document.write( "Question 1195430: The first term of an AP is 3. Given that the sum of the first 6 terms is 48 and that the sum of all the terms is 168 .calculate the common difference,the number of terms in the AP and the last term \n" ); document.write( "
Algebra.Com's Answer #827907 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "The sum of an AP is the number of terms, multiplied by the average of the first and last terms. Given first term 3 and sum the first six terms 48, the average is 48/6 = 8. If the 6th term is x, then \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "So the 6th term is 13. \n" ); document.write( "That 6th term, 13, is the first term, 3, plus the common difference, d, 5 times: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "So the AP has first term 3 and common difference 2. \n" ); document.write( "Now let n be the number of terms in the whole sequence. The last term is the first term, 3, plus the common difference, d, (n-1) times: \n" ); document.write( " \n" ); document.write( "The sum of all the terms, 168, is the number of terms, n, times the average of the first and last terms: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The number of terms in the whole sequence is 12. \n" ); document.write( "The last (12th) term is 3+11(2)=25. \n" ); document.write( "ANSWERS: \n" ); document.write( "common difference: 2 \n" ); document.write( "number of terms: 12 \n" ); document.write( "last term: 25 \n" ); document.write( "CHECK: \n" ); document.write( "(Sum = number of terms times average of first and last) \n" ); document.write( "12((3+25)/2)=12(28/2)=12*14=168 \n" ); document.write( " \n" ); document.write( " |