Question

4th term of GP is 40 and 10th term is 2560.find the 7th term​

Answer 1

Answer:

320

Step-by-step explanation:

Formula to find nth term of a G.P = arⁿ⁻¹

So 4th term =40 

 

 ⇒⇒ ar³=40

10th term =2560

⇒⇒⇒ar⁹=2560 

Now dividing ar⁹/ar³

          

ar⁹     2560

---- = ---------

ar³     40

r⁶ = 256/4

r⁶= 64

 

r⁶=2⁶

⇒⇒ r=2

 

ar³=40

a=40/r³

a=40/8=5

7th term = ar⁶= 5*2⁶=5*64=320.

Answer 2

The 7th term is 320.

Step-by-step explanation:

To Find

7th term of the GP whose 4th and 10th term is given.

Formula Used

$T_{n}=ar^{n-1}$

Given

4th term of GP= 40

10th term of GP= 2560

We know that General Expression for any n term of GP is given by

$T_{n}=ar^{n-1}$ where, n= any general term, a= first term and r= common difference.

We know 4th term of this GP

So we can write it by putting the given values as

40= a$r^{4-1}$

or,40= a$r^{3}$_______equation(1)

Also, we have 10th we can write it similarly as,

2560=a$r^{n-1}$

or,2560=a$r^{9}$_______equation(2)

On dividing equation (2) from equation(1), we have,

$\frac{2560}{40} = \frac{ar^{9} }{ar^{3} }$

On dividing and canceling out we get,

$64=r^{6}$

so the value of r is

∴r= 2.

Putting r=2 in equation(1)the

a$r^{3}$= 40

Putting the value of r

a($2^{3}$)=40

cube of 2 is 8, so we have

8a=40

this gives us

∴a= 5

Now we know the value of a and r so we can get the value of 7th term easily.

By putting 7th term in the general expression and values of a=5 and r= 2 we get,

$T_{7}= 5(2)^{7-1}$

on solving further we have,

$T_{7}=5(2)^{6}$

value of $2^{6}$ is 64,

$T_{7}=5X64$

multiplying the remaining term we get

$T_{7}=320$

So, the 7th term is 320.