# Video: Finding the Arithmetic Sequence under a Given Condition

Find the four numbers which form an arithmetic sequence given that the sum of the four terms equals 362, and the sum of the squares of the first and fourth terms exceeds the sum of the squares of the second and third terms by 900.

06:51

### Video Transcript

Find the four numbers which form an arithmetic sequence, given that the sum of the four numbers equals 362. And the sum of the squares of the first and fourth terms exceeds the sum of the squares of the second and third terms by 900.

We do know that we have an arithmetic sequence. But thereβs quite a lot about the sequence that we donβt currently know. In these cases, itβs sensible to form equations with what we do know. We know that the sum of the four terms is 362. And we have some further information about the squares of individual terms. Weβll begin by dealing with the sum of the four terms. The formula for the sum of the first π terms over an arithmetic sequence with first term π and common difference π is π over two multiplied by two π plus π minus one multiplied by π. Letβs substitute what we know about our sequence into this formula.

Weβre finding the sum of the four terms. So π is equal to four. And we know this gives us a value of 362. So 362 must be equal to four divided by two multiplied by two π plus four minus one multiplied by π. Four divided by two is two. And four minus one is three. So we can see that 362 is equal to two multiplied by two π plus three π. And in fact, weβll simplify this equation by dividing both sides by two. And we can see that 181 is equal to two π plus three π. We now have one equation in terms of π and π. We still donβt have enough information to solve this for π and π. So letβs look into the next piece of information.

We know that the sum of the squares of the first and fourth terms, thatβs π’ one squared plus π’ four squared, exceeds the sum of the squares of the second and third terms. Thatβs π’ two squared plus π’ three squared by 900. So we can say that π’ two squared plus π’ three squared plus 900 must be equal to π’ one squared plus π’ four squared. And this time weβll use the formula for the πth term of the arithmetic sequence. Itβs π plus π minus one multiplied by π. Letβs use this to form expressions for the four terms. π’ one is π plus one minus one π. And one minus one is zero. So π’ one is π. And in fact, this makes a lot of sense. We said that π is the first term in the sequence. So π’ one must be π. π’ two is π plus two minus one π, which is just π plus π. And we can repeat this for the next two terms. π’ three is π plus two π. And π’ four is π plus three π.

Remember, weβre interested in the square of each of these terms. So we can say that π’ one squared is equal to π squared. π’ two squared is π plus π squared and so on. Letβs distribute the brackets representing π’ two squared, π’ three squared, and π’ four squared. Before distributing, itβs much more sensible to write it as π plus π multiplied by π plus π. Similarly, with π’ three squared, we write it as π plus two π multiplied by π plus two π. And π’ four squared is π plus three π multiplied by π plus three π. And we can use a FOIL method or a grid method to expand these brackets. Letβs look at the FOIL method. F stands for first. We multiply the first term in each bracket to get π squared. O stands for outer. π multiplied by π is ππ. I is inner. And we get another ππ. And L is last. Thatβs π squared. We can simplify by collecting like terms. And we see that π plus π all squared is π squared plus two ππ plus π squared.

We can repeat this process. And we see that π’ three squared is π squared plus four ππ plus four π squared. And π’ four squared is π squared plus six ππ plus nine π squared. We can now form an equation by finding the sum of π’ one squared and π’ four squared and the sum of π’ two squared and π’ three squared. π’ one squared plus π’ four squared is two π squared plus six ππ plus nine π squared. And the sum of the squares of the second and third terms is two π squared plus six ππ plus five π squared. And of course, weβve got this 900. We should be able to see now that we can simplify this somewhat. We can subtract two π squared from both sides of the equation. And we can subtract six ππ. And that leaves us with nine π squared is equal to five π squared plus 900.

Our next step is to subtract five π squared from both sides. We get four π squared is equal to 900. We divide by four. Four π squared is equal to 225. And then we take the square root of both sides of the equation. And we get π is equal to plus or minus 15. Now, usually at this point, weβd look to make a decision as to whether π is 15. The common difference is 15 or negative 15. Actually though, weβre just being asked to find the four numbers which form the sequence. So we can either choose π is positive 15. And weβll get one starting number. Or π is negative 15. And weβll get an alternative starting number. Either way though, when we calculate the next three terms, they will be the same whether we choose π is equal to positive 15 or negative 15. So letβs choose π is equal to 15.

Weβre going to substitute it into the equation we formed earlier. Three multiplied by 15 is 45. So 181 is equal to two π plus 45. To solve this equation for π, we subtract 45 from both sides to get 136 is equal to two π. And we divide by two. π, the first term in our sequence, is 68. Once we know that we can find the second term by adding 15 to this number. Thatβs 83. We then add 15 again to get the third term. We see that the third term is 98. And we need to add one more 15 to get 113. And we see that the four numbers which form the arithmetic sequence we require are 68, 83, 98, and 113.

And letβs briefly check what wouldβve happened had we chosen a common difference of negative 15. This time, the equation wouldβve become 181 equals two π minus 45. We add 45 to both sides. And then we divide by two. And this time, we get π to be 113. Then, since the common difference is negative 15, we take 15 away. And we see that the next term is 98. We take it away again. And we get a next term of 83. And we take it away one more time. And we get 98. We get the exact same four numbers.