Question

In an ap if a =4 ,and n=7 , a n= 4 then the value of d is

Answer 1

Answer:

0.

Explanation :

Let first term be a , common difference be d,

We have, formula

$\tt a_n = a +( n - 1)d.$

$\tt \implies4 = 4 + (7 - 1)d.$

$\tt \implies \cancel4 = \cancel4 + (7 - 1)d.$

$\tt \implies0= (7 - 1)d.$

$\tt \implies 0 = 6d.$

$\tt \implies d = 0.$

Some information :

$\blacksquare \tt {n}^{th} \: a_n = a +( n - 1)d$

$\tt \blacksquare sum \: of \: {n}^{th}$

$\tt s_n = \frac{n}{2}[ 2a + (n - 1)d].$

Answer 2

Given:

a = 4

n = 7

an = 4

To find:

common difference (d)

Answer:

we know that

an= a+ (n-1)d

=> 4= 4 + (7-1) d

=> 4-4 = (6)d

=> 0/6 = d

=> d = 0

Thus the common difference of the AP is 0.