Question

the second term is 8 and the fifth term is -27. Find a1 in this geometric sequence.

Answer 1

Answer:

$a_{1} = \frac{-16}{3}$

Now the Geometric sequence

          $\frac{-16}{3} , 8 ,-12 ,18 ,-27$

Step-by-step explanation:

Step(i):-

Given second term is '8'

                           $t_{n}= a r^{n-1}$

                           $t_{2}= a r^{2-1}$

                         $a r^{2-1} = 8$

                         a r = 8 ...(i)

And also given  fifth term is -27

                       $t_{5}= a r^{5-1}$

                       $a r^{4} = -27$ ...(ii)

Step(ii):-

   Now  simplify

                   $\frac{a r^{4} }{ar} = \frac{-27}{8}$

                  $r^{3} = \frac{-27}{8}$

                  $r^3 =\frac{(-1)^{3} (3)^{3} }{(2)^{3} }$

                 $(r^3)^{\frac{1}{3} } =(\frac{(-1)^{3} (3)^{3} }{(2)^{3} })^{\frac{1}{3} }$

On simplification, we get

                  $r = \frac{-3}{2}$

  now from(i)

              a r = 8

              $a(\frac{-3}{2} ) = 8$

              $a = \frac{-16}{3}$

            $a_{1} = \frac{-16}{3}$

Now the Geometric sequence

          $\frac{-16}{3} , 8 ,-12 ,18 ,-27$