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Find the geometric sequence and the sum of the first six terms given the second term is four times the fourth one, the sum of the fourth and seventh terms is two, and all terms are positive.

Letβs begin by recalling what we understand by a geometric sequence. A geometric sequence is found by multiplying successive terms by a single constant called the common ratio. And the πth term of a geometric sequence is given by π sub π equals π sub one times π to the power of π minus one, where π sub one is the value of the first term and π is that common ratio.

Now weβre told that the second term in our geometric sequence is four times the size of the fourth one. Well, the second term is π sub two. And using the πth term rule, we get π sub one times π to the power of two minus one or simply π sub one π. The fourth term is π sub one times π to the power of four minus one, which we can say is π sub one times π cubed.

Since we know the second term is four times the size of the fourth one, we can say that π two is equal to four π four. But then we can replace π two and π four with π one π and π one π cubed. So we find that π one π equals four π one π cubed.

Now in fact, we can divide through by π sub one. And we find that π is equal to four π cubed. Next, weβre going to divide through by four π. On the left-hand side, that leaves us with one-quarter. And on the right-hand side, four divided by four is one and π cubed divided by π is π squared.

Finally, we square-root both sides. Now the square root of one-quarter is one-half. Usually though, when weβre square-rooting, we consider both the positive and negative square root. In this question though, weβre told that all terms are positive. And thereβs no way to multiply a positive by a negative number and get another positive. So the common ratio itself must be positive.

Now that we know the value of the common ratio, letβs move on to the second bit of information in this question. Weβre told that the sum of the fourth and seventh terms is two. That is, π sub four plus π sub seven equals two. Now we know that π sub four is π one π cubed. And then π sub seven would be π one times π to the power of seven minus one or π to the sixth power.

But of course, we now know that π is equal to one-half. So letβs replace π with one-half in this equation. When we do, we get π one times a half cubed plus π one times a half to the sixth power equals two. Well, one-half cubed is one-eighth and one-half to the sixth power is one over 64. And then weβre going to add the fractions on the left-hand side by creating a common denominator.

Now the common denominator here is 64. And we achieve that by multiplying both the numerator and denominator of our first fraction by eight. So we get eight π one over 64 plus one π one over 64 equals two. But then we can add the numerator. So we get nine π one over 64 equals two.

Weβre going to divide both sides of this equation by nine over 64, remembering that when we divide by a fraction, we multiply by its reciprocal. On the left-hand side, that leaves us with π one. And on the right, we get two times 64 over nine. And if we think about two as two over one, we get 128 over nine as π one. And thatβs the first term in our sequence. So the first term is 128 over nine.

We know that the second term is found by multiplying 128 over nine by the common ratio. Thatβs one-half. And so thatβs 64 over nine.

The third term is found by multiplying this by one-half again or by multiplying 128 over nine by one-half squared. And that gives us 32 over nine.

So weβve worked out the first few terms of our sequence. The next thing that we need to do is work out the sum of the first six terms. Letβs clear some space.

Now whilst we could list out the first six terms and then add them, there is a Formula we can use. The sum of the first π terms of a geometric sequence is π times one minus π to the πth power over one minus π. Now π and π sub one are interchangeable. Iβve written π because this is generally how this formula is represented. In this case, we want the sum of the first six terms. So thatβs π sub six. We know the first term π is 128 over nine. So we multiply this by one minus the common ratio β thatβs one half β to the sixth power over one minus one-half. That gives us 128 over nine times 63 over 64 divided by one-half.

But we can simplify. First of all, we cross-cancel by dividing through by nine. And then we know that 128 and 64 can both be divided by 64. The numerator then becomes two times seven, which is 14. And weβre dividing that by one-half. And of course, dividing by one-half is the same as multiplying by two. So we find that the sum of the first six terms is simply 28.

And so the geometric sequence is 128 over nine, 64 over nine, 32 over nine, and so on. And the sum of the first six terms is 28.