in an A.p 6th term is half the 4th term and the 3rd term is 15.How many terms are needed to give a sum that 16 equal to 66.
To solve this problem, we'll use the formula for the nth term of an arithmetic progression (AP):
\[a_n = a_1 + (n-1)d\]
where \(a_n\) is the nth term, \(a_1\) is the first term, \(n\) is the number of terms, and \(d\) is the common difference.
Let's break down the problem step by step:
Step 1: Write down what we know.
In the given AP, the 6th term (\(a_6\)) is half of the 4th term (\(a_4\)), and the 3rd term (\(a_3\)) is 15.
Step 2: Express the given information as equations.
From the given information, we can write two equations:
\[a_6 = \frac{1}{2}a_4\]
\[a_3 = 15\]
Step 3: Find the common difference.
To find the common difference (\(d\)), subtract the 3rd term from the 4th term:
\(d = a_4 - a_3\)
Since we know that \(a_3 = 15\), we can substitute it into the equation:
\[d = a_4 - 15\]
Step 4: Express \(a_6\) in terms of \(a_1\) and \(d\).
Using the formula for the nth term of an AP, we can write \(a_6\) as:
\[a_6 = a_1 + 5d\] (since \(n = 6\) in this case)
Step 5: Substitute the expressions for \(a_6\) and \(a_4\) into the first equation.
Substituting the values of \(a_6\) and \(a_4\), we get:
\[a_1 + 5d = \frac{1}{2}(a_1 + 3d)\]
Step 6: Solve for \(a_1\) in terms of \(d\).
Solving the equation, we find:
\[a_1 = 7d\]
Step 7: Substitute the value of \(a_1\) into the second equation.
Substituting \(a_1 = 7d\) into the equation \(a_3 = 15\), we get:
\[7d + 2d = 15\]
\[9d = 15\]
\[d = \frac{15}{9}\]
\[d = \frac{5}{3}\]
Step 8: Find the number of terms needed to give a sum of 66.
To find the number of terms (\(n\)) needed to give a sum of 66, we'll use the formula for the sum of an arithmetic series:
\[S_n = \frac{n}{2}(a_1 + a_n)\]
Substituting the values we know, we get:
\[66 = \frac{n}{2}\left(7\left(\frac{5}{3}\right) + 7\left(\frac{5}{3}\right)(n-1)\right)\]
Simplifying this equation and solving for \(n\), we find:
\[n = 6\]
Therefore, 6 terms are needed to give a sum equal to 66.