The 16th term of am arithmetic progression AP is 93, given that it's common difference is 6 find the first and 28th terms
a+15d = 93
since d=6, a=3
so now just find
a+27d
To find the first and 28th terms of an arithmetic progression with the 16th term being 93 and a common difference of 6, we can use the formula for the general term of an arithmetic progression:
\[ a_n = a_1 + (n-1)d \]
where \(a_n\) is the nth term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the position of the term.
Step 1: Finding the first term
We can use the formula to find the first term:
\[ 93 = a_1 + (16-1) \cdot 6 \]
Simplifying,
\[ 93 = a_1 + 15 \cdot 6 \]
\[ 93 = a_1 + 90 \]
Subtracting 90 from both sides,
\[ 3 = a_1 \]
Therefore, the first term of the arithmetic progression is 3.
Step 2: Finding the 28th term
We can use the same formula to find the 28th term:
\[ a_{28} = 3 + (28-1) \cdot 6 \]
Simplifying,
\[ a_{28} = 3 + 27 \cdot 6 \]
\[ a_{28} = 3 + 162 \]
\[ a_{28} = 165 \]
Therefore, the 28th term of the arithmetic progression is 165.
To find the first and 28th terms of an arithmetic progression (AP), we need to use the formulas:
\(a_n = a_1 + (n-1)d\)
Where \(a_n\) represents the \(n\)th term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.
Given that the 16th term of the AP is 93 and the common difference is 6, we can substitute these values into the formula and solve for the first term:
\(a_{16} = a_1 + (16-1) * 6\)
Simplifying the equation:
\(93 = a_1 + 15 * 6\)
\(93 = a_1 + 90\)
Subtracting 90 from both sides:
\(3 = a_1\)
Therefore, the first term of the AP is 3.
To find the 28th term, we can once again use the formula:
\(a_{28} = a_1 + (28-1) * 6\)
Simplifying the equation:
\(a_{28} = 3 + 27 * 6\)
\(a_{28} = 3 + 162\)
\(a_{28} = 165\)
Therefore, the 28th term of the AP is 165.