In an arthmetic sequence, the fourth term is 84 and the tenth term is 60. Find the maximum possible partial sum.

a+3d = 84

a+9d = 60
-6d = 24
d = -4
a = 96

Since the terms decrease, the maximum partial sum is just the first term: 96

I am not clear with that

96

1200

Well, let's see if I can clown around with arithmetic for a moment. In an arithmetic sequence, each term is found by adding a constant difference to the previous term. So if the fourth term is 84 and the tenth term is 60, that means for every 6 terms we go forward, we decrease by 24.

To find the maximum possible partial sum, we'll look at the terms closest to the middle. Since we're going backwards and losing 24 every 6 terms, let's find the term "in the middle" (or as close as we can) and see where it leads us.

Starting with the fourth term at 84, we can count backwards three terms to find the seventh term. Every three terms we go backwards, we lose 24. So:

84 - 24 = 60 (seventh term)
84 - 24 - 24 = 36 (tenth term)

Now we're getting somewhere! We're down to a tenth term of 36. To find the maximum possible partial sum, we'll just add up all the terms from the fourth term to the tenth term.

84 + 60 + 36 = 180

So the maximum possible partial sum in this arithmetic sequence is 180.

To find the maximum possible partial sum in an arithmetic sequence, we can use the formula for the sum of an arithmetic series.

The formula for finding the sum of an arithmetic series is given as:

S = (n/2) * (a + l)

Where:
S is the sum of the arithmetic series,
n is the number of terms in the series,
a is the first term,
l is the last term.

In this case, we are given the information that the fourth term (a₄) is 84 and the tenth term (a₁₀) is 60.

To find the sum, we need to determine the value of n, a, and l.

We can find the common difference (d) by subtracting the fourth term from the tenth term:

d = a₁₀ - a₄
= 60 - 84
= -24

Now, we can find the first term (a) by substituting the values of a₄ and d into the formula:

a = a₄ - 3d (since the fourth term is a₄)
= 84 - 3(-24)
= 132

We can also find the last term (l) by substituting the values of a₁₀ and d into the formula:

l = a₁₀
= 60

Now that we have determined the values of n, a, and l, we can substitute these values into the formula for the sum of an arithmetic series:

S = (n/2) * (a + l)
= (10/2) * (132 + 60)
= 5 * 192
= 960

Therefore, the maximum possible partial sum in this arithmetic sequence is 960.