Question

Progession
The sixth term of an arithmetic progession is 265 and the sum of the first 5 terms is 1445. Find the minimun value of n so that the sum of the first n terms is negative.

Answer 1

@ParthKohli

Answer 2

@hartnn

Answer 3

Do you know the formulae?

you can form 2 equations with a1 and d using the given info....

Answer 4

yes
\[265=a+5d-First~eqn\]\[289=a+2d-Second~eqn\]

Answer 5

\[First~term,a=305,common~difference,d=-8\]
is it correct?

Answer 6

yes! good work :)

Answer 7

but i'm not sure what do next...i'm stuck at here :(

Answer 8

okay,what should we do next?

Answer 9

lets setup the sum inequality

we want the sum of 'n' terms to be negative,

\((n/2) (2a + (n-1)d) < 0 \)

plug in a and d values in this

Answer 10

\[\frac{ n }{ 2 }(2(305)-(n-1)(-8))\]

Answer 11

yup, keep on simplifying it

Answer 12

\[-4n^2+309n<0\]

Answer 13

Remember : n cannot be negative

Answer 14

4n^2-309n>0

Answer 15

how did u get 309 ?

Answer 16

n (305 - (-4)*(n-1)) < 0

n < 0 or (305 - (-4)*(n-1) < 0

discarding n<0

305 +4(n-1) < 0
301 + 4n < 0
....

Answer 17

got that?

Answer 18

yup

Answer 19

n>-305/4

Answer 20

the formula has a + in between

Answer 21

(305 + (-4)*(n-1) < 0

Answer 22

(305 + 4 - 4n) < 0

309 -4n < 0
4n > 309/4

Answer 23

is n>77.25

Answer 24

**
n > 309/4
n > 77.25

Answer 25

yes

Answer 26

so answer is n=78

Answer 27

\(\huge \checkmark \)

Answer 28

Thnx @hartnn :)

Answer 29

welcome ^_^