Question

square root of product of 4th and 6th term of GP is 7. find the series​

Answer 1

Answer:

question is incomplete question is that what you doing now

Answer 2

Answer:

$a, a^{3/4}\sqrt[4]{7} ,\sqrt{7a} , 7^{3/4}\sqrt[4]{a}, 7, ..............$

Step-by-step explanation:

Nth term of the GP is given by

$T_n = ar^{n-1}$

Given that,

$\sqrt{T_4T_6} =7\\\implies ar^4 = 7\\\implies r =\sqrt[4]{ \frac{7}{a} }$

so general GP gfor any arbitary constant a is given by

$a, a^{3/4}\sqrt[4]{7} ,\sqrt{7a} , 7^{3/4}\sqrt[4]{a}, 7, ..............$