मराठी

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 - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Let A and d be the first term and common difference respectively of an A.P. and x and R be the first term and common ratio respectively of the G.P.

∴ A + (p – 1)d = a   .....(i)

A + (q – 1)d = b   .....(ii)

And A + (r – 1)d = c   ......(iii)

For G.P., we have

xRp–1 = a  .....(iv)

xRq–1 = b  .....(v)

And xRr–1 = c   .....(vi)

Subtracting equation (ii) from equation (i) we get

(p – q)d = a – b   ......(vii)

Similarly, (q – r)d = b – c   ......(viii)

And (r – p)d = c – a   ......(ix)

Now we have to prove that

ab–c . bc–a . ca–b = 1

L.H.S. ab–c . bc–a . ca–b

= `[x"R"^(p - 1)]^((q - r)d) * [x"R"^(q - 1)]^((r - p)d) * [x"R"^(r - 1)]^((p - q)d)`  ....[From (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix)]

= `x^((q - r)d) * "R"^((p - 1) (q - r)d) * x^((r - p)d) * "R"^((q - 1) (r - p)d) * x^((p - q)d) * "R"^((r - 1)(p - q)d)`

= `x^((q - r)d + (r - p)d) "R"^((p - 1)(q - r)d + (q - 1)(r - p)d + (r - 1)(p - q)d)`

= `x^((q-r + r - p + p - q)d) * "R"^((pq - pr - q + r + qr - pq - r + p + pr + pr - qr - p + q)d)`

= `x^((0)d) * "R"^((0)d)`

= `x^0 * "R"^0`

= 1 R.H.S.

L.H.S. = R.H.S. 

Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Sequences and Series - Exercise [पृष्ठ १६२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 9 Sequences and Series
Exercise | Q 16 | पृष्ठ १६२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Which term of the following sequence:

`1/3, 1/9, 1/27`, ...., is `1/19683`?


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.


Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn


Find:

the 10th term of the G.P.

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

 


Find : 

nth term of the G.P.

\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]


Find :

the 10th term of the G.P.

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, . . .\]


Which term of the G.P. :

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}, . . . \text { is }\frac{1}{512\sqrt{2}}?\]


Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?


Which term of the G.P. :

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


Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?

 

If \[\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}\] (x ≠ 0), then show that abc and d are in G.P.


Find the sum of the following geometric series:

 0.15 + 0.015 + 0.0015 + ... to 8 terms;


Evaluate the following:

\[\sum^{11}_{n = 1} (2 + 3^n )\]


Find the sum of the following series:

7 + 77 + 777 + ... to n terms;


The sum of n terms of the G.P. 3, 6, 12, ... is 381. Find the value of n.


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).


Find the sum of 2n terms of the series whose every even term is 'a' times the term before it and every odd term is 'c' times the term before it, the first term being unity.


Find the sum of the following serie to infinity:

8 +  \[4\sqrt{2}\] + 4 + ... ∞


The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of the succeeding terms. Find the G.P.


The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.


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)


The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio `(3+2sqrt2):(3-2sqrt2)`.


If A1, A2 be two AM's and G1G2 be two GM's between and b, then find the value of \[\frac{A_1 + A_2}{G_1 G_2}\]


If A be one A.M. and pq be two G.M.'s between two numbers, then 2 A is equal to 


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


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


Find: `sum_("r" = 1)^10 5 xx 3^"r"`


Express the following recurring decimal as a rational number:

`0.bar(7)`


If the first term of the G.P. is 16 and its sum to infinity is `96/17` find the common ratio.


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


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:

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:

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


Let `{a_n}_(n = 0)^∞` be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all n ≥ 0. Then, `sum_(n = 2)^∞ a^n/7^n` is equal to ______.


If the sum of an infinite GP a, ar, ar2, ar3, ...... . is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, .... is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×