हिंदी

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.

Advertisements
Advertisements

प्रश्न

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.

Advertisements

उत्तर

Let the first term of an A.P. be a and its common difference be d.

\[a_1 + a_2 + a_3 = 15\]

\[ \Rightarrow a + \left( a + d \right) + \left( a + 2d \right) = 15\]

\[ \Rightarrow 3a + 3d = 15 \]

\[ \Rightarrow a + d = 5 . . . . . . . (i)\]

\[\text { Now, according to the question }: \]

\[a + 1, a + d + 3 \text { and  }a + 2d + 9 \text { are in G . P }  . \]

\[ \Rightarrow \left( a + d + 3 \right)^2 = \left( a + 1 \right)\left( a + 2d + 9 \right)\]

\[ \Rightarrow \left( 5 - d + d + 3 \right)^2 = \left( 5 - d + 1 \right) \left( 5 - d + 2d + 9 \right) \left[ \text { From } (i) \right] \]

\[ \Rightarrow \left( 8 \right)^2 = \left( 6 - d \right)\left( 14 + d \right)\]

\[ \Rightarrow 64 = 84 + 6d - 14d - d^2 \]

\[ \Rightarrow d^2 + 8d - 20 = 0\]

\[ \Rightarrow \left( d - 2 \right)\left( d + 10 \right) = 0\]

\[ \Rightarrow d = 2, - 10\]

\[\text { Now, putting } d = 2, - 10 \text { in equation (i), we get, a } = 3, 15,\text {  respectively } . \]

\[\text { Thus, for } a = 3 \text { and  }d = 2, \text { the A . P . is } 3, 5, 7 . \]

\[\text { And, for a = 15 and d = - 10, the A . P . is }15 , 5, - 5 . \]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Geometric Progression - Exercise 20.5 [पृष्ठ ४५]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.5 | Q 4 | पृष्ठ ४५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7thterm.


If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that `a^(q - r) b^(r-p) c^(p-q) = 1`.


Find the value of n so that  `(a^(n+1) + b^(n+1))/(a^n + b^n)` may be the geometric mean between a and b.


If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) 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.


Show that the sequence <an>, defined by an = \[\frac{2}{3^n}\], n ϵ N is a G.P.


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.

\[\frac{1}{a^3 x^3}, ax, a^5 x^5 , . . .\]


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


If 5th, 8th and 11th terms of a G.P. are p. q and s respectively, prove that q2 = ps.


Find the sum of the following geometric series:

1, −a, a2, −a3, ....to n terms (a ≠ 1)


Find the sum of the following geometric series:

`sqrt7, sqrt21, 3sqrt7,...` to n terms


Find the sum of the following serie:

5 + 55 + 555 + ... to n terms;


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 \[\frac{1}{r^n}\].


How many terms of the G.P. `3, 3/2, 3/4` ..... are needed to give the sum `3069/512`?


If S1, S2, ..., Sn are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S1 + S2 + 2S3 + 3S4 + ... (n − 1) Sn = 1n + 2n + 3n + ... + nn.


Let an be the nth term of the G.P. of positive numbers.

Let \[\sum^{100}_{n = 1} a_{2n} = \alpha \text { and } \sum^{100}_{n = 1} a_{2n - 1} = \beta,\] such that α ≠ β. Prove that the common ratio of the G.P. is α/β.


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

 (a + b + c + d)2 = (a + b)2 + 2 (b + c)2 + (c + d)2


If a, b, c are in G.P., prove that the following is also in G.P.:

a3, b3, c3


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 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 pth, qth and rth terms of a G.P. re x, y, z respectively, then write the value of xq − r yr − pzp − q.

 

 

 


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.


Find four numbers in G.P. such that sum of the middle two numbers is `10/3` and their product is 1


The number of bacteria in a culture doubles every hour. If there were 50 bacteria originally in the culture, how many bacteria will be there at the end of 5th hour?


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.

`sqrt(5)`, −5, `5sqrt(5)`, −25, ...


Determine whether the sum to infinity of the following G.P.s exist, if exists find them:

`2, 4/3, 8/9, 16/27, ...`


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


Insert two numbers between 1 and −27 so that the resulting sequence is a G.P.


Select the correct answer from the given alternative.

Which term of the geometric progression 1, 2, 4, 8, ... is 2048


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:

In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term


Answer the following:

Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.


Let A1, A2, A3, .... be an increasing geometric progression of positive real numbers. If A1A3A5A7 = `1/1296` and A2 + A4 = `7/36`, then the value of A6 + A8 + A10 is equal to ______. 


If the expansion in powers of x of the function `1/((1 - ax)(1 - bx))` is a0 + a1x + a2x2 + a3x3 ....... then an is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×