Home
Class 11
MATHS
Which term of the G.P., 2, 8, 32, . . ....

Which term of the G.P., `2, 8, 32, . . .` up to n terms in `131072`?

Text Solution

Verified by Experts

let the nth herm be `131072`
so `ar^(n-1) = 131072`
`2*4^(n-1) = 131072`
`= 2*4^8`
`4^(n-1)= 4^8`
`n-1= 8`
`n=9`
Promotional Banner

Topper's Solved these Questions

  • SEQUENCES AND SERIES

    NCERT|Exercise MISCELLANEOUS EXERCISE|32 Videos

  • SEQUENCES AND SERIES

    NCERT|Exercise EXERCISE 9.4|10 Videos

  • RELATIONS AND FUNCTIONS

    NCERT|Exercise EXERCISE 2.3|5 Videos

  • SETS

    NCERT|Exercise EXERCISE 1.5|7 Videos

Similar Questions

Explore conceptually related problems

Which term of the G.P:: 2,2,sqrt(2),4, is 128?

Which term of the progression 2, 8, 32,… is 131072 ?

nth term of a G.P. is a + (n – 1)d.

Find the 7 th term of the G.P.4,8,16,dots

Which term of the geometric sequences, (i) 2,8,32,....... is 131072? (ii) sqrt(3),3,3sqrt(3),……..is 729? (iii) 1/3,1/9,1/27…………….. is 1/19683 ?

If the nth term of a GP 2,8,32,... is 131072, then the value of n is

Find the number of terms in the G.P. 1, 2, 4, 8, ... 4096.

The nth term of a G.P. is 3cdot2^(n). Find the sum of 8 terms of the G.P.