Question

the 4th term from the end of the A P 2,5, 8...35 is?

Answer 1

Answer:

23

Step-by-step explanation:

tn= a+(n-1)d

35=2+(n-1)3

35-2=(n-1)3

33=(n-1)3

33/3= n-1

11 = n-1

n= 11+1 = 12

4th term from last is 8 therefore

t_8 =  2+(8-1) 3

    = 2+7*3 = 2+21= 23

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Answer 2

The fourth term from last is 23

Step-by-step explanation:

AP : 2,5,8,......., 35

First term = a = 2

Common difference = 5-2 = 8-5 = 3

Formula of nth term : $a_n = a+(n-1)d$

35=2+(n-1)3

35-2=(n-1)3

33=(n-1)3

$\frac{33}{3}= n-1$

11 = n-1

n= 11+1 = 12

4th term from last = 12-4 = 8

a_8=2+(8-1)(3)=23

Hence The fourth term from last is 23

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The 4th term from the end of the AP -11,-8,-5...........,49 is

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