What is the 32nd term of the arithmetic sequence where a1 = 4 and a6 = 34

Question

parthkohli · Answer

Firstly, the common difference is 5. You can see that we are adding 5 again and again.
Secondly, to find out the nth term, we must use this formula:

$\Large \color{MidnightBlue}{\Rightarrow a_{n} = a_{1} + (n - 1)d }$

parthkohli · Answer

d is the common difference.
a1 is the first term.
an is the nth term of the arithmetic sequence.

apoorvk · Answer

so the first term = a = 4 ...(i)
6th term = a + 5d = 34 ...(ii)

from 'i' and 'ii', 
d=6

now nth term of an AP is --> a + (n-1)d.
a=4, d= 6, and n=32, since you need the 32nd term. substitute and evaluate.

anonymous · Answer

a6 = a1+5d => 34 = 4+5d => d=6
and a32 = a1+31d = 4+31*6 = 190

parthkohli · Answer

Oh oh... I got the common difference wrong.

parthkohli · Answer

Did I or did you??

anonymous · Answer

so its 190?

anonymous · Answer

yup

parthkohli · Answer

Yes, the common difference is 6.

anonymous · Answer

sweet! okayy well how do i do this one?
What is the 35th term of the arithmetic sequence where a1 = 13 and a17 = –67

apoorvk · Answer

same way. it's a new AP. so, just the values of the first term, and common difference would change. form equations, and find out the 'a', 'd'. your n=35

anonymous · Answer

a17 =a1+16d => -67 = 13 + 16d => d =-5
then a35 = a1 + 34d = 13 + 34*(-5) =-157

anonymous · Answer

ahhh your the best thank you so much!