# The third term of a GP is 4 , find the product of its first five terms .( answer = 1024 )

let the terms be a /r^2 , a/r , a ,ar, ar^2

now product =  a/r^2 * a/r * a* ar* ar^2

= a^5

= 4 ^5

= 1024

• 17

A3 =4 = ar2

product of first five terms ....

a(ar)(ar2 )(ar3 )(ar4 )= a5 r10

= (ar2)5

= (4)5

=1024

• 104
Karan Muvval is ​absolutely right.
• 2 • 1 • 1 • 5
Karan muval is absolutely right
• 0
1024
• -3
1024

• 0
1024

• 3
halooo

• 4 Hope this helps.........
• 4
1024

• 0
Let the GP be a , ar , ar2 , ar3 ....
T3 = ar=4
We need to find product of first five tems, so
T1 * T2 * T3 * T4 * T5 = a * ar * ar2 * ar3 * ar4

=a5r10

=(ar2)5
=(4)5
= 1024
• 0 • 2
Pls find below • 2
1024

• -1 • 4
A3=4=ar^2 a(ar)(ar^2)(ar^3)(ar^4) Product of first 5 terms are a^5r^10= (ar^2)^5 = 4^5 Done. OK love u will u be my girlfriend 8421023221
• -1
LET THE FIVE TERMS BE  a / r 2 , a / r , a , a r , a r 2 ;
THIRD TERM IS 4 (i.e) a = 4;
a / r* a / r * a * a r * a r 2 = a5 = 4 5 =1024
• 1
If seventh term like between 130 and 140. Find common diff
• 1
Ahiy uip
• 1 • 2
a straight line passes through two three and the portion of the line intercepted between the axis is bisected at this point find its equation
• 1 • 0
Find the some of the Gp • 0
A3=4=ar2
product of first five terms
a.ar.ar2.ar3.ar4= a5r10
= (ar2)5
= 45
=1024
• 0
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