38
Find three consecutive terms in A.P. whose sum is $21$ and their product is $231$.
Show SolutionHide Solution↓
OR
(b) Let the three consecutive terms in A.P. be $a - d, a, a + d$
therefore $(a - d) + a + (a + d) = 21$
$a = 7$ (1/2 Mark)
and $a (a^2 - d^2) = 231 \Rightarrow d^2 = 16$
$d = \pm 4$ (1/2 Mark)
When $d = 4$ the terms are $3, 7, 11$
When $d = -4$ the terms are $11, 7, 3$ (1 Mark)
(b) Let the three consecutive terms in A.P. be $a - d, a, a + d$
therefore $(a - d) + a + (a + d) = 21$
$a = 7$ (1/2 Mark)
and $a (a^2 - d^2) = 231 \Rightarrow d^2 = 16$
$d = \pm 4$ (1/2 Mark)
When $d = 4$ the terms are $3, 7, 11$
When $d = -4$ the terms are $11, 7, 3$ (1 Mark)