Find, in terms of 𝑛, the general term of an arithmetic sequence whose sixth term is 46 and the sum of the third and tenth term is 102.
If the first term of the arithmetic sequence is 𝑎 and the common difference is 𝑑, then the 𝑛th term or general term is given by 𝑎 plus 𝑛 minus one multiplied by 𝑑. In this example, we’re told that the sixth term is 46. This means that 𝑎 plus five 𝑑 is equal to 46. We were also told that the sum of the third and tenth term is 102. This means that 𝑎 plus two 𝑑 plus 𝑎 plus nine 𝑑 is equal to 102. Simplifying the left-hand side of this equation gives us two 𝑎 plus 11𝑑 equals 102.
We now have a pair of simultaneous equations. Multiplying the first equation by two gives us two 𝑎 plus 10𝑑 is equal to 92. Our next step is to subtract equation one from equation two. Two 𝑎 minus two 𝑎 is zero, 11𝑑 minus 10𝑑 is equal to one 𝑑 or 𝑑, and 102 minus 92 is equal to 10. Therefore, 𝑑, the common difference, is equal to 10.
Substituting this value of 𝑑 back into equation one gives us two 𝑎 plus 10 multiplied by 10 equals 92. 10 multiplied by 10 is 100. So we have two 𝑎 plus 100 equals 92. If we subtract 100 from both sides of this equation, we’re left with two 𝑎 equals negative eight. Finally, dividing both sides of this equation by two gives us a value for 𝑎, the first term, of negative four.
If the sixth term of our arithmetic sequence is 46 and the third term plus the tenth term is equal to 102, then the first term 𝑎 equals negative four and the common difference of the sequence is equal to 10. The general term can therefore be calculated by substituting these values into the formula 𝑇 𝑛 is equal to negative four plus 10 multiplied by 𝑛 minus one. Expanding the parenthesis or bracket gives us 10𝑛 minus 10 and negative four minus 10 is negative 14. Therefore, 𝑇 𝑛, the general term, is equal to 10𝑛 minus 14.
An arithmetic sequence whose sixth term is 46 and sum of the third and tenth term is 102 has general term 10𝑛 minus 14.