The 14th term of an AP 96 while the 25th term is 173. Find
a) 19th term
b) sum of 13th and 56th term
c) product of 6th and 13th term
Given that the 14th term of the AP is 96 and the 25th term is 173.
We can use the formula for the nth term of an AP:
An = A1 + (n - 1)d
Where An is the nth term, A1 is the first term, n is the term number, and d is the common difference.
a) To find the 19th term:
A19 = A1 + (19 - 1)d
96 = A1 + 18d
b) To find the sum of the 13th and 56th term:
A13 = A1 + (13 - 1)d
A56 = A1 + (56 - 1)d
Sum = A13 + A56
c) To find the product of the 6th and 13th term:
A6 = A1 + (6 - 1)d
A13 = A1 + (13 - 1)d
Product = A6 * A13
Now we have a system of equations to solve:
A1 + 18d = 96 (equation 1)
A1 + 12d + A1 + 55d = sum (equation 2)
A1 + 5d * A1 + 12d = product (equation 3)
From equation 1, we can solve for A1 in terms of d:
A1 = 96 - 18d
Substituting this value into equation 2 and 3:
(96 - 18d) + 12d + (96 - 18d) + 55d = sum
(96 - 18d) + 5d * (96 - 18d) + 12d = product
Simplifying:
192 - 6d + 48d = sum
288 - 7d * 192 - 11d = product
192 + 42d = sum (equation 4)
294 - 18d = product (equation 5)
Now we can solve for d by subtracting equation 5 from equation 4:
(192 + 42d) - (294 - 18d) = 0
192 + 42d - 294 + 18d = 0
60d - 102 = 0
60d = 102
d = 102/60
d = 1.7
Now we can substitute d back into equation 1 to solve for A1:
A1 + 18(1.7) = 96
A1 + 30.6 = 96
A1 = 96 - 30.6
A1 = 65.4
Now we can find the values of the different terms:
a) To find the 19th term:
A19 = A1 + (19 - 1)d
A19 = 65.4 + (19 - 1)(1.7)
A19 = 65.4 + 18(1.7)
A19 = 65.4 + 30.6
A19 = 96
b) To find the sum of the 13th and 56th term:
A13 = A1 + (13 - 1)d
A13 = 65.4 + (13 - 1)(1.7)
A13 = 65.4 + 12(1.7)
A13 = 65.4 + 20.4
A13 = 85.8
A56 = A1 + (56 - 1)d
A56 = 65.4 + (56 - 1)(1.7)
A56 = 65.4 + 55(1.7)
A56 = 65.4 + 93.5
A56 = 158.9
Sum = A13 + A56
Sum = 85.8 + 158.9
Sum = 244.7
c) To find the product of the 6th and 13th term:
A6 = A1 + (6 - 1)d
A6 = 65.4 + (6 - 1)(1.7)
A6 = 65.4 + 5(1.7)
A6 = 65.4 + 8.5
A6 = 73.9
A13 = A1 + (13 - 1)d
A13 = 65.4 + (13 - 1)(1.7)
A13 = 65.4 + 12(1.7)
A13 = 65.4 + 20.4
A13 = 85.8
Product = A6 * A13
Product = 73.9 * 85.8
Product = 6334.62
Therefore,
a) The 19th term is 96.
b) The sum of the 13th and 56th term is 244.7.
c) The product of the 6th and 13th term is 6334.62.
To find the solution, we can use the formula for the nth term of an arithmetic progression (AP):
a_n = a + (n-1)d
where:
a_n is the nth term of the AP
a is the first term of the AP
d is the common difference between consecutive terms
Let's solve the given problem step by step:
Step 1: Find the common difference (d) using the given information:
a_14 = a + (14-1)d = 96
a_14 = a + 13d = 96 ----------(eq.1)
a_25 = a + (25-1)d = 173
a_25 = a + 24d = 173 ----------(eq.2)
Step 2: Solve the system of equations (eq.1 and eq.2) to find the values of a and d:
Subtract eq.1 from eq.2 to eliminate a:
(a + 24d) - (a + 13d) = 173 - 96
11d = 77
d = 7
Substitute the value of d into either eq.1 or eq.2 to find a:
a + 13(7) = 96
a + 91 = 96
a = 96 - 91
a = 5
Therefore, the first term (a) is 5 and the common difference (d) is 7.
Now we can answer each question:
a) To find the 19th term, we can substitute the values of a and d into the formula:
a_19 = a + (19-1)d
a_19 = 5 + (19-1)7
a_19 = 5 + 18*7
a_19 = 5 + 126
a_19 = 131
Therefore, the 19th term of the AP is 131.
b) To find the sum of the 13th and 56th terms, we can use the formula:
S_n = (n/2) * (2a + (n-1)d)
Sum of terms = S_13 + S_56
Sum of terms = (13/2) * (2a + (13-1)d) + (56/2) * (2a + (56-1)d)
Sum of terms = 13/2 * (2*5 + 12*7) + 56/2 * (2*5 + 55*7)
Sum of terms = 13/2 * (10 + 84) + 56/2 * (10 + 385)
Sum of terms = 13/2 * 94 + 28 * 395
Sum of terms = 611 + 11060
Sum of terms = 11671
Therefore, the sum of the 13th and 56th terms is 11671.
c) To find the product of the 6th and 13th terms, we can substitute the values of a and d into the formula:
Product of terms = a_6 * a_13
Product of terms = (a + (6-1)d) * (a + (13-1)d)
Product of terms = (5 + 5*7) * (5 + 12*7)
Product of terms = 10 * 89
Product of terms = 890
Therefore, the product of the 6th and 13th terms is 890.