Question

Find G.P. where 7th term is 320 and 10th term is 2560

Answer 1

a(n)=a(1)xr^n-1
a(7)=a(1)xr^7
320=a(1)xr^7.....................1
and 2560=a(1)xr^10.......................2
divide 2 by 1,
we get 8=r^3
i.e. r=2
by putting r=2 in 1,
we get 320=a(1)x128
i.e. a(1)=320/128=5/2
a(n)=a(1)xr^n-1
we get, a(1)=5
a[2] = 10
a[3] = 20
i.e. the required gp is 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560,........

Answer 2

Answer:

So , the GP is

5 10 20 40 80 160 320 640 1280 2560.

Step-by-step explanation:

Let the first term of the GP is a .

The common difference of the GP is d.

The rules of the $n^{th}$ term of a GP is $a*d^{n-1}$.

According to the question

$a*d^{n-1}$ $= 320$

$or , a * d^{6} = 320$.......... (i)

$a*d^{n-1} = 2560\\or , a * d^{9} = 2560$.........(ii)

(ii) is divided by (i)

$d^{3} = 8\\or , d = 2.$

$a * 2^{6} = 320\\or , a = 5.$

So , the GP is

5 10 20 40 80 160 320 640 1280 2560.

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