Question

How many terms of the AP 16,14,12,.....are needed to give the sum 60?Explain why do we get two answers.​

Answer 1

let n terms needed to get sum as 60

a=16

d=-2

Sn=60

n=?

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

60=n/2(2*16+(n-1)(-2))

120=n(32-2n+2)

120=32n-2n^2+2n

120=34n-2n^2

2n^2-34n+120=0

n^2-17n+60=0

n^2-12n-5n+60=0

n(n-12)-5(n-12)=0

(n-12)(n-5)=0

n=12or 5

we got two answers bcoz the sum of 5 terms is equal to sum of 12 terms as it is decreasing ap and new negative terms are added are cut down by positive terms and becomes 0 so nothing more is added

Answer 2

Step-by-step explanation:

a =16

d =-2

60=sn

60= n÷2(2a + ( n-1) d)

60= n÷2(2(16) + (n-1) -2)

60= n÷2(32-2n+2)

60 = n÷2×2(16-n+1 )

60 =n(16-n+1)

60=17n-n

${n}^{2} - 17n + 60$

b

${b }^{2} - 4ac = d$

17×17-4×1×60=49

$- 17 + \sqrt{49} \div 2 \times 1$

$- 17 - \sqrt{49} \div 2 \times 1$

-17+or-7÷2

=5,-12

ans is 5 which possible not-12