Question

if the 7th term of AP 1/7 find its 68term​

Answer 1

Answer:

The 68th term is $[\frac{1}{7}+61d]$.

Step-by-step explanation:

The nth term of AP is, $T_{n}=a+(n-1)d$

Here a = first term and d = common difference.

It is provided that the 7th term is $\frac{1}{7}$.

Then compute the value of a as follows:

$\frac{1}{7} =a+(7-1)d\\\frac{1}{7} =a+6d\\\frac{1}{7} -6d=a$

The 68th term is:

$T_{68}=\frac{1}{7}-6d +(68-1)d\\=\frac{1}{7}-6d+67d\\=\frac{1}{7}+61d$

Thus, the 68th term is $[\frac{1}{7}+61d]$.

Answer 2

Answer:

a+6d=1/7

7a+42d=1

7a=1-42d

a=1-42d/7

a68=a+(n-1)d

n=68

68 term=a+67d

1-42x/7+67d

d should be guven