In an A.P. (arithmetic progression), the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
+128475 Let the last term be 1 + 4(n -1) where n is the total number of terms
Sum of an arithmetic series =
(first term + last term ) (number of terms /2)
So we have
( 1 +1 + 4(n - 1) ) (n/2) = 120 simplify
( 4n -2) (n/2) =120
(2 n -1) n =120
2n^2 - n =120
2n^2 - n -120 = 0
(2n + 15) ( n - 8) = 0
The second factor set to 0 and solved for n gives us our answer = 8 terms
