Question

Write the first six terms of an AP in which
(d) a=4, d =7/3​

Answer 1

Answer:

Step-by-step explanation:

let the no.s be

a,a+d,a+2d,a+3d,a+4d,a+5d

4,4+7/3,4+2*7/3,4+3*7/3,4+4*7/3,4+5*7/3

4,19/3,26/3,11,40/3,47/3

these are the six terms of the following AP hope it helps u

thank you

Answer 2

Answer:The First Six terms are:-

4

19/3

26/3

11

40/3 and

47/3.

ExplanaTion:-

Given:-

First term (a) = 4.

Common Difference (d) =  

To Find:-

The first six terms of that AP.

FormulaUsed:-

Where,

a = First term.

n = Number of terms.

d = Common Difference.

So Here,

a = 4.

d =  

And we have to find,

So lets the values in above formula:-

↦ a2 = a + (2 - 1)d.

↦ a2 = a + d.

↦ a2 = 4 + 7/3.

↦ a2 = 12 + 7/3.

[Taking LCM].

↦ a2 = 19/3.

Similarly,

↦ a3 = a + 2d.

↦ a3 = 4 + 2(7/3).

↦ a3 = 4 + 14/3.

↦ a3 = 12 + 14/3.

↦ a3 = 26/3.

And,

↦ a4 = a + 3d.

↦ a4 = 4 + 3(7/3)

↦ a4 = 12 + 21/3.

↦ a4 = 33/3.

↦ a4 = 11.

Also,

↦ a5 = a + 4d.

↦ a5 = 4 + 4(7/3).

↦ a5 = 12 + 28/3.

↦ a5 = 40/3.

Finally,

↦ a6 = a + 5d.

↦ a6 = 4 + 5(7/3).

↦ a6 = 12 + 35/3.

↦ a6 = 47/3.

So the terms are 4, 19/3, 26/3, 11, 40/3 And 47/3.