How many terms from the progression 3, 5, 7, 9, ... must be taken in order that their sum will be 2600? Date of Exam November 1994 Subject Mathematics, Surveying and Transportation Engineering Problem How many terms from the progression 3, 5, 7, 9, ... must be taken in order that their sum will be 2600? A. 80 C. 50 B. 60 D. 70 Answer Key Click here to show or hide the answer key [ C ] Solution Click here to expand or collapse this section 3, 5, 7, 9, ... ← is an AP with d = 2 $S = \dfrac{n}{2}\left[ 2a_1 + (n - 1)d \right]$ $2600 = \dfrac{n}{2}\left[ 2(3) + (n - 1)(2) \right]$ $2600 = 3n + n(n - 1)$ $n^2 + 2n - 2600 = 0$ $(n - 50)(n + 52) = 0$ $n = 50 ~ \text{and} ~ -52$ Use $n = 50$ ← answer Category Algebra Progression Arithmetic Progression Sum of Arithmetic Progression Number of Terms in an Arithmetic Progression Log in or register to post comments
