Question

At what value of parameter "a" are there values of x such that the numbers \[5^{1+x}+5^{1-x},\frac{a}{2},25^{x}+25^{-x}\] form an Arithmetic Progression for real x.

Answer 1

If this a, b and c are in AP then:

\[b = \frac{a + c}{2}\]

Answer 2

Plug in the values there...

Answer 3

@waterineyes

Answer 4

Take the first step don't go for second before the first..


Just plug in the values there and show me..

Answer 5

\[\frac{a}{2}= \frac{5^{1+x}+5^{1-x}+25^{x}+25^{-x}}{2}\]

Answer 6

Multiply by 2 both the sides..

Answer 7

\[a= {5^{1+x}+5^{1-x}+25^{x}+25^{-x}}\]

Answer 8

You can write 25 as:

\[25 = 5^2\]

Answer 9

\[25^x = 5^{2x}\]

Answer 10

\[{a}= {5^{1+x}+5^{1-x}+5^{2x}+5^{-2x}}\]

Answer 11

Yes great..

And you can also write:

\[5^{1+x} = 5 \cdot 5^x\]

Answer 12

\[{a}= {5.5^{x}+5.5^{-x}+5^{2x}+5^{-2x}}\]

Answer 13

Can you take \(5^x\) common there??

Answer 14

??

Answer 15

*Factor out..

Answer 16

\[{a}= 5^x({5+5.5^{-1}+5^{2}+5^{-2}})\]

Answer 17

\[{a}= 5^x({5+1+25+\frac{1}{25}}) \implies a = 5^x(31 + \frac{1}{25})\]

Answer 18

776/25

Answer 19

Yes..

Answer 20

\[a = 5^x(\frac{776}{5^2}) \implies a = 5^{x-2} \times 776\]

Answer 21

then.

Answer 22

a belongs to [12,inf) how??

Answer 23

Wait..

Answer 24

No my brain is not working..

Don't worry..

@mukushla can you help here..??

Answer 25

\[t=5^x>0\] then\[a=5(t+\frac{1}{t})+t^2+\frac{1}{t^2}\]

Answer 26

complete the square or take derivative.

Answer 27

i dont know derivatives.

Answer 28

pls explain using completing the sq.

Answer 29

wait

Answer 30

ok

Answer 31

ok i understood

Answer 32

continue.......

Answer 33

\[t+\frac{1}{t}=(\sqrt{t}-\frac{1}{\sqrt{t}})^2+2 \ge2\]\[t^2+\frac{1}{t^2}=(t-\frac{1}{t})^2+2 \ge2\]

Answer 34

ok

Answer 35

\[a\ge 5(2)+2=12\]

Answer 36

Wow @mukushla ..

Answer 37

:)