the 15th term and the 50th term of the sequence.
3. We need to find the 79th partial sum of the sequence.

STEP 2

1. Determine the common difference of the sequence.
2. Find the first term of the sequence.
3. Calculate the 79th partial sum using the formula for the sum of an arithmetic sequence.

STEP 3

The formula for the n n -th term of an arithmetic sequence is given by: an=a1+(n1)d a_n = a_1 + (n-1) \cdot d where a1 a_1 is the first term and d d is the common difference.
Given: a15=33 a_{15} = 33 a50=103 a_{50} = 103
We can set up two equations: a1+14d=33 a_1 + 14d = 33 a1+49d=103 a_1 + 49d = 103

STEP 4

Subtract the first equation from the second to find the common difference d d : (a1+49d)(a1+14d)=10333(a_1 + 49d) - (a_1 + 14d) = 103 - 33 35d=7035d = 70 d=2d = 2

STEP 5

Substitute d=2 d = 2 back into the first equation to find a1 a_1 : a1+142=33a_1 + 14 \cdot 2 = 33 a1+28=33a_1 + 28 = 33 a1=5a_1 = 5

STEP 6

The formula for the sum of the first n n terms of an arithmetic sequence is: Sn=n2(2a1+(n1)d) S_n = \frac{n}{2} \cdot (2a_1 + (n-1) \cdot d)
Substitute n=79 n = 79 , a1=5 a_1 = 5 , and d=2 d = 2 into the formula: S79=792(25+(791)2)S_{79} = \frac{79}{2} \cdot (2 \cdot 5 + (79-1) \cdot 2) S79=792(10+782)S_{79} = \frac{79}{2} \cdot (10 + 78 \cdot 2) S79=792(10+156)S_{79} = \frac{79}{2} \cdot (10 + 156) S79=792166S_{79} = \frac{79}{2} \cdot 166 S79=7983S_{79} = 79 \cdot 83 S79=6557S_{79} = 6557
The 79th partial sum of the arithmetic sequence is:
6557 \boxed{6557}

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