Question

♡●♡●》☆☆please solve this question☆☆《《 ■♡■♡ =》A person is to count 4500 currency notes. Let a n denote the number of notes he counts in the n th minute. If a1 = a 2 = .... = a 10 = 150 and a 10 , a 11 ,... are in an AP with common difference – 2, then the time taken by 
him to count all notes is ❤❤❤❤ dont spam give a brainliest answer​

Answer 1

Answer:

Let he take n minutes to count the notes

If a1 = a 2 = .... = a 10 = 150

then a1+a2+........+a10=1500

hence a11+a12+.....+an=3000

\frac{n-10}{2}(a11+an)=3000

\frac{n-10}{2}(148+148+(n-11)(-2))=3000

n2-169n+4590=0

n=135,34

so all the notes get counted in 34 minutes

Answer 2

Answer:

a1+a2=......=a10

a1+a2+......+a10=1500

hence a11+a12+.......+an=3000

{n-10}{2}{a11+an}=3000

{n-10}{2}{148+148+(n-11(-2))=3000

n=2-16 9n+4590=0

n=135,34

notes get counted is 34 minutes