Question

The sum of the first 7 terms of an arithmetic series is 329. The common difference is 14.
a) Find the value of the first term.
b) 69,800 is the sum of the first n terms of the sequence. Find n.

The sixth term of an arithmetic sequence is 49 and the fifteenth term is 130.
a) Find the common difference for this sequence. (I got this, it's d=9)
b) Find the first term.
c) How many f the terms of this sequence have a value which is less than 300?

Answer 1

329/7 is your first term

Answer 2

@Holly00d1248 Gotcha. So 47

Answer 3

yep

Answer 4

@dellzasaur nah....how it could be 47....huh

Answer 5

@rishavraj He said to divide 329/7 to get the first term for the first problem

Answer 6

nah its not the way.....u need to find the first term???i mean u need help in part a??

Answer 7

u remember
sum of "n" terms = (n/2)*[2a + (n-1)d]
a - first term
d - common difference

Answer 8

@rishavraj Yeah

Answer 9

plug the values and get "a" i.e first term....:)

Answer 10

@rishavraj I got -13/2? I'm not sure

Answer 11

don't take it otherwise but how u managed to get a negative value ...that also fraction...lol
see sum of n terms = 329
n = 7
d= 14
now solve

Answer 12

\[329 = (7/2)*[2a + (7 - 1)14]\]

Answer 13

@rishavraj 35?

Answer 14

nope a = 5.....solve it again.....
its
94 = 2a + 84
get "a"....remember "a" is the first term

Answer 15

@rishavraj Ok I got it now

Answer 16

@dellzasaur need help in other parts too???

Answer 17

@rishavraj Please! I really need help on b of the first question and b and c of the second question. I think I may be able to do b on the second one but i just want to make sure

Answer 18

@dellzasaur see now its
69800 = (n/2)*[2a + (n - 1)d]
a = 5
d = 14
get "n"...:))

Answer 19

@rishavraj Thank you :) I'm working it out but I'm not sure what to do with the n's in the equation. Can you show me one or two steps of it so I can see? I can probably go from there!

Answer 20

Would I divide everything by 2?

Answer 21

or multiply the 69800 by 2? I dont think that's right

Answer 22

u will get fking quadratic equation....:((
\[69800 = (n/2)*[10 + (n - 1)14]\]

Answer 23

@dellzasaur
see for question no 2
its
\[a _{6} = a + (n - 1)d\]

Answer 24

and
\[a _{15} = a + (15 - 1)d\]

Answer 25

\[a _{6} = a + (5 - 1)d\]
u having two equation and two variables...get "a" and "d"

Answer 26

@rishavraj the d is 9 right?

Answer 27

I got 13 and 4 for a, not entirely sure

Answer 28

@dellzasaur in question 2
common difference will be 9....and u first term = 4...as u said....