The 9th and 2nd term of an A.P are 29 and 55 respectively,find the sum of its first 70 terms
The general form of an arithmetic progression (A.P.) is given by:
\[a_n = a_1 + (n-1) d\]
where \(a_n\) is the \(n\)th term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.
Given that the 9th term is 29, we can substitute \(a_n = 29\) and \(n = 9\) into the equation:
\[29 = a_1 + (9-1) d\]
\[29 = a_1 + 8d\]
Similarly, given that the 2nd term is 55, we can substitute \(a_n = 55\) and \(n = 2\) into the equation:
\[55 = a_1 + (2-1) d\]
\[55 = a_1 + d\]
Now we have a system of two equations with two unknowns, \(a_1\) and \(d\):
\[29 = a_1 + 8d\]
\[55 = a_1 + d\]
Solving this system of equations, we can subtract the second equation from the first to eliminate \(a_1\):
\[29 - 55 = (a_1 + 8d) - (a_1 + d)\]
\[-26 = 7d\]
\[d = -4\]
Substituting this value of \(d\) into the second equation, we can solve for \(a_1\):
\[55 = a_1 + (-4)\]
\[59 = a_1\]
Now we have the values of \(a_1 = 59\) and \(d = -4\).
To find the sum of the first 70 terms, we can use the formula for the sum of an arithmetic series:
\[S_n = \frac{n}{2} (2a_1 + (n-1) d)\]
Substituting \(n = 70\), \(a_1 = 59\), and \(d = -4\), we get:
\[S_{70} = \frac{70}{2} (2(59) + (70-1)(-4))\]
\[S_{70} = 35(118 + 69(-4))\]
\[S_{70} = 35(118 - 276)\]
\[S_{70} = 35(-158)\]
\[S_{70} = -5530\]
The sum of the first 70 terms of the given arithmetic progression is -5530.