Let's assume that the first term of the arithmetic progression is "a" and the common difference is "d".
We know that the sum of the 8 terms is given by the formula:
S8 = (n/2)(2a + (n-1)d), where n is the number of terms.
Plugging in the given values, we have:
160 = (8/2)(2a + (8-1)d)
160 = 4(2a + 7d)
40 = 2a + 7d -- Equation (1)
Similarly, the sum of the 20 terms is given by:
S20 = (n/2)(2a + (n-1)d)
Plugging in the given values, we have:
880 = (20/2)(2a + (20-1)d)
880 = 10(2a + 19d)
88 = 2a + 19d -- Equation (2)
To find the 43rd term (a):
We can assume the 43rd term is T43.
Using the formula for the nth term of an arithmetic progression:
Tn = a + (n-1)d, we have:
T43 = a + (43-1)d
T43 = a + 42d
To find the sum of the 12 terms:
Using the formula for the sum of an arithmetic progression:
S12 = (n/2)(2a + (n-1)d)
Plugging in the given values, we have:
S12 = (12/2)(2a + (12-1)d)
S12 = 6(2a + 11d)
Now, we can solve the equations:
From Equation (1): 40 = 2a + 7d
From Equation (2): 88 = 2a + 19d
Multiply Equation (1) by 2: 80 = 4a + 14d
Subtract Equation (2) from the above equation:
80 - 88 = (4a + 14d) - (2a + 19d)
-8 = 2a - 5d
-8/2 = a - (5/2)d
-4 = a - (5/2)d -- Equation (3)
Now, substitute Equation (3) into the expression T43 = a + 42d:
T43 = (-4) + 42d
T43 = -4 + 42d
So, the 43rd term is -4 + 42d, where d is the common difference.
To find the sum of the 12 terms, substitute Equation (3) into the expression S12 = 6(2a + 11d):
S12 = 6(2(-4) + 11d)
S12 = 6(-8 + 11d)
S12 = -48 + 66d
Therefore, the 43rd term is -4 + 42d and the sum of the 12 terms is -48 + 66d.
The sum of 8term of an a.p 160while the sum of 20term is 880 find.
a. the 43 term
b.the sum of 12 term
1 answer