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?

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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