Advertisements
Advertisements
Question
Find the third term from the end of the G.P.
`2/27, 2/9, 2/3, .........,162.`
Advertisements
Solution
Given G.P. : `2/27, 2/9, 2/3, .........,162`
Here,
Common ratio, r =`(2/9)/(2/27)` = 3
Last term, l = 162
∴ 3rd term from an end =`1/r^2`
= `162/(3)^2`
= `162/9`
= 18
RELATED QUESTIONS
Find, which of the following sequence from a G.P. :
8, 24, 72, 216, .............
If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that : (q – r) log a + (r – p) log b + (p – q) log c = 0
Q 7
If a, b and c are in G.P., prove that : log a, log b and log c are in A.P.
If a, b and c are in A.P, a, x, b are in G.P. whereas b, y and c are also in G.P.
Show that : x2, b2, y2 are in A.P.
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that `1/x + 1/y = 2/b`
Find the sum of G.P. :
`1 - 1/2 + 1/4 - 1/8 + ..........` to 9 terms.
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?
Q 8
