What is the 32nd term of the arithmetic sequence where a1 = 4 and a6 = 39 ?

Question

muhammad_nauman_umair · Answer

a6=a1+5d
here
d=a6-a1/5
as 
a6=39
and a1=1
d=38/5

and 
a32=a1+31d
a32=1+31(38/5)

anonymous · Answer

A) 209
B) 213
C) 217
D) 221 those are the possible answers

callisto · Answer

Hmm.. the above procedures are correct, but not the number.
a6=a1+5d
d=a6-a1/5, where a6 = 39 and a1=4
d=(39-4)/5 = 7

a32=a1+(32-1)d
a32=4+31(7) = Can you work it out now?

anonymous · Answer

yup 221 correct..

muhammad_nauman_umair · Answer

sorry ...
for my mistake

campbell_st · Answer

1st find d the common difference 

use 

$$T _{n} = a + (n - 1)d$$

n = 6 T6 = 39 and a = 4

$$39 = 4 + (6 - 1)d$$

$$35 = 5d$$

d = 7

to find the 32 term, use the above formula with a = 4 d = 7 and n = 32



then just evaluate it