Question

Check whether 32 is present or not in the G.P.√2,2,2√2,4,........

Answer 1

Answer:

yes

Step-by-step explanation:

a=√2

r=2/√2

=√2

l =32

l= a(rn -1)/r-1

l=√2(2n-1)/2-1

32=2n

2*2*2*2*2*2=2n

12 = n

Answer 2

Step-by-step explanation:

Given GP is √2 , 2 ,2√2...

We know that,

n th term of GP

$= a \times {r}^{n - 1}$

Here,

Common ratio (r) = t(n)/t(n-1)

= t(2)/t(1)

= 2/√2

= √2

first term (a) = √2

$32 = \sqrt{2} \times {( \sqrt{2}) }^{n - 1} \\ {2}^{5} = ( {2}^{ \frac{1}{2} } ) \times ({2}^{\frac{1}{2} } ) ^{n - 1} \\ \\ w e\: know \: that \: ( {a}^{m} ) ^{n} = {a}^{m \times n} \\ \\ {2}^{5} = ({2}^{ \frac{1}{2} } ) \times ({2}^{ \frac{n - 1}{2} }) \\ \\ we \: know \: that \: {a}^{m} \times {a}^{n} = {a}^{m + n} \\ \\ {2}^{5} = {2}^{ \frac{1}{2} + \frac{n - 1}{2} } \\ \\ {2}^{5} = {2}^{ \frac{1 + n - 1}{2} } \\ \\ {2}^{5} = {2}^{ \frac{n}{2} } \\ \\ here \: the \: bases \: are \: equal \: then \: the \: powers \: are \: also \: equal. \\ \\ 5 = \frac{n}{2} \\ \\ n \: = 5 \times 2 \\ \\ n = 10.$

Hence 32 is present in the given GP.