An arithmetic progression has first term a, common difference 3. the Nth term is 128 and the sum of the first 2N terms is 9842. How do I find N and a?

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To solve this problem we need the formula for `a_n,` the `n`-th term of an arithmetic progression, and the formula for `S_n,` the sum of its `n` terms.

These formulas are:  `a_n = a_1 + d*(n-1)`  and  `S_n = n/2 (a_1 + a_n).` Here `a_1` is the first term...

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Hello!

To solve this problem we need the formula for `a_n,` the `n`-th term of an arithmetic progression, and the formula for `S_n,` the sum of its `n` terms.

These formulas are:  `a_n = a_1 + d*(n-1)`  and  `S_n = n/2 (a_1 + a_n).` Here `a_1` is the first term and `d` is the common difference.

In our problem `d = 3,` `a_N = 128` and `S_(2N) = 9842.` The unknowns are `a = a_1` and `N.` Substitute them into the above formulas:

`a_N = a + 3(N - 1) = 128,`   `a_(2N) = a + 3(2N - 1),`  

`S_(2N) = (2N)/2 (a + a_(2N)) = N(a + a + 3(2N - 1)) = 9842.`

 

This way we obtained two equations for `a` and `N,`

`a + 3(N - 1) = 128`  and  `N(2a + 3(2N - 1)) = 9842,`

let's solve them. Express `a = 128 - 3(N - 1) = 131 - 3N` from the first equation and substitute it to the second:

`N (2(131 - 3N) + 3(2N-1)) = 9842.`

Simplify the expression in the parentheses:  `N*259 = 9842,`  or  `N = 9842/259 = 38.`

Recall that  `a=131-3N` and obtain  `a=131-3*38=131-114=17.`

The answer: a=17 and N=38.

 

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