Saved Bookmarks
| 1. |
Find the sum of the first 30 terms of an AP\xa0whose nth term is 2-3n?\xa0 |
| Answer» an\xa0= 2-3nTo form an A.P. from it, substitute different values of\xa0na1\xa0= 2-3(1) = 2-3 = -1a2\xa0= 2-3(2) = 2-6 = -4a3\xa0= 2-3(3) = 2-9 = -7a4\xa0= 2-3(4) = 2-12 = -10-1 , -4 , -7 , -10 forms an A.P.The first term a = -1, common difference d = -4-(-1) = -4+1 = -3 , number of terms n =30S30 =\xa0\\({ n \\over 2}\\)\\({[ 2a+ (n-1) d]}\\)= 15 [ -2 + 29 X -3]= 15[-89]S30= -1335 | |