Question Video: Finding the Arithmetic Sequence under a Certain Condition | Nagwa Question Video: Finding the Arithmetic Sequence under a Certain Condition | Nagwa

# Question Video: Finding the Arithmetic Sequence under a Certain Condition Mathematics • Second Year of Secondary School

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Determine the arithmetic sequence in which πββ + πββ = 500 and πβ + πββ + πββ = 138.

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### Video Transcript

Determine the arithmetic sequence in which π sub 50 plus π sub 28 equals 500 and π sub three plus π sub 11 plus π sub 35 equals 138.

We begin by recalling that an arithmetic sequence is a sequence with a common difference between consecutive terms. The general term or πth term in an arithmetic sequence can be found using the formula π sub π is equal to π sub one plus π minus one multiplied by π, where π sub one is the first term in the sequence and π is the common difference. This means that π sub 50, the 50th term of the sequence, can be written π sub one plus 50 minus one multiplied by π, which simplifies to π sub one plus 49π. The 28th term, π sub 28, can be written as π sub one plus 27π. As the sum of the 50th and 28th terms is 500, we have π sub one plus 49π plus π sub one plus 27π is equal to 500. This simplifies to two π sub one plus 76π is equal to 500. Dividing both sides by two, we have π sub one plus 38π is equal to 250. And subtracting 38π from both sides, π sub one is equal to 250 minus 38π. We will call this equation one.

The second equation tells us that the sum of the third, 11th, and 35th terms in the sequence is 138. The third term can be written as π sub one plus two π, the 11th term as π sub one plus 10π, and the 35th term as π sub one plus 34π. These three expressions sum to give us 138. Collecting like terms on the left-hand side, we have three π sub one plus 46π is equal to 138. If we call this equation two, we can now substitute our expression for π sub one into this equation. This gives us three multiplied by 250 minus 38π plus 46π is equal to 138. Distributing the parentheses or expanding the brackets gives us 750 minus 114π.

We now have an equation that we can solve to find the common difference π. Collecting like terms and moving the terms containing π to the right-hand side, we have 750 minus 138 is equal to 114π minus 46π. The left-hand side simplifies to 612, and the right-hand side is 68π. We can then divide through by 68, giving us π is equal to nine. The common difference of the arithmetic sequence is nine.

We can now substitute this value of π into equation one to calculate the first term. π sub one is equal to 250 minus 38 multiplied by nine. This is equal to negative 92. The first term in the arithmetic sequence is negative 92.

The arithmetic sequence where the sum of the 50th term and 28th term is 500 and the sum of the third, 11th, and 35th terms is 138 is negative 92, negative 83, negative 74, and so on. This sequence has a first term of negative 92 and a common difference of nine.

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