In an AP the 6th term is half the 4th term and the 3rd term is 15.
a. Find the fist term and the common difference.
b. How many times are needed to give a sum that is less than 65?
"the 6th term is half the 4th term" ---> a+5d = (1/2)(a+3d)
2a + 10d = a + 3d
a + 7d = 0
"the 3rd term is 15. " ---> a + 2d = 15
subtract those two equations ...
5d = -15
d = -3
sub into
a+ 2d = 15 ---> a - 6 = 15
a = 21
a = 21 , b = -3
let the number of terms be n
(n/2)[42 - (n-1)(-3) < 65
n [ 42 + 3n - 3 ] < 130
3n^2 + 39n -130 < 0
Consider 3n^2 + 39n - 130 = 0
n = 2.75 or a negative
but n must be a whole number
let n=3
21 + 18 + 15 < 65
3 terms are needed
If 3rd tarm of an AP is four times the first team and 6th term is 17 find the AP
Arithmetic progression :
an = a1 + ( n - 1 ) * d
In this case :
a6 = a1 + ( 6 - 1 ) * d
a6 = a1 + 5 d
a4 = a1 + ( 4 - 1 ) *d
a4 = a1 + 3 d
a6 = ( 1 / 2 ) a4
a6 = ( 1 / 2 ) ( a1 + 3 d )
a6 = a1 / 2 + 3 d / 2
a6 = a6
a1 + 5 d = a1 / 2 + 3 d / 2 Substract 5 d to both sides
a1 + 5 d - 5 d = a1 / 2 + 3 d / 2 - 5 d
a1 = a1 / 2 + 3 d / 2 - 10 d / 2
a1 = a1 / 2 - 7 d / 2 Substract a1 / 2 to both sides
a1 - a1 / 2 = a1 / 2 - 7 d / 2 - a1 / 2
a1 / 2 = - 7 d / 2 Multiply both sides by 2
a1 = - 7 d
a3 = a1 + ( 3 - 1 ) * d
a3 = a1 + 2 d
a3 = 15
15 = a1 + 2 d
15 = - 7 d + 2 d
15 = - 5 d Divide both sides by - 5
15 / - 5 = d
- 3 = d
d = - 3
a1 = - 7 d
a1 = - 7 * ( - 3 ) = 21
a1 = 21
a2 = a1 + ( 2 - 1 ) * d = a1 + d = 21 + ( - 3 ) = 21 - 3 = 18
a3 = a1 + ( 3 - 1 ) * d = a1 + 2 d = 21 + 2 * ( - 3 ) = 21 + ( - 6 ) = 21 - 6 = 15
a4 = a1 + ( 4 - 1 ) * d = a1 + 3 d = 21 + 3 * ( - 3 ) = 21 + ( - 9 ) = 21 - 9 = 12
a5 = a1 + ( 5 - 1 ) * d = a1 + 4 d = 21 + 4 * ( - 3 ) = 21 + ( - 12 ) = 21 - 12 = 9
a6 = a1 + ( 6 - 1 ) * d = a1 + 5 d = 21 + 5 * ( - 3 ) = 21 + ( - 15 ) = 21 - 15 = 6
AP:
21 , 18 , 15 , 12 , 9 , 6
b.
The sum of the n members of a arithmetic progression :
Sn = ( n / 2 ) ( a1 + an )
In this case :
65 < ( n / 2 ) ( a1 + an )
65 < ( n / 2 ) * ( 21 + an ) Multiply both sides by 2
2 * 65 < 2 * ( n / 2 ) * ( 21 + an )
135 < n * ( 21 + an )
Sn = ( n / 2 ) ( a1 + an )
For n = 6
S6 = ( 6 / 2 ) ( 21 + 6 )
S6 = 3 * 27 = 81
81 > 65 That is not solution
For n = 5
S5 = ( 5 / 2 ) ( 21 + 9 )
S5 = ( 5 / 2 ) * 30 = 150 / 2 = 75
75 > 65 That is not solution
For n = 4
S4 = ( 4 / 2 ) ( 21 + 12 )
S4 = 2 * 33 = 66
66 > 65 That is not solution.
For n = 3
S3 = ( 3 / 2 ) ( 21 + 15 )
S3 = ( 3 / 2 ) * 36 = 108 / 2 = 54
54 < 65 That is solution.
a1 + a2 + a3 = 21 + 18 + 15 = 54
nth term of 9,12,15,18
To solve this problem, we can use the formula for the nth term of an arithmetic progression (AP), which is given by:
\[a_n = a_1 + (n-1)d\]
where:
- \(a_n\) represents the nth term
- \(a_1\) represents the first term
- \(d\) represents the common difference
- \(n\) represents the position of the term in the sequence
a. First, let's find the first term and the common difference.
We are given the information that the 6th term is half the 4th term. We can write this as:
\[a_6 = \frac{1}{2}a_4\]
Substituting the formula for the nth term, we have:
\[a_1 + 5d = \frac{1}{2}(a_1 + 3d)\]
Now we can solve for \(a_1\) in terms of \(d\):
\[2(a_1 + 5d) = a_1 + 3d\]
\[2a_1 + 10d = a_1 + 3d\]
\[a_1 = -7d\]
We are also given that the 3rd term (\(a_3\)) is equal to 15:
\[a_1 + 2d = 15\]
Now we can substitute the expression we found for \(a_1\) into this equation:
\[-7d + 2d = 15\]
\[-5d = 15\]
\[d = -3\]
Now we can find the first term (\(a_1\)) using the equation \(a_1 = -7d\):
\[a_1 = -7(-3) = 21\]
So, the first term is 21 and the common difference is -3.
b. To find how many terms are needed to give a sum less than 65, we can use the formula for the sum of an arithmetic progression:
\[S_n = \frac{n}{2}(a_1 + a_n)\]
We want to find the minimum number of terms, \(n\), such that \(S_n < 65\). We already know the values of \(a_1\) and \(d\) from part a.
We can solve this inequality to find the value of \(n\):
\[\frac{n}{2}(a_1 + a_n) < 65\]
\[\frac{n}{2}(21 + (21 + (n-1)(-3))) < 65\]
Simplifying further:
\[\frac{n}{2}(21 + 21 - 3n + 3) < 65\]
\[\frac{n}{2}(45 - 3n) < 65\]
Now we can solve this inequality.