please help

The 11th term of an arithmetic progression is 1. the sum of the first 10 terms is 120.
find the 4th term :S

formula for the...
sum of terms=1/2n(a+l)=1/2n{2a+(n-1)d}

thanks

Question

please help

The 11th term of an arithmetic progression is 1. the sum of the first 10 terms is 120.
find the 4th term :S

formula for the...
sum of terms=1/2n(a+l)=1/2n{2a+(n-1)d}

thanks

yuki · Answer

okay

yuki · Answer

$${11(a_1 + a_{11}) \over 2} = 120 + 1 $$

yuki · Answer

so$${a_1 + 1} = 22$$

thus a_1 = 21

yuki · Answer

Then you can say$$a_{11} = a_1 + 10d$$

$$a_11 = 1 = 21 + 10d$$

yuki · Answer

so d = -2

yuki · Answer

that means that the 4th term is 

21+3(-2) = 15

yuki · Answer

:)

yuki · Answer

did it make sense ?

anonymous · Answer

but you didnt even use 4 once :S???
and i dont really understand where 21+3(-2)=15 comes from :(

yuki · Answer

ok,

$$a_n = a_1 + d(n-1)$$

so$$a_4 = a_1 + (-2)(4-1) = 21 + 3(-2)$$

yuki · Answer

the idea of arithmetic sequence is that you are
"Adding the same number for all sequence"

anonymous · Answer

thanks :)

yuki · Answer

np :)

anonymous · Answer

sorry but i first thought that i knew where 11(a1+a11)2=120+1
came from but i dont :S

yuki · Answer

Hope you read this.

yuki · Answer

the first ten numbers add up to 120, so

10(a1+a10) /2= 120

now if you add a11 to it, it is 

120 + 1

but adding a11 to the sum from a1 to a10  is as same as adding from a1 to a11

so 11(a1 + a11)/2 = 121