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How many terms of the A.P. $3, 5, 7, 9, ...$ must be added to get the sum $80$?
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Solution:
(a) Here $a = 3, d = 2, S_n = 80$ (1 Mark)
$\frac{n}{2}[6 + (n - 1)2] = 80$ (1/2 Mark)
$n^2 + 2n - 80 = 0$
$n = 8, -10$ (rejected)
$n = 8$ (1/2 Mark)
(a) Here $a = 3, d = 2, S_n = 80$ (1 Mark)
$\frac{n}{2}[6 + (n - 1)2] = 80$ (1/2 Mark)
$n^2 + 2n - 80 = 0$
$n = 8, -10$ (rejected)
$n = 8$ (1/2 Mark)