In an arithmatic sequence, sum of 9 terms is 279 and sum of 20 terms is 1280.
Find 5th term ?
Find 16th term ?
Write the sequence.
S9 = 279
F5 = 279/9 = 31
S20 = 1280
S20-S9 = 1280-279 =1001
=[sum of F11 to F20]
F15 =1001/11 = 91.
F15-F5 = 91-31 = 60 = 10 d
d = 60/10 = 6
F16 = 91 + 6 =97
F1 =31-4d =31-24= 7
so, 7,13,19,25,..........
To find the 5th and 16th terms in an arithmetic sequence, we need to first find the common difference (d) and the first term (a1).
Step 1: Use the formula for the sum of an arithmetic sequence, which is given by Sn = (n/2)(2a1 + (n-1)d), where Sn is the sum of n terms.
Given that the sum of 9 terms is 279 and the sum of 20 terms is 1280, we can set up the following equations:
For 9 terms: 279 = (9/2)(2a1 + 8d)
For 20 terms: 1280 = (20/2)(2a1 + 19d)
Step 2: Simplify the equations and solve for a1 and d.
For the first equation, dividing both sides by 9 gives: 31 = 2a1 + 8d (equation 1)
For the second equation, dividing both sides by 20 gives: 64 = 2a1 + 19d (equation 2)
Step 3: Solve the system of equations by elimination or substitution to find the values of a1 and d.
Equation 2 minus Equation 1 gives: 64 - 31 = (2a1 + 19d) - (2a1 + 8d)
Simplifying, we get: 33 = 11d
Dividing both sides by 11, we find: d = 3
Substituting the value of d into Equation 1, we have: 31 = 2a1 + 8(3)
31 = 2a1 + 24
2a1 = 31 - 24
2a1 = 7
Dividing both sides by 2, we find: a1 = 3.5
Step 4: Use the values of a1 and d to find the 5th and 16th terms.
To find the terms in the sequence, we can use the formula for the nth term of an arithmetic sequence, given by an = a1 + (n-1)d.
For the 5th term (n = 5): a5 = 3.5 + (5-1)3
a5 = 3.5 + 4*3
a5 = 3.5 + 12
a5 = 15.5
For the 16th term (n = 16): a16 = 3.5 + (16-1)3
a16 = 3.5 + 15*3
a16 = 3.5 + 45
a16 = 48.5
Step 5: Write out the arithmetic sequence.
The arithmetic sequence can be written as: 3.5, 6.5, 9.5, 12.5, 15.5, ...
So, the 5th term is 15.5 and the 16th term is 48.5.
Sum of a certain number of terms of an arithmetic sequence:
Sn = n ( a1 + an ) / 2
In this case:
S9 = 9 ( a1 + a9 ) / 2 = 279
S20 = 20 ( a1 + a20 ) / 2 = 1280
9 ( a1 + a9 ) / 2 = 279 Multiply both sides by 2
9 ( a1 + a9 ) = 279 * 2
9 a1 + 9 a9 = 558
On the other side:
an = a1 ( n - 1 ) d
in this case: n = 9, n - 1 = 8
a9 = a1 + 8 d
9 a1 + 9 a9 = 558
9 a1 + 9 ( a1 + 8 d ) = 558
9 a1 + 9 a1 + 9 * 8 d = 558
9 a1 + 9 a1 + 72 d = 558
18 a1 + 72 d = 558
20 ( a1 + a20 ) / 2 = 1280 Multiply both sides by 2
20 ( a1 + a20 ) = 1280 * 2
20 a1 + 20 a20 = 2560
an = a1 ( n - 1 ) d
In this case: n = 20, n - 1 = 19
a20 = 20 a1 + 19 d
20 a1 + 20 a20 = 2560
20 a1 + 20 ( a1 + 19 d ) = 2560
20 a1 + 20 a1 + 20 * 19 d = 2560
20 a1 + 20 a1 + 380 d = 2560
40 a1 + 380 d = 2560
Now you must solve system of 2 equations with 2 unknows:
18 a1 + 72 d = 558
40 a1 + 380 d = 2560
The solutions are :
a1 = 7
d = 6
an = a1 + ( n - 1 ) d
a5 = 7 + ( 5 - 1 ) * 6
a5 = 7 + 4 * 6
a5 = 7 + 24
a5 = 31
an = a1 + ( n - 1 ) d
a16 = 7 + ( 16 - 1 ) * 6
a16 = 7 + 15 * 6
a16 = 7 + 90
a16 = 97
Your sequence:
7, 13, 19 , 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121