Question

how many terms of the ap -6, -11/2, -5 are needed to give the sum -25

Answer 1

Answer:

$a = 4 \: \: and \: \: \: \: d = - \frac{11}{2} - ( - 6) = \frac{1}{2}$

We know,

$s = \frac{n}{2} (2a + (n - 1)d)$

And s= -25

so, putting the values,

$- 25 = \frac{n}{2} (2 \times ( - 6) + (n - 1) \times \frac{1}{2} )$

$- 25 \times 2 = n( - 12 + \frac{n}{2} - \frac{1}{2} )$

$- 50 = n( \frac{ - 24 + n - 1}{2} )$

$- 100 = - 25n \: + {n}^{2}$

${n}^{2} - 25n + 100 = 0$

${n}^{2} - 20n - 5n+ 100 = 0$

$so \: n = 20 \: or \: n = 5$

If n = 5

$s = \frac{5}{2} (2 \times ( - 6) + (5 - 1) \times \frac{1}{2} )$

$s = \frac{5}{2} ( - 12 + 2)$

$s = - 25$

So, the number of terms needed to get a sum of -25 is 5terms