Question

how many terms of the A.P. is16,14, 12,..... are needed to give the sum 60?explain why we get two answer?​

Answer 1

Answer:

Step-by-step explanation:

16,14,12,..........

Sn=60

a=16

d= -2

Sn=n/2{2a+(n-1)d}

60=n{32+(-2n+1)}

120=32n-2n^2+n

-2n^2+33n-120=0

Solve this equation and you will get th value of n

{2n^2-(16+15)n+120}

2n^2-16n-15n+120

2n(n-8)-15(n-8)

(2n-15)(n-8)

n=8

HOPE IT WILL HELP YOU

Answer 2

we have given. a = 16 , d = -2

Sn = 60 , n= ?

so,

$sn \: = \frac{n}{2} \times( 2a + ( n - 1)d)$

$60 = \frac{n}{2} \times (32 + (n - 1) - 2)$

$120 = n(32 - 2n + 2) \\ 120 = 34n - 2 {n}^{2} + 2 \\ 2 {n}^{2} - 34n + 118 = 0 \\ {n}^{2} - 17n + 59 = 0$

so, this is not possible to get sum = 60

as there is no possible roots