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
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.