Question

What are next term of an A.P is √7,√28,√63...…..?

Answer 1

Answer:

$T_{4} = 4\sqrt{7}$  

$T_{5} = 5\sqrt{7}$

$T_{6} = 6\sqrt{7}$

Step-by-step explanation:

The given series is-

$\sqrt{7}, \ \sqrt{28}, \ \sqrt{63}.........$

Here, a = √7

difference = $T_{2}-T_{1}$

d = √28 - √7

  = 2√7 - √7

  = √7(2 - 1)

d = √7

The next term of A.P. =?

We know that-

$T_{n} = a+(n-1)d$

So, $T_{4} = \sqrt{7}+(4-1)\sqrt{7}$

               $= \sqrt{7} + 3 \times\sqrt{7}$

               $= \sqrt{7}(1+3)$

$T_{4} = 4\sqrt{7}$  

$T_{5} = a+(5-1)d$

        $= \sqrt{7} + (5-1)\sqrt{7}$

        $= \sqrt{7} + 4\sqrt{7}$

$T_{5} = 5\sqrt{7}$

SImilarly, $T_{6} = \sqrt{7} + (6-1)\sqrt{7}$

                         $= \sqrt{7} + 5\sqrt{7}$

$T_{6} = 6\sqrt{7}$