The sum of 8 terms of an aritmetic progression is 160. While the sum of 20 term is 880. Find 43 term and the sum of 12 term
Let's denote the first term of the arithmetic progression as 'a' and the common difference as 'd'.
The formula for the sum of the first n terms of an arithmetic progression is:
Sn = (n/2)(2a + (n-1)d)
Given that the sum of 8 terms is 160, we can form the equation:
160 = (8/2)(2a + (8-1)d)
160 = 4(2a + 7d)
40 = 2a + 7d .....(1)
Similarly, the sum of 20 terms is 880, which gives us the equation:
880 = (20/2)(2a + (20-1)d)
880 = 10(2a + 19d)
88 = 2a + 19d .....(2)
Now, we have two equations with two variables (a and d). We can solve this system of equations to find their values.
Multiplying equation (1) by 19 and equation (2) by 7, we get:
760 = 38a + 133d .....(3)
616 = 14a + 133d .....(4)
Subtracting equation (4) from equation (3), we have:
760 - 616 = (38a + 133d) - (14a + 133d)
144 = 24a
a = 6
Substituting the value of a into equation (1):
40 = 2(6) + 7d
40 = 12 + 7d
7d = 28
d = 4
Therefore, the common difference (d) is 4.
To find the 43rd term, we can use the formula for the nth term of an arithmetic progression:
an = a + (n-1)d
a43 = 6 + (43-1)4
a43 = 6 + 168
a43 = 174
Hence, the 43rd term is 174.
To find the sum of the 12 terms, we can use the formula for the sum of the first n terms of an arithmetic progression:
S12 = (12/2)(2a + (12-1)d)
S12 = 6(2(6) + (12-1)4)
S12 = 6(12 + 44)
S12 = 6(56)
S12 = 336
Therefore, the sum of the 12 terms is 336.