Question

for an AP whose 3rd term is 5 and 7th term is 9, the value of d is?

Answer 1

Answer:

tn=a+(n-1)d

Hence,

t3=a+2d

i.e. a+2d=5....(1)

t7=a+6d

i.e a+6d=9

a+2d+4d=9

But from (1) a+2d=5

5+4d=9

4d=9-5

4d=4

d=4/4

d=1

Hence, Common different in 1.

Answer 2

Answer:

3rd term of the AP: a+2d = 5 …(1)

7th term: a+6d = 9 …(2)

Subtract (1) from (2)

4d = 4, or

d = 1 and so a = 3.

The AP is 3,4,5,6,….

Step-by-step explanation: