Question

if the third term of a GP is 5/ 2 and its 8th term is 5/64 then find the sum of its first 10 terms ​

Answer 1

Answer:

Ar^2=5/2 ar^7=5/64

Therefore r =1/2 (divide the above two)

Find a by putting r in anyone equation

A=10

S10=a(1-r^n)/1-r

Cause r is smaller than 1

Therefore sum is 20(1-1/2^10)

Sum = 19.98046875

Step-by-step explanation:

Answer 2

The sum of the first 10 terms ​of a GP $=20(1-\dfrac{1}{2}^{10})$

Step-by-step explanation:

Given,

The third term of a GP ($a_{3}$) = $\dfrac{5}{2}$ and

The 8th term of a GP ($a_{8}$) = $\dfrac{5}{64}$

To find, the sum of the first 10 terms ​of a GP = ?

We know that,

The nth term of a GP

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

$ar^2$ = $\dfrac{5}{2}$                            .............. (1)

$ar^7$ = $\dfrac{5}{64}$                          .............. (2)

Dividing equation (2) by (1), we get

$\dfrac{ar^7}{ar^2} =\dfrac{\dfrac{5}{64}}{\dfrac{5}{2}}$

⇒ $r^5 =\dfrac{1}{32}$

⇒$r^5 =(\dfrac{1}{2})^5$

⇒ $r =\dfrac{1}{2}$

Put $r =\dfrac{1}{2}$ in equation (1), we get

$a(\dfrac{1}{2})^2$ = $\dfrac{5}{2}$    

a = 10  

The sum of the first 10 terms ​of a GP

$=\dfrac{10(1-\dfrac{1}{2}^{10})}{1-\dfrac{1}{2}}$ [ ∵ The sum of the first nth terms ​of a GP =  $\dfrac{a(1-r^n)}{1-r}$]

$=20(1-\dfrac{1}{2}^{10})$

∴ The sum of the first 10 terms ​of a GP $=20(1-\dfrac{1}{2}^{10})$