The sum of 8th term of an A. P is 160 while to sum of the 20th term is 880. Find the 43rd term

Let a and d be the first term and common difference, respectively. Then

the 8th term is a+7d
the 20th term is a+19d

The sum of the first 8 terms is
(a)+(a+d)+(a+2d)+...+(a+7d) = 8a+28d

The sum of the first 20 terms is
(a)+(a+d)+(a+2d)+...+(a+19d) = 20a+190d

So

8a + 28d = 160
20a+ 190d = 880

40a + 140d = 800
40a + 380d = 1760
240d = 960
d = 4

8a + 112 = 160
8a=48
a =
6

The first term is 6 and the common difference is 4.

The 43rd term is a+42d = 6+42(4) = 6+168 = 174

The sum of the first 12 terms is
(a)+(a+d)+(a+2d)+...+(a+11d) = 12a+66d = 12(6)+66(4) = 72+264 = 336

so the perpendicular line goes through the given pointyou would get a perpendicular line at some random location

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.