Advertisements
Advertisements
Question
For a G.P. If t3 = 20 , t6 = 160 , find S7
Solution
t3 = 20, t6 = 160
tn = arn–1
∴ t3 = ar3–1 = ar2
∴ ar2 = 20
∴ a = `20/"r"^2` ...(i)
Also, t6 = ar5
ar5 = 160
∴ `(20/"r"^2)"r"^5` = 160 ...[From (i)]
∴ r3 = `160/20` = 8
∴ r = 2
Substituting the value of r in (i) we get
a = `20/2^2` = 5
Now, Sn = `("a"("r"^"n"- 1))/("r" - 1)`, for r > 1
∴ S7 = `(5(2^7 - 1))/(2 - 1)`
= 5(128 – 1)
= 635
APPEARS IN
RELATED QUESTIONS
For what values of x, the numbers `-2/7, x, -7/2` are in G.P?
Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn – 1 and A, AR, AR2, … `AR^(n-1)` form a G.P, and find the common ratio
If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 .
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.
Find the sum of the following geometric series:
x3, x5, x7, ... to n terms
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.
Find the rational number whose decimal expansion is \[0 . 423\].
Find the rational numbers having the following decimal expansion:
\[3 . 5\overline 2\]
Find the rational numbers having the following decimal expansion:
\[0 . 6\overline8\]
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.
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.
If a, b, c, d are in G.P., prove that:
\[\frac{1}{a^2 + b^2}, \frac{1}{b^2 - c^2}, \frac{1}{c^2 + d^2} \text { are in G . P } .\]
If (a − b), (b − c), (c − a) are in G.P., then prove that (a + b + c)2 = 3 (ab + bc + ca)
If a, b, c are in G.P., then prove that:
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.
If a, b, c are in A.P., b,c,d are in G.P. and \[\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\] are in A.P., prove that a, c,e are in G.P.
If a, b, c are in A.P. and a, x, b and b, y, c are in G.P., show that x2, b2, y2 are in A.P.
Insert 5 geometric means between \[\frac{32}{9}\text{and}\frac{81}{2}\] .
The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is
For the G.P. if r = − 3 and t6 = 1701, find a.
The numbers 3, x, and x + 6 form are in G.P. Find x
The numbers 3, x, and x + 6 form are in G.P. Find nth term
For the following G.P.s, find Sn.
p, q, `"q"^2/"p", "q"^3/"p"^2,` ...
For the following G.P.s, find Sn
0.7, 0.07, 0.007, .....
Find the sum to n terms of the sequence.
0.5, 0.05, 0.005, ...
Find `sum_("r" = 0)^oo (-8)(-1/2)^"r"`
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 –
Select the correct answer from the given alternative.
Sum to infinity of a G.P. 5, `-5/2, 5/4, -5/8, 5/16,...` is –
Select the correct answer from the given alternative.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively. (a, b > 0)
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:
For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P.
Answer the following:
For a G.P. if t2 = 7, t4 = 1575 find a
Answer the following:
If p, q, r, s are in G.P., show that (p2 + q2 + r2) (q2 + r2 + s2) = (pq + qr + rs)2
Answer the following:
If p, q, r, s are in G.P., show that (pn + qn), (qn + rn) , (rn + sn) are also in G.P.
If a, b, c, d are in G.P., prove that a2 – b2, b2 – c2, c2 – d2 are also in G.P.
If pth, qth, and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab–c . bc – a . ca – b = 1
