Question

Prove that the product of first n terms of a gp , whose first term is 'a' and last term is 'l' is (al)^(n/2)​

Answer 1

Step-by-step explanation:

Geometric progression

1

st

term =a

n

th

term =b

P= product of n terms

G.P.=a,ar,ar

2

.....ar

n−1

→b=ar

n−1

→P=(a)(ar)(ar

2

)...(ar

n−1

)

→P=a

n

(1×r×r

2

×...r

n−1

)

→P=a

n

r

(1+2+3+...+n+1)

→P=a

n

r

2

(n−1)n

Now,

→P

2

=a

2n

r

n(n−1)

→P

2

=(a

2

r

n−1

)

n

→P

2

=(a⋅b)

n

Hence

P

2

=(a⋅b)

2

Answer 2

To prove:

The product of first n terms of a gp , whose first term is 'a' and last term is 'l' is (al)^(n/2)

Proof:

In an Geometric Progression,

• 1st term = a
• nth term = l
• p = product of n terms

The geometric progression,

• G. P = a, ar, ar²,... ar^(n−1)
• l = ar^ (n-1)
• P=(a)(ar)(ar²)...(ar²(n−1))
• P=a^n (1×r×r²×...r^(n−1) )
• P=a^n .r^(1+2+3+...+n+1)
• P=a^n.r^( (n-1)n/2)

Squaring on both sides,

• P²=a^2n.r ^n(n−1)
• P²=(a².r^n−1)^n
• P²=(a.l)^n
• P= (a.l)^(n/2)

Hence, the product of first n terms of a gp , whose first term is 'a' and last term is 'l' is (al)^(n/2).

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