Advertisements
Advertisements
Question
For a G.P. a = 2, r = `-2/3`, find S6
Solution
a = 2, r = `-2/3`
Sn = `("a"(1 - "r"^"n"))/(1 - "r")`, for r < 1
S6 = `(2[1 - (-2/3)^6])/(1 - (-2/3)`
= `(2[1 - (2/3)^6])/(5/3)`
= `6/5[(729 - 64)/3^6]`
= `6/5[665/729]`
S6 = `266/243`
APPEARS IN
RELATED QUESTIONS
Which term of the following sequence:
`2, 2sqrt2, 4,.... is 128`
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Find :
the 8th term of the G.P. 0.3, 0.06, 0.012, ...
Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?
If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal, find the value of n.
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 three numbers in G.P. whose sum is 65 and whose product is 3375.
The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.
Find the sum of the following geometric progression:
1, −1/2, 1/4, −1/8, ... to 9 terms;
Find the sum of the following geometric progression:
4, 2, 1, 1/2 ... to 10 terms.
Find the sum of the following geometric series:
\[\frac{a}{1 + i} + \frac{a}{(1 + i )^2} + \frac{a}{(1 + i )^3} + . . . + \frac{a}{(1 + i )^n} .\]
How many terms of the series 2 + 6 + 18 + ... must be taken to make the sum equal to 728?
The 4th and 7th terms of a G.P. are \[\frac{1}{27} \text { and } \frac{1}{729}\] respectively. Find the sum of n terms of the G.P.
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 \[\frac{1}{r^n}\].
If Sp denotes the sum of the series 1 + rp + r2p + ... to ∞ and sp the sum of the series 1 − rp + r2p − ... to ∞, prove that Sp + sp = 2 . S2p.
Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.
If a, b, c are in G.P., prove that:
\[a^2 b^2 c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3\]
If a, b, c are in G.P., prove that:
\[\frac{(a + b + c )^2}{a^2 + b^2 + c^2} = \frac{a + b + c}{a - b + c}\]
If the fifth term of a G.P. is 2, then write the product of its 9 terms.
Write the product of n geometric means between two numbers a and b.
If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]
Check whether the following sequence is G.P. If so, write tn.
7, 14, 21, 28, …
For the following G.P.s, find Sn
3, 6, 12, 24, ...
For a G.P. if a = 2, r = 3, Sn = 242 find n
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`-3, 1, (-1)/3, 1/9, ...`
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`1/5, (-2)/5, 4/5, (-8)/5, 16/5, ...`
Express the following recurring decimal as a rational number:
`2.3bar(5)`
Find : `sum_("n" = 1)^oo 0.4^"n"`
Select the correct answer from the given alternative.
The common ratio for the G.P. 0.12, 0.24, 0.48, is –
Answer the following:
Find the sum of the first 5 terms of the G.P. whose first term is 1 and common ratio is `2/3`
Answer the following:
For a sequence , if tn = `(5^("n" - 2))/(7^("n" - 3))`, verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Answer the following:
If for a G.P. t3 = `1/3`, t6 = `1/81` find r
In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. is ______.
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in ______.
If the expansion in powers of x of the function `1/((1 - ax)(1 - bx))` is a0 + a1x + a2x2 + a3x3 ....... then an is ______.
