Question

what least no of the term of the sequence 17,15 4/5,14 3/5 should be taken so that the sum is negative

Answer 1

Solution - 
               Given - 
                               17, 15 4/5, 14 3/5 
                 We can see it is an arithmetic sequence and 
                 Common difference d = -1 1/5 
                 

Answer 2

Answer:

Required minimum term is 30 terms.

Step-by-step explanation:

Given : Sequence $17,15 \frac{4}{5},14\frac{3}{5}$

To find : What least no of the term of the sequence should be taken so that the sum is negative?

Solution :

The first term of the sequence is a=17

The common difference of the sequence is

$15 \frac{4}{5}-17=\frac{79}{5}-17=\frac{79-85}{5}=-\frac{6}{5}$

Sum is negative i.e, $S_n<0$

The sum of the arithmetic sequence is

$S_n=\frac{n}{2}(a+(n-1)d)$

$\frac{n}{2}(2(17)+(n-1)\frac{-6}{5})<0$

$\frac{n}{2}(17+\frac{85}{5}+\frac{6}{5}-\frac{6}{5})<0$

$17+\frac{91-6n}{5}<0$

$\frac{91-6n}{5}<-17$

$91-6n<-85$

$176<6n$

$n>\frac{176}{6}$

$n>29$

Therefore, The minimum number of terms is 30 which make the sum negative.