मराठी

Find: the Ninth Term of the G.P. 1, 4, 16, 64, ... - Mathematics

Advertisements
Advertisements

प्रश्न

Find:
the ninth term of the G.P. 1, 4, 16, 64, ...

उत्तर

Here,

\[\text { First term, } a = 1 \]

\[\text { Common ratio }, r = \frac{a_2}{a_1} = \frac{4}{1} = 4\]

\[ \therefore 9th\text {  term } = a_9 = a r^{(9 - 1)} = 1(4 )^8 = 4^8 = 65536\]

\[\text { Thus, the 9th term of the given GP is } 65536 .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Geometric Progression - Exercise 20.1 [पृष्ठ १०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 20 Geometric Progression
Exercise 20.1 | Q 3.1 | पृष्ठ १०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the sum to indicated number of terms in the geometric progressions 1, – a, a2, – a3, ... n terms (if a ≠ – 1).


Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.


If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 .


if `(a+ bx)/(a - bx) = (b +cx)/(b - cx) = (c + dx)/(c- dx) (x != 0)` then show that a, b, c and d are in G.P.


If a and b are the roots of are roots of x2 – 3x + p = 0 , and c, d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17 : 15.


Find:

the 10th term of the G.P.

\[- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .\]

 


Which term of the G.P. :

\[2, 2\sqrt{2}, 4, . . .\text {  is }128 ?\]


If the pth and qth terms of a G.P. are q and p, respectively, then show that (p + q)th term is \[\left( \frac{q^p}{p^q} \right)^\frac{1}{p - q}\].


Find the sum of the following geometric progression:

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


How many terms of the series 2 + 6 + 18 + ... must be taken to make the sum equal to 728?


Find the sum :

\[\sum^{10}_{n = 1} \left[ \left( \frac{1}{2} \right)^{n - 1} + \left( \frac{1}{5} \right)^{n + 1} \right] .\]


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 serie to infinity:

\[\frac{1}{3} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^4} + \frac{1}{3^5} + \frac{1}{56} + . . . \infty\]


Find the rational numbers having the following decimal expansion: 

\[0 . \overline3\]


Find the rational numbers having the following decimal expansion: 

\[0 . 6\overline8\]


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


If a, b, c are in G.P., prove that:

a (b2 + c2) = c (a2 + b2)


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

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


The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio \[(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})\] .


If (p + q)th and (p − q)th terms of a G.P. are m and n respectively, then write is pth term.


If the first term of a G.P. a1a2a3, ... is unity such that 4 a2 + 5 a3 is least, then the common ratio of G.P. is


The value of 91/3 . 91/9 . 91/27 ... upto inf, is 


If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]


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 


Check whether the following sequence is G.P. If so, write tn.

3, 4, 5, 6, …


The numbers 3, x, and x + 6 form are in G.P. Find 20th term.


For the following G.P.s, find Sn.

`sqrt(5)`, −5, `5sqrt(5)`, −25, ...


Express the following recurring decimal as a rational number:

`2.3bar(5)`


The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of the areas of all the squares


The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of the perimeters of all the squares


Select the correct answer from the given alternative.

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:

For a G.P. a = `4/3` and t7 = `243/1024`, find the value of r


Answer the following:

Find three numbers in G.P. such that their sum is 35 and their product is 1000


Answer the following:

Find five numbers in G.P. such that their product is 243 and sum of second and fourth number is 10.


Answer the following:

For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P.


At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was Rs 1250, find the value at the end of 5 years.


The sum or difference of two G.P.s, is again a G.P.


The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×