Question

How many terms of the A.P. 16, 14, 12, ... are needed to give the sum 60 ? Explain

why do we get two answers.​

Answer 1

Step-by-step explanation:

a=16 d=-2

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

60= n/2 [2×16 + n-1 (-2)]

120= n [32 -2n+2]

120 = n [ 34-2n]

120 = 34n - 2n^2

By middle term split

2n^2 -34n + 120 = 0

Aage khud krlo yrr

Answer 2

Answer:

Here,

a=16

d=14-16=2

Sn=60

So,

applying the formula,

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

60=n/2[2(16) +(n-1) -2]

60=n/2[32-2n+2]

60=n[16-n+1]

60=n[17-n]

60=17n-n^2

n^2-17n-60=0

n^2-20n+3n+60=0

n(n-20) -3(n-20) =0

(n-3) (n-20) =0

n=3 or n=20

So, it has two solutions

because sum of first 20 terms or sum of first 3 will come the same i.e, 60