Question Video: Finding an Arithmetic Sequence given the Value of a Term in That Sequence | Nagwa Question Video: Finding an Arithmetic Sequence given the Value of a Term in That Sequence | Nagwa

# Question Video: Finding an Arithmetic Sequence given the Value of a Term in That Sequence Mathematics • Second Year of Secondary School

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Find the arithmetic sequence for which πββ = 279 and πββ is the additive inverse of πββ.

03:50

### Video Transcript

Find the arithmetic sequence for which π sub 43 is equal to 279 and π sub 11 is the additive inverse of π sub 13.

We begin by recalling that an arithmetic sequence is a sequence with common difference between consecutive terms. We are told in this question that the 43rd term in the sequence is equal to 279. The 11th term of the sequence is the additive inverse of the 13th term. This means that these two terms sum to zero. The 11th term plus the 13th term equals zero.

The general term or πth term of 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. The 43rd term can therefore be written as π sub one plus 43 minus one multiplied by π. This simplifies to π sub one plus 42π, and we know this is equal to 279. We will call this equation one.

Using the general formula, the 11th term can be written as π sub one plus 10π. The 13th term is π sub one plus 12π. This gives us a second equation, π sub one plus 10π plus π sub one plus 12π equals zero. Collecting like terms, we have two π sub one plus 22π is equal to zero. We can then divide through by two, giving us π sub one plus 11π equals zero. We will call this equation two.

And we now have a pair of simultaneous equations, and we can use these to calculate the values of π and π sub one. We will subtract equation two from equation one. π sub one minus π sub one is equal to zero and 42π minus 11π is 31π. The left-hand side simplifies to 31π, and this is equal to 279. Dividing both sides of this equation by 31 gives us π is equal to nine. We can now substitute this value of π into one of our equations. In this case, weβll choose equation two. This gives us π sub one plus 11 multiplied by nine is equal to zero. 11 multiplied by nine is 99. We can then subtract this from both sides of our equation such that π sub one is equal to negative 99. The first term of the arithmetic sequence is negative 99 and the common difference is nine.

We can therefore conclude that the arithmetic sequence for which the 43rd term is 279 and the 11th term is the additive inverse of the 13th term is negative 99, negative 90, negative 81, and so on.

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