Question

The third element of G.Pis twice the second element and the fifth
element is 32. Find the G.P.​

Answer 1

Step-by-step explanation:

Given :-

The third element of G.Pis twice the second element and the fifth element is 32.

To find :-

Find the G.P. ?

Solution :-

Given that :

Let the first term of the GP be a

Let the Common ratio be r

The general term

The third term of the GP = an = a×r^(n-1)

Second term of the GP = a2

=> a2 = a×r^(2-1)

=> a2 = ar

The third term of the GP

a3 = a×r^(3-1)

a3 = ar²

Given that

The third element of G.Pis twice the second element

=> a3 = 2×a2

=> ar² = 2×ar

=>a×r×r = 2×a×r

=>(a×r×r)/(a×r) = 2

=> r = 2

Common ratio of the GP = 2

Given that

Fifth term of the GP = 32

=> a5 = 32

=>a×r^(5-1) = 32

=> ar⁴ = 32

=> a(2)⁴ = 32

=> a×2×2×2×2 = 32

=> a×16 = 32

=> a = 32/16

=> a = 2

First term of the GP = 2

We know that

The general form of a GP = a, ar ,ar² ,ar³,...

a = 2

ar= 2×2 = 4

ar² = 2×2² = 2×4 = 8

ar³ = 2×2³ = 2×8 = 16

The GP : 2, 4, 8, 16, ...

Answer:-

The required GP for the given problem is 2,4,8,16,...

Check:-

Third term = 8

= 2×4

= 2× Second term

Third term is twice the second term

Fifth term =32

Verified the given relations in the given problem.

Check :-

• The general form of a GP = a, ar ,ar²,...

• The general term of a GP = a×r^(n-1)

Where,

• a = First term

• r = Common ratio

• n = number of terms

• Common ratio = an / an-1

• an = nth term

• an-1 = (n-1)th term