The 9th and 2nd term of an A.P are 29 and 55 respectively,find the sum of its first 70 terms

Answer 1

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.