Advertisements
Advertisements
Question
If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that p − q, q − r, r − s are in G.P.
Advertisements
Solution
\[\text { Here, } a_p = a + \left( p - 1 \right)d\]
\[ a_q = a + \left( q - 1 \right)d\]
\[ a_r = a + \left( r - 1 \right)d\]
\[ a_s = a + \left( s - 1 \right)d\]
\[\text { It is given that }a_p , a_q , a_r \text { and } a_s\text { are in G . P } . \]
\[ \therefore \frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_q - a_r}{a_p - a_q} = \frac{q - r}{p - q} . . . . . . . (i)\]
\[\text { Similarly }, \frac{a_r}{a_q} = \frac{a_s}{a_r} = \frac{a_r - a_s}{a_q - a_r} = \frac{r - s}{q - r} . . . . . . . (ii)\]
\[\text { Using }\left( i \right) \text { and }\left( ii \right): \]
\[\frac{q - r}{p - q} = \frac{r - s}{q - r}, \]
\[\text { Therefore }, p - q, q - r\text { and } r - s \text { are in G . P } .\]
APPEARS IN
RELATED QUESTIONS
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
Which term of the following sequence:
`sqrt3, 3, 3sqrt3`, .... is 729?
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 ninth term of the G.P. 1, 4, 16, 64, ...
Find:
the 10th term of the G.P.
\[- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .\]
Find :
the 12th term of the G.P.
\[\frac{1}{a^3 x^3}, ax, a^5 x^5 , . . .\]
Find the 4th term from the end of the G.P.
Find three numbers in G.P. whose sum is 38 and their product is 1728.
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:
4, 2, 1, 1/2 ... to 10 terms.
If S1, S2, S3 be respectively the sums of n, 2n, 3n terms of a G.P., then prove that \[S_1^2 + S_2^2\] = S1 (S2 + S3).
If a and b are the roots of x2 − 3x + p = 0 and c, d are the roots x2 − 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q − p) = 17 : 15.
Prove that: (91/3 . 91/9 . 91/27 ... ∞) = 3.
Express the recurring decimal 0.125125125 ... as a rational number.
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 are in G.P., prove that:
a (b2 + c2) = c (a2 + b2)
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:
(b + c) (b + d) = (c + a) (c + d)
If a, b, c, d are in G.P., prove that:
(a2 + b2), (b2 + c2), (c2 + d2) 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 the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z 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.
The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio `(3+2sqrt2):(3-2sqrt2)`.
If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is
If second term of a G.P. is 2 and the sum of its infinite terms is 8, then its first term is
If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then
If A be one A.M. and p, q be two G.M.'s between two numbers, then 2 A is equal to
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
Let x be the A.M. and y, z be two G.M.s between two positive numbers. Then, \[\frac{y^3 + z^3}{xyz}\] is equal to
In a G.P. if the (m + n)th term is p and (m − n)th term is q, then its mth term is
For the G.P. if r = `1/3`, a = 9 find t7
For what values of x, the terms `4/3`, x, `4/27` are in G.P.?
For a G.P. if S5 = 1023 , r = 4, Find a
Find the sum to n terms of the sequence.
0.5, 0.05, 0.005, ...
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`-3, 1, (-1)/3, 1/9, ...`
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
Insert two numbers between 1 and −27 so that the resulting sequence is a G.P.
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.
