English

Find the Sum of the Following Geometric Progression: 1, 3, 9, 27, ... to 8 Terms;

Advertisements
Advertisements

Question

Find the sum of the following geometric progression:

1, 3, 9, 27, ... to 8 terms;

Advertisements

Solution

 Here, a = 1 and r = 3.

\[\therefore S_8 = a\left( \frac{r^8 - 1}{r - 1} \right) \]

\[ = 1 \left( \frac{3^8 - 1}{3 - 1} \right) \]

\[ = \frac{6561 - 1}{2}\]

\[ = 3280\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 20: Geometric Progression - Exercise 20.3 [Page 27]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.3 | Q 1.2 | Page 27

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Which term of the following sequence:

`sqrt3, 3, 3sqrt3`, .... is 729?


Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015,…


Show that one of the following progression is a G.P. Also, find the common ratio in case:

\[a, \frac{3 a^2}{4}, \frac{9 a^3}{16}, . . .\]


Find :

the 12th term of the G.P.

\[\frac{1}{a^3 x^3}, ax, a^5 x^5 , . . .\]


Which term of the G.P. :

\[\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . \text { . is } \frac{1}{19683} ?\]


If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal, find the value of n.


In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.


Find the sum of the following series:

9 + 99 + 999 + ... to n terms;


Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is \[\frac{1}{r^n}\].


Find the sum of the following series to infinity:

`1/3+1/5^2 +1/3^3+1/5^4 + 1/3^5 + 1/56+ ...infty`


Find the rational numbers having the following decimal expansion: 

\[0 .\overline {231 }\]


Show that in an infinite G.P. with common ratio r (|r| < 1), each term bears a constant ratio to the sum of all terms that follow it.


If a, b, c are in G.P., prove that log a, log b, log c are in A.P.


Find k such that k + 9, k − 6 and 4 form three consecutive terms of a G.P.


Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers.


If a, b, c are in G.P., prove that:

\[\frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2}\]


If a, b, c are in G.P., prove that:

(a + 2b + 2c) (a − 2b + 2c) = a2 + 4c2.


If a, b, c, d are in G.P., prove that:

\[\frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b}\]


If a, b, c, d are in G.P., prove that:

 (a + b + c + d)2 = (a + b)2 + 2 (b + c)2 + (c + d)2


If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a − b, d − c are in G.P.


Insert 6 geometric means between 27 and  \[\frac{1}{81}\] .


If in an infinite G.P., first term is equal to 10 times the sum of all successive terms, then its common ratio is 


In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is 


For the G.P. if a = `7/243`, r = 3 find t6.


Which term of the G.P. 5, 25, 125, 625, … is 510?


Find four numbers in G.P. such that sum of the middle two numbers is `10/3` and their product is 1


For a G.P. if S5 = 1023 , r = 4, Find a


For a G.P. If t4 = 16, t9 = 512, find S10


Determine whether the sum to infinity of the following G.P.s exist, if exists find them:

`-3, 1, (-1)/3, 1/9, ...`


Find : `sum_("r" = 1)^oo 4(0.5)^"r"`


Insert two numbers between 1 and −27 so that the resulting sequence is a G.P.


The sum of 3 terms of a G.P. is `21/4` and their product is 1 then the common ratio is ______.


Answer the following:

Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666, ...


Answer the following:

Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.


For a, b, c to be in G.P. the value of `(a - b)/(b - c)` is equal to ______.


For an increasing G.P. a1, a2 , a3 ........., an, if a6 = 4a4, a9 – a7 = 192, then the value of `sum_(i = 1)^∞ 1/a_i` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×