Question

What is the 41st term of the arithmetic sequence where a1 = 18 and a15 = -38 ?

Answer 1

is the sequence is arithmetic then the terms (from the second) would be
\[\large a_2=a_1+d \]
\[\large a_3=a_2+d=a_1+2d \]
\[\large a_4=a_3+d=a_1+3d \]
so in general the n-th term would be
\[\large a_n=a_{n-1}+d=a_1+(n-1)d \]

Answer 2

u have to compute the value of \(d\)

Answer 3

howw?

Answer 4

@helder_edwin

Answer 5

u have a1 and a15, so
\[\large a_{15}=a_{14}+d=a_1+14d \]

Answer 6

solve for d

Answer 7

oh

Answer 8

-138?

Answer 9

what is that?

Answer 10

umm the answer?

Answer 11

u mean d=-138 or a41=-138

Answer 12

a41

Answer 13

let's see
\[\large a_{15}=-38\qquad a_1=18 \]
so
\[\large -38=18+14d\Rightarrow d=-4 \]
so
\[\large a_{41}=a_1+40d=18+40(-4)=-142 \]

Answer 14

damm

Answer 15

thanks

Answer 16

u r welcome