What least no of terms of the sequence 17,15 (4/3),14 (3/5)........ should be taken so that the sum is negative

Question

First prove it as an ap and find n

Answer 1

I think this should be 17, 15⅘, 14⅗ so a (first term)=17 and d (common difference)=1⅕.

The sum to n terms is an+n(n-1)d/2.

The sum to n terms is 17n-(3/5)n(n-1)=17n-0.6n²+0.6n=17.6n-0.6n².

When this goes negative, 17.6n-0.6n²<0 so 17.6<0.6n and n>29.3.

So n=30. This is the minimum number of terms necessary to make the sum negative. The actual sum is -12 and the 30th term is -17⅘.