The sum of first 8th terms of an A.P is 160 while the sum of 20 term is 880.Find a. The 43rd term b.The sum of 12 term
8/2 (2a+7d) = 160
20/2 (2a+19d = 880)
solve for a and d, and then you want
a+42d
S12 = 12/2 (2a+11d)
The 9th term of an A.P is 52 while the 12th term is 20find the of its first 20 term
To find the value of 'a' and the 43rd term, we first need to find the common difference 'd' of the arithmetic progression (A.P).
We know that the sum of the first 'n' terms of an A.P is given by the formula:
Sn = (n/2) * (2a + (n-1)d)
where:
Sn is the sum of the first 'n' terms,
a is the first term, and
d is the common difference.
Given that the sum of the first 8 terms is 160, we can set up the equation:
160 = (8/2) * (2a + (8-1)d)
160 = 4 * (2a + 7d)
40 = 2a + 7d (Equation 1)
Similarly, we can set up another equation using the sum of the 20 terms:
880 = (20/2) * (2a + (20-1)d)
880 = 10 * (2a + 19d)
88 = 2a + 19d (Equation 2)
Now, we can solve the system of equations 1 and 2 to find the values of 'a' and 'd'.
From Equation 1, we can express 'a' in terms of 'd':
-7d = 2a - 40
2a = 7d + 40
a = (7d + 40)/2
Substituting this value of 'a' into Equation 2:
88 = 2[(7d + 40)/2] + 19d
88 = 7d + 40 + 19d
88 - 40 = 26d
48 = 26d
d = 48/26
d = 24/13
Now, substitute the value of 'd' back into the expression for 'a':
a = (7*(24/13) + 40)/2
a = (168/13 + 40)/2
a = (168/13 + (40 * 13) / 13) / 2
a = (168/13 + 520/13) / 2
a = 688/26
a = 344/13
So, the first term 'a' of the A.P is 344/13.
To find the 43rd term, we can use the formula:
an = a + (n - 1)d
Substituting the values of 'a' and 'd':
a43 = (344/13) + (43 - 1) * (24/13)
a43 = 344/13 + 42 * 24/13
a43 = 344/13 + 1008/13
a43 = (344 + 1008)/13
a43 = 1352/13
Therefore, the 43rd term of the A.P is 1352/13.
To find the sum of the first 12 terms, we can again use the formula for the sum of the first 'n' terms:
S12 = (12/2) * (2a + (12-1)d)
S12 = 6 * (2(344/13) + (12-1)(24/13))
S12 = 6 * (688/13 + 11(24/13))
S12 = 6 * (688/13 + 264/13)
S12 = 6 * (952/13)
S12 = (6 * 952)/13
S12 = 5712/13
Therefore, the sum of the first 12 terms of the A.P is 5712/13.
To find the values of a) the 43rd term and b) the sum of the 12 terms of the arithmetic progression (A.P.), we can use the formulas for the sum of the first n terms and the nth term of an A.P.
Given information:
Sum of the first 8 terms: S8 = 160
Sum of the 20 terms: S20 = 880
Let's start by using the formula for the sum of the first n terms of an A.P:
S = n/2 * (2a + (n - 1)d)
where S is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.
Using this equation for S8:
160 = 8/2 * (2a + (8 - 1)d)
160 = 4 * (2a + 7d)
40 = 2a + 7d ......(Equation 1)
Using this equation for S20:
880 = 20/2 * (2a + (20 - 1)d)
880 = 10 * (2a + 19d)
88 = 2a + 19d ......(Equation 2)
Now, we have two equations (Equations 1 and 2) with two variables (a and d). We can solve these equations simultaneously to find the values of a and d.
Multiplying Equation 1 by 2:
80 = 4a + 14d ......(Equation 3)
Now, subtract Equation 3 from Equation 2:
88 - 80 = (2a + 19d) - (4a + 14d)
8 = -2a + 5d
Simplifying, we get:
2a - 5d = -8 ......(Equation 4)
We have two equations with two unknowns (Equations 3 and 4). We can solve them using any suitable method, such as substitution or elimination.
Let's solve these equations using substitution:
From Equation 3:
4a + 14d = 80 ......(Equation 3)
Solving Equation 3 for a:
4a = 80 - 14d
a = (80 - 14d) / 4
a = 20 - (7/2)d
......(Equation 5)
Substituting the value of a from Equation 5 into Equation 4:
2(20 - (7/2)d) - 5d = -8
40 - 7d - 5d = -8
-12d = -48
d = 4
Substituting the value of d = 4 into Equation 5:
a = 20 - (7/2) * 4
a = 20 - 14
a = 6
Therefore, the first term (a) is 6 and the common difference (d) is 4.
a) To find the 43rd term:
Using the formula for the nth term of an A.P:
An = a + (n - 1)d
Substituting the values we found:
A43 = 6 + (43 - 1) * 4
A43 = 6 + 168
A43 = 174
Therefore, the 43rd term of the A.P. is 174.
b) To find the sum of the first 12 terms (S12):
Using the formula for the sum of the first n terms of an A.P:
S12 = 12/2 * (2a + (12 - 1)d)
Substituting the values we found:
S12 = 12/2 * (2 * 6 + (12 - 1) * 4)
S12 = 6 * (12 + 11 * 4)
S12 = 6 * (12 + 44)
S12 = 6 * 56
S12 = 336
Therefore, the sum of the first 12 terms of the A.P. is 336.