Question Video: Finding the General Term of an Arithmetic Sequence | Nagwa Question Video: Finding the General Term of an Arithmetic Sequence | Nagwa

Question Video: Finding the General Term of an Arithmetic Sequence Mathematics • Second Year of Secondary School

The fifth term of an arithmetic sequence is 50 and the tenth term is 25 times greater than the second term. Find the general term 𝑇_𝑛.

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Video Transcript

The fifth term of an arithmetic sequence is 50 and the tenth term is 25 times greater than the second term. Find the general term 𝑇 of 𝑛.

The general term 𝑇 𝑛 of any arithmetic sequence is given by 𝑎 plus 𝑛 minus one multiplied by 𝑑, where 𝑎 is the first term of the sequence and 𝑑 is the common difference. In this example, the fifth term is equal to 50. This means that 𝑎 plus four 𝑑 is equal to 50. We were also told that the tenth term is 25 times greater than the second term. So the tenth term is equal to 25 multiplied by the second term.

This means that 𝑎 plus nine 𝑑 is equal to 25 multiplied by 𝑎 plus 𝑑. Expanding the bracket or parenthesis gives us 25𝑎 plus 25𝑑. Simplifying this equation by grouping like terms gives us 24𝑎 plus 16𝑑 is equal to zero. We now have two simultaneous equations which we can solve to calculate the values of 𝑎 and the values of 𝑑.

If we divide equation two by four, we’re left with six 𝑎 plus 4𝑑 equals zero. We now have two equations six 𝑎 plus four 𝑑 equals zero and 𝑎 plus four 𝑑 is equal to 50. Subtracting these two equations gives us five 𝑎 is equal to negative 50 as six 𝑎 minus 𝑎 is five 𝑎, four 𝑑 minus four 𝑑 is zero, and zero minus 50 is negative 50. If we divide both sides of this equation by five, we’re left with a value of 𝑎 of negative 10. The first term in our sequence is negative 10.

Substituting this value into the equation 𝑎 plus four 𝑑 equals 50 gives us negative 10 plus four 𝑑 is equal to 50. Adding 10 to both sides of this new equation gives us four 𝑑 is equal to 60. And finally, dividing both sides by four gives us a value of 𝑑 equal to 15. The common difference in the sequence is 15.

The general term 𝑇𝑛 was given by 𝑎 plus 𝑛 minus one multiplied by 𝑑. We now need to substitute in our value of 𝑎 negative 10 and a value of 𝑑 15. This gives us negative 10 plus 𝑛 minus one multiplied by 15. Expanding the bracket gives us 15𝑛 minus 15. Therefore, 𝑇𝑛 is equal to negative 10 plus 15𝑛 minus 15. This gives us a final answer of 𝑇𝑛 equals 15𝑛 minus 25.

The general term of a sequence with fifth term 50 and tenth term 25 times greater than the second term is 15𝑛 minus 25.

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