The 4th term and the 19th term of an ap are -3 and 12 respectively. Find the common difference, 5th term and the number of terms which will give a sum of 135
Step 1: Find the common difference (d).
We are given that the 4th term (a4) is -3 and the 19th term (a19) is 12.
a4 = a1 + (4-1)d
-3 = a1 + 3d
a19 = a1 + (19-1)d
12 = a1 + 18d
Solve these two equations to find the value of d.
-3 = a1 + 3d ...(1)
12 = a1 + 18d ...(2)
Subtract equation (1) from equation (2):
12 - (-3) = a1 + 18d - (a1 + 3d)
15 = 18d - 3d
15 = 15d
d = 15/15
d = 1
Therefore, the common difference (d) is 1.
Step 2: Find the 5th term (a5).
Using the formula for the nth term of an arithmetic sequence:
an = a1 + (n-1)d
Substitute the values:
a5 = a1 + (5-1)d
a5 = a1 + 4d
Since we know that d = 1, substitute that value:
a5 = a1 + 4(1)
a5 = a1 + 4
a5 = a1 + 4
So, the 5th term (a5) is equal to a1 + 4.
Step 3: Determine the number of terms for a sum of 135.
The formula for the sum of the first n terms of an arithmetic sequence is:
Sn = (n/2)(2a1 + (n-1)d)
We need to find the value of n when Sn = 135.
135 = (n/2)(2a1 + (n-1)d)
Substitute the values we know:
135 = (n/2)(2a1 + (n-1)(1))
135 = (n/2)(2a1 + n - 1)
Simplify the equation:
135 = a1n + (n^2 - n)/2
270 = 2a1n + n^2 - n
Rearrange the equation to form a quadratic equation:
n^2 + (2a1 - 1)n - 270 = 0
Solve the quadratic equation to find the value of n.
Once we find the value of n, we will substitute it into the equation of the nth term to find the 5th term (a5).
To find the common difference (d) of an arithmetic progression (AP), we can use the formula:
\(d = \frac{{a_{n+1} - a_n}}{{n+1 - n}}\),
where \(a_n\) is the nth term of the AP.
Given that the 4th term (a₄) is -3 and the 19th term (a₁₉) is 12, we can use these values to find the common difference.
Substituting the values into the formula:
\(d = \frac{{a_{19} - a_4}}{{19 - 4}}\),
\(d = \frac{{12 - (-3)}}{{19 - 4}}\),
\(d = \frac{{15}}{{15}}\),
\(d = 1\).
So, the common difference (d) is 1.
To find the 5th term (a₅), we can use the formula:
\(a_n = a + (n-1) \cdot d\),
where \(a\) is the first term of the AP and \(d\) is the common difference.
Substituting \(a = -3\) and \(d = 1\) into the formula:
\(a_5 = -3 + (5-1) \cdot 1\),
\(a_5 = -3 + 4\),
\(a_5 = 1\).
So, the 5th term (a₅) is 1.
To find the number of terms (n) that will give a sum of 135, we can use the formula for the sum of an AP:
\(S = \frac{{n}}{2} \cdot (2a + (n-1) \cdot d)\),
where \(S\) is the sum of the AP, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.
Substituting \(S = 135\), \(a = -3\), and \(d = 1\) into the formula:
\(135 = \frac{{n}}{2} \cdot (2(-3) + (n-1) \cdot 1)\),
\(270 = n \cdot (-6 + n - 1)\),
\(270 = -6n + n^2 - n\),
\(n^2 - 8n + 270 = 0\).
To solve this quadratic equation, we can either factorize it or use the quadratic formula. By factoring the equation, we find:
\((n - 5)(n - 45) = 0\).
This gives us two solutions: n = 5 or n = 45.
Therefore, the number of terms (n) that will give a sum of 135 is either 5 or 45.
I need the answer
Since the two given terms are 15 terms apart,
15d = 12 - (-3) = 15
d = 1
Now you have
T5 = t4 + d = -3 + 1 = -2
Now jsut figure a, and solve for n in
n/2 (2a + (n-1)d) = 135