In an arithmetic series, the 15th term is 43 and the sum of the first 15 terms is 120. Add the first 20 terms of the series.

To solve this problem, first we need to find the common difference of the arithmetic series by using the 15th term.

To find the common difference (d), we can use the formula:

nth term = a + (n - 1)d

Given: 15th term = 43, n = 15

43 = a + (15 - 1)d
43 = a + 14d ...........(1)

Next, we can use the formula for the sum of an arithmetic series:

Sum of n terms = (n/2)(a + l)

Given: Sum of first 15 terms = 120, n = 15

120 = (15/2)(a + a + 14d)
120 = 7.5(2a + 14d)
16 = 2a + 14d ...........(2)

Now we have a system of equations (equation 1 and equation 2) that we can solve simultaneously.

To solve, we can multiply equation 1 by 7 and equation 2 by 2 to eliminate the variable 'd':

7 * (43 = a + 14d) -> 301 = 7a + 98d
2 * (16 = 2a + 14d) -> 32 = 4a + 28d

Next, we can subtract the two equations to eliminate 'a':

(301 - 32) = (7a - 4a) + (98d - 28d)
269 = 3a + 70d

To isolate 'a', we can divide the entire equation by 3:

269/3 = (3a + 70d)/3
89 = a + 70/3d ...........(3)

Now that we have the values for 'a' and 'd', we can find the 20th term (using n = 20) and then find the sum of the first 20 terms.

20th term = a + (n - 1)d
20th term = 89 + (20 - 1)d
20th term = 89 + 19d

Sum of first 20 terms = (20/2)(a + 20th term)
Sum of first 20 terms = 10(a + 89 + 19d)
Sum of first 20 terms = 10(89 + 19d + a)

Now we substitute the value of 'a' from equation 3:

Sum of first 20 terms = 10(89 + 19d + (a + 70/3d))
Sum of first 20 terms = 10(89 + 19d + 89 + 70d/3)
Sum of first 20 terms = 10(178 + 22d + 70d/3)
Sum of first 20 terms = 10(178 + 22d + (210d/3))

Simplifying further:

Sum of first 20 terms = 10(178 + (22d + 70d)/3)
Sum of first 20 terms = 10[(178 + 92d)/3]
Sum of first 20 terms = (10/3)(178 + 92d)

Therefore, the sum of the first 20 terms of the arithmetic series is (10/3)(178 + 92d).

To find the sum of an arithmetic series, we can use the formula:

Sn = n/2 * (a1 + an)

Where Sn is the sum of the series, n is the number of terms, a1 is the first term, and an is the nth term.

Given that the 15th term is 43 and the sum of the first 15 terms is 120, we can determine the value of a1 and n.

We know that the 15th term (an) is 43, so we can substitute that into the formula:

120 = 15/2 * (a1 + 43)

Simplifying, we have:

120 = 7.5 * (a1 + 43)

Dividing both sides by 7.5:

16 = a1 + 43

Subtracting 43 from both sides:

-27 = a1

Now that we know a1, we can find the sum of the first 20 terms (S20) by substituting n = 20 and a1 = -27 into the formula:

S20 = 20/2 * (-27 + a20)

To find a20, we can use the formula for the nth term of an arithmetic series:

an = a1 + (n-1) * d

Where d is the common difference.

Since this is an arithmetic series, the difference between consecutive terms is constant. We can find d by subtracting a1 from a2:

d = a2 - a1

To find a2, we can use the formula:

a2 = a1 + d

Substituting a1 = -27:

a2 = -27 + d

Now we can substitute a2 = -27 + d and a1 = -27 into the formula for an:

43 = -27 + (15-1) * d

Solving for d:

43 = -27 + 14d

70 = 14d

d = 5

Now we know the value of d, so we can find a20:

a20 = a1 + (20-1) * d

a20 = -27 + 19 * 5

a20 = -27 + 95

a20 = 68

Finally, we can substitute n = 20, a1 = -27, and a20 = 68 into the formula for S20:

S20 = 20/2 * (-27 + 68)

Simplifying, we have:

S20 = 10 * 41

S20 = 410

Therefore, the sum of the first 20 terms of the series is 410.

a+14d = 43

15/2 (2a+14d) = 120
(a,d) = (-27,5)

S20 = 20/2 (2a+19d) = 410