Question

Pls solve question no. 9,12 & 14.. N pls answer it fast...Its emergency..

Answer 1

Heya user!!!!!

Here's yr answers.......

9.Solution.......

$a + (n - 1)d \\ = > 18 + (n - 1)(31 \div 2 - 18 \div 2) = - 47 \\ = > 18 + (n - 1) \times - 5 \div 2 = - 4 \\ = > - 5 \div 2(n - 1) = - 65 \\ = > n - 1 = 26 \\ = > n = 27$
12. Solution...........
$a + (n - 1)d = 3 \\ = > 5 \div 6 + (n - 1)(1 - 5 \div 6) = 3 \\ = > 5 \div 6 + (n - 1)(1 \div 6) = 3 \\ = > (n - 1) \times 1 \div 6 = 13 \div 6 \\ = > n = 14$
14.Solution...............
$the \: 41st \: term \\ = a + (41 - 1) \times d \\ = 8 + 40 \times 6 \\ = 248 \\ 72 \: more \: than \: 248 = 320 \\ now \\ 8 + (n - 1)6 = 320 \\ = > 8 + 6n - 6 = 320 \\ = > 6n = 318 \\ = > n = 53$
Here r the solutions...Hope they help!!!!