Question

the 4th and 7th the terms of an ap are 3,36 find the ap anf its

Answer 1

here you go ☆☆

▪a4= 3
▪a7 = 36

▪a4 = a+ 3d
▪a7 = a+ 6d

a + 6d = 36
a + 3d = 3
-_-____-______
3d = 33

d = 33/3

▪d = 11

a + 3 × 11 = 3

a + 33 = 3

a = 3 - 33

▪a = -30

□AP = -30,-19,-8.........

hope it helps you........

Answer 2

T4=3 [given ]

So,

a+(n-1)d=3

a+3d=3.........<1>

Again,

T7=36. [given]

So,

a+(n-1)d=36

a+6d=36........<2>

Now, solving equation 1&2 we get,

3d=33

Therefore d=11....{common difference }

Putting value of d in <1> we get,

a=-30

Thus required AP is....

a,a+d,a+2d,a+3d....

-30,-19,-8,3....

Hope it helps you...