Question

50th term is 238, 93rd term is 539. state the general term

Answer 1

of AP?

Answer 2

what?

Answer 3

of Arithmetic sequence ??

Answer 4

general term of what?

Answer 5

yes arithmetic

Answer 6

the general 'n' term is
\[T_n=a+((n-1)d)\]
so T50=238=a+49d ----->(1)
T93=539=a+92d ----->(2)
can u solve these two equations to get a and d ??

Answer 7

u can start by subtracting (1) from (2)

Answer 8

so a=49d-a+92d ?

Answer 9

*a+49d-a+92d ?

Answer 10

yes,thats your right side,where a gets cancelled
but whats your left side??

Answer 11

and *92d-49d

Answer 12

oh, so um im not sure

Answer 13

ok,subtracting the equations u get
539-238=(92-49)d
so
\[d=\frac{539-238}{92-49}=??\]

Answer 14

258?

Answer 15

how 258?

Answer 16

because I got 301/43 = 258?

Answer 17

301/43 is actually only 7

Answer 18

oh worry yeah I got 7

Answer 19

so u have d=7
now can put d=7 in any equation and find the value of a

Answer 20

238=a+49*7
a=343-238=??

Answer 21

105?

Answer 22

yup :)
so u have a=105,d=7
and your general term will be :
\[T_n=105+7(n-1)\]

Answer 23

yup :)
so u have a=105,d=7
and your general term will be :
\[T_n=105+7(n-1)\]

Answer 24

no the answer should be tn=7n-112

Answer 25

but i dont know how to get that

Answer 26

oh,sorry
a would be -105
so
\[T_n=-105+7(n-1)=-105+7n-7=7n-112\]
ok ?

Answer 27

thank you :)