An Ap has 15 terms and a common difference of -3.find it's first and last term of its sum of -120

Answer 1

We know that the formula for the sum of an arithmetic series is given by:

S = (n/2)(2a + (n-1)d)

Where S is the sum of the series, a is the first term, d is the common difference, and n is the number of terms.

In this case, S = -120, d = -3, and n = 15. We need to find the value of a.

-120 = (15/2)(2a + (15-1)(-3))
-120 = (15/2)(2a + 14(-3))
-120 = (15/2)(2a - 42)

Now, let's simplify the equation:

-120 = (15/2)(2a - 42)
-240 = 15(2a - 42)
-240 = 30a - 630
30a = -240 + 630
30a = 390
a = 390/30
a = 13

So, the first term of the arithmetic series is 13.

To find the last term, we can use the formula for the nth term of an arithmetic series:

an = a + (n-1)d

In this case, an is the last term, a is the first term (13), and d is the common difference (-3).

an = 13 + (15-1)(-3)
an = 13 + 14(-3)
an = 13 - 42
an = -29

Therefore, the last term of the arithmetic series is -29.