Q1 |
Find the 20th and nthterms of the G.P. |
Ans: |
Here, a = First Term = ![5 over 2](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=f3efef5f99ecb0b6a36ec7add667024d.png)
r = common ratio = ![fraction numerator begin display style 5 over 4 end style over denominator begin display style 5 over 2 end style end fraction equals 1 half](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=1f68f337df090f6c1d0a76bc5ed5d8ff.png)
![a subscript 20 space space end subscript equals space a r to the power of 20 minus 1 end exponent equals 5 over 2 open parentheses 1 half close parentheses to the power of 19 equals fraction numerator 5 over denominator open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of 19 end fraction equals 5 over open parentheses 2 close parentheses to the power of 20
a subscript n space equals space a r to the power of n minus 1 end exponent equals 5 over 2 open parentheses 1 half close parentheses to the power of n minus 1 end exponent equals fraction numerator 5 over denominator open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of n minus 1 end exponent end fraction equals 5 over open parentheses 2 close parentheses to the power of n](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=8fd457c81607e07b9830d7ae500d04f0.png) |
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Q2 |
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. |
Ans: |
Common ratio, r = 2
Let a be the first term of the G.P.
∴ a8 = ar 8–1 = ar7
⇒ ar7 = 192
a(2)7 = 192
a(2)7 = (2)6 (3)
![rightwards double arrow a equals fraction numerator open parentheses 2 close parentheses to the power of 6 cross times 3 space over denominator open parentheses 2 close parentheses to the power of 7 end fraction equals 3 over 2
s o
a subscript 12 space equals a r to the power of 12 minus 1 end exponent space equals space open parentheses 3 over 2 close parentheses open parentheses 2 close parentheses to the power of 11 equals open parentheses 3 close parentheses open parentheses 2 close parentheses to the power of 10 equals 3072](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=2620935faf469808de3edd1f4ff2ebe9.png)
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Q3 |
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. |
Ans: |
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a5 = a r5–1 = a r4 = p … (1)
a8 = a r8–1 = a r7 = q … (2)
a11 = a r11–1 = a r10 = s … (3)
Dividing equation (2) by (1), we obtain
![fraction numerator a r to the power of 7 over denominator a r to the power of 4 end fraction equals q over p
r cubed equals q over p space space space space space space... left parenthesis 4 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=fc2b37a5f2860663ea461d8b5e9f15e6.png)
Dividing equation (3) by (2), we obtain
![fraction numerator a r to the power of 10 over denominator a r to the power of 7 end fraction equals s over q
rightwards double arrow r cubed equals s over q space space space space space space space... left parenthesis 5 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=ae65e09f8c53db609deaffb55cd14f2a.png)
Equating the values of r3 obtained in (4) and (5), we obtain
![q over p equals s over q
rightwards double arrow q squared equals p s](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=88bca06894083b68a48b239bcdbddf28.png)
Thus, the given result is proved.
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Q4 |
The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term. |
Ans: |
Let a be the first term and r be the common ratio of the G.P.
∴ a = –3
It is known that, an = arn–1
∴a4 = ar3 = (–3) r3
a2 = a r1 = (–3) r
According to the given condition,
(–3) r3 = [(–3) r]2
⇒ –3r3 = 9 r2
⇒ r = –3
a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187
Thus, the seventh term of the G.P. is –2187.
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Q5 |
Which term of the following sequences:
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Ans: |
(a) The given sequence is ![2 comma space 2 square root of 2 comma 4 comma space...](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=d0586b3a54eb2de7b1987412568aa929.png)
Here, a = 2 and r = ![fraction numerator 2 square root of 2 over denominator 2 end fraction equals square root of 2](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=cdd8f88fc523b2a176bb95e48c0f9b04.png)
Let the nth term of the given sequence be 128.
![a subscript n space end subscript equals space a r to the power of n minus 1 end exponent
rightwards double arrow open parentheses 2 close parentheses open parentheses square root of 2 close parentheses to the power of n minus 1 end exponent space equals space 128
rightwards double arrow open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 2 close parentheses to the power of 7
rightwards double arrow open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction plus 1 end exponent equals open parentheses 2 close parentheses to the power of 7
s o space fraction numerator n minus 1 over denominator 2 end fraction plus 1 equals 7
rightwards double arrow fraction numerator n minus 1 over denominator 2 end fraction space equals space 6
rightwards double arrow n space minus 1 space equals space 12
rightwards double arrow n equals 13](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=41416d6f04d183017279e169db0a521a.png)
Thus, the 13th term of the given sequence is 128.
(b) The given sequence is ![square root of 3 space comma space 3 comma space 3 square root of 3 space comma space... space i s space 729 ?](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=b4560864e92d551c999aac06b8c0aed0.png)
Here, a= ![square root of 3 space a n d space r space equals space fraction numerator 3 over denominator square root of 3 end fraction space equals space square root of 3](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=570bbc03eb908fafe60b87232d5a60ef.png)
Let the nth term of the given sequence be 729.
![a subscript n equals a r to the power of n minus 1 end exponent
s o
a r to the power of n minus 1 end exponent space equals space 729
rightwards double arrow open parentheses square root of 3 close parentheses space open parentheses square root of 3 close parentheses to the power of n minus 1 end exponent space equals space 729
rightwards double arrow open parentheses 3 close parentheses to the power of 1 half end exponent open parentheses 3 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6
rightwards double arrow open parentheses 3 close parentheses to the power of 1 half plus fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6
s o
1 half plus fraction numerator n minus 1 over denominator 2 end fraction equals 6
rightwards double arrow fraction numerator 1 plus n minus 1 over denominator 2 end fraction equals 6
rightwards double arrow n equals 12](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=43c7dea43e3209b677ae600787ae1351.png)
Thus, the 12th term of the given sequence is 729.
(c) The given sequence is ![1 third comma space 1 over 9 comma space 1 over 27 comma space... space](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=9f49ac25c1944008eccc7f877912a350.png)
Here,
a=![1 third space space a n d space r space equals space 1 over 9 obelus divided by 1 third equals 1 third](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=37b52cb17825202cfa8b180e5afabd6a.png)
Let the nth term of the given sequence be .
![a subscript n space equals space a r to the power of n minus 1 end exponent
s o
a r to the power of n minus 1 end exponent equals 1 over 19683
rightwards double arrow open parentheses 1 third close parentheses open parentheses 1 third close parentheses to the power of n minus 1 end exponent equals 1 over 19683
rightwards double arrow open parentheses 1 third close parentheses to the power of n equals open parentheses 1 third close parentheses to the power of 9
rightwards double arrow n equals 9](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=ddac4cea1b7e3a4d0c7a140918d0cf2e.png)
Thus, the 9th term of the given sequence is .
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Q6 |
For what values of x, the numbers are in G.P? |
Ans: |
The given numbers are .
Common ratio = ![fraction numerator x over denominator begin display style fraction numerator minus 2 over denominator 7 end fraction end style end fraction equals fraction numerator minus 7 x over denominator 2 end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=5e149606f33aafcd52bb9633fe4cc829.png)
Also, common ratio = ![fraction numerator minus 7 over denominator begin display style 2 over x end style end fraction equals fraction numerator minus 7 over denominator 2 x end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=c62aa2719afb0c8bcf8fd72ce7f16d35.png)
so
![fraction numerator minus 7 x over denominator 2 end fraction equals fraction numerator minus 7 over denominator 2 x end fraction
rightwards double arrow x squared equals fraction numerator minus 2 space cross times 7 over denominator minus 2 space cross times 7 end fraction space equals 1
rightwards double arrow x space equals space square root of 1 space
rightwards double arrow x space equals space plus-or-minus 1 space](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=48a0931092f2fc17e674c3341353bc49.png)
Thus, for x = ± 1, the given numbers will be in G.P.
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Q7 |
Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 … |
Ans: |
The given G.P. is 0.15, 0.015, 0.00015, …
Here, a = 0.15 and r = ![fraction numerator 0.015 over denominator 0.15 end fraction equals 0.1](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=66689cbb1cdf887669328adaa852eba5.png)
![S subscript n space end subscript equals space fraction numerator a space open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction
s o
S subscript 20 space equals space fraction numerator 0.15 open square brackets 1 minus open parentheses 0.1 close parentheses to the power of 20 close square brackets over denominator 1 minus 0.1 end fraction
space space space space space space space space space equals space fraction numerator 0.15 over denominator 0.9 end fraction open square brackets 1 minus open parentheses 0.1 close parentheses to the power of 20 close square brackets
space space space space space space space space space equals space 15 over 90 space open square brackets 1 space minus space open parentheses 0.1 close parentheses to the power of 20 close square brackets
space space space space space space space space space equals space 1 over 6 open square brackets 1 minus open parentheses 0.1 to the power of 20 close parentheses close square brackets space](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=b0820a1e8768b2ae60033f3511464348.png)
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Q8 |
Find the sum to n terms in the geometric progression |
Ans: |
The given G.P. is ![square root of 7 space comma space square root of 21 space comma space 3 square root of 7 space comma space...](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=3b795b0552c993246dd0acccd6a67abd.png)
Here, a = ![square root of 7](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=63542b2939883ce8423513412a544f8b.png)
![r space equals fraction numerator square root of 21 over denominator square root of 7 end fraction space equals space square root of 3
S subscript n space space end subscript equals space fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction
S o
S subscript n space equals space fraction numerator square root of 7 space open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus square root of 3 end fraction
space space space space space space equals space fraction numerator square root of 7 space open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus square root of 3 end fraction cross times fraction numerator 1 plus square root of 3 over denominator 1 space plus space square root of 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses B y space r a t i o n a l i z i n g close parentheses
space space space space space space space equals space fraction numerator square root of 7 space open parentheses 1 space plus square root of 3 close parentheses open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus 3 end fraction
space space space space space space space space equals space fraction numerator minus square root of 7 space open parentheses 1 space plus space square root of 3 close parentheses over denominator 2 end fraction open square brackets 1 minus open parentheses 3 close parentheses to the power of n over 2 end exponent close square brackets
space space space space space space space space equals space fraction numerator square root of 7 space open parentheses 1 plus square root of 3 close parentheses over denominator 2 end fraction open square brackets open parentheses 3 close parentheses to the power of n over 2 end exponent minus 1 close square brackets](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=9137ab0b4bbdac546bd4343e701fb260.png)
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Q9 |
Find the sum to n terms in the geometric progression 1,-a, a2,-a3, ... (if a ≠ -1) |
Ans: |
The given G.P. is 1,-a, a2,-a3, .........
Here, first term = a1 = 1
Common ratio = r = – a
![S subscript n equals fraction numerator a subscript 1 open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction
s o
S subscript n equals fraction numerator 1 open square brackets 1 minus open parentheses minus a close parentheses to the power of n close square brackets over denominator 1 minus open parentheses minus a close parentheses end fraction space equals space fraction numerator open square brackets 1 minus open parentheses minus a close parentheses to the power of n close square brackets over denominator 1 plus a end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=ad3167f3c190d8c5335708c5c88a1103.png) |
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Q10 |
Find the sum to n terms in the geometric progression x3, x5, x7 ... (if x ≠ ±1) |
Ans: |
The given G.P. is x3, x5, x7 ........
Here, a = x3 and r = x2
![S subscript n space end subscript equals space fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction equals fraction numerator x cubed open square brackets 1 minus open parentheses x squared close parentheses to the power of n close square brackets over denominator 1 minus x squared end fraction equals fraction numerator x cubed open parentheses 1 minus x to the power of 2 n end exponent close parentheses over denominator 1 minus x squared end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=5274ed2647fad91602ba22989287b7fc.png) |
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Q11 |
Evaluate
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Ans: |
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Q12 |
The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms. |
Ans: |
Let be the first three terms of the G.P.
![a over r space plus space a space plus space a r space equals space 39 over 10 space space space space space space... space left parenthesis 1 right parenthesis
open parentheses a over r close parentheses open parentheses a close parentheses open parentheses a r close parentheses space equals space 1 space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=12000cdb9a56127ab562a0234cc1752e.png)
From (2), we obtain
a3 = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
![1 over r space plus space 1 space plus space r space equals space 39 over 10
rightwards double arrow 1 space plus space r space plus space r squared space equals space 39 over 10 r
rightwards double arrow 10 space plus space 10 r space plus space 10 r squared space minus 39 r space equals 0
rightwards double arrow 10 r squared space minus 29 r space plus space 10 space equals 0
rightwards double arrow space 10 r squared space minus space 25 r space minus space 4 r space plus space 10 space equals space 0
rightwards double arrow 5 r space open parentheses 2 r space minus 5 close parentheses space minus space 2 open parentheses 2 r space minus space 5 close parentheses space equals space 0
rightwards double arrow open parentheses 5 r space minus space 2 close parentheses open parentheses 2 r minus 5 close parentheses space equals space 0
rightwards double arrow r space equals space 2 over 5 space o r space 5 over 2
T h u s comma space t h e space t h r e e space t e r m s space o f space G. P. space a r e space 5 over 2 comma space 1 space comma space a n d space 2 over 5. space space](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=b529f7ec444f775234e95c938ca4e2a4.png)
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Q13 |
How many terms of G.P. 3, 32, 33, … are needed to give the sum 120? |
Ans: |
The given G.P. is 3, 32, 33, …
Let n terms of this G.P. be required to obtain the sum as 120.
![S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r minus 1 end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=02904ed8cb681a8eed9b269a3ebc47b0.png)
Here, a = 3 and r = 3
∴ ![S subscript n space equals space 120 space equals space fraction numerator 3 space open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fraction
rightwards double arrow 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 2 end fraction
rightwards double arrow fraction numerator 120 space cross times 2 over denominator 3 end fraction space equals space 3 to the power of n space minus 1
rightwards double arrow 3 to the power of n space minus 1 space equals space 80
rightwards double arrow 3 to the power of n space equals space 81
rightwards double arrow 3 to the power of n space equals space 3 to the power of 4](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=d0626e284243d7097c9d54c768306a47.png)
∴ n = 4
Thus, four terms of the given G.P. are required to obtain the sum as 120.
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Q14 |
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. |
Ans: |
Let the G.P. be a, ar, ar2, ar3, …
According to the given condition,
a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128
⇒ a (1 + r + r2) = 16 … (1)
ar3(1 + r + r2) = 128 … (2)
Dividing equation (2) by (1), we obtain
![fraction numerator a r cubed open parentheses 1 plus r plus r squared close parentheses over denominator a open parentheses 1 plus r plus r squared close parentheses end fraction space equals space 128 over 16
rightwards double arrow space r cubed space equals space 8
therefore space r space equals space 2](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=77c7dd288bbfa2e994452d8b55e267a1.png)
Substituting r = 2 in (1), we obtain
⇒ a (7) = 16
![rightwards double arrow a space equals space 16 over 7
S subscript n space end subscript equals space fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction
rightwards double arrow S subscript n space equals space fraction numerator 16 space over denominator 7 end fraction fraction numerator open parentheses 2 to the power of n space minus 1 close parentheses over denominator 2 minus 1 end fraction space equals space 16 over 7 space open parentheses 2 to the power of n space minus 1 close parentheses](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=f77314b2b20e226ee05bd76c6f1dbbbe.png)
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Q15 |
Given a G.P. with a = 729 and 7th term 64, determine S7. |
Ans: |
a = 729
a7 = 64
Let r be the common ratio of the G.P.
It is known that, an = a rn–1
a7 = ar7–1 = (729)r6
⇒ 64 = 729 r6
![rightwards double arrow r to the power of 6 space equals space 64 over 729
rightwards double arrow r to the power of 6 space space equals space open parentheses 2 over 3 close parentheses to the power of 6
rightwards double arrow r space equals space 2 over 3](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=9e21cf0c4263b23279a2d80c6629b5ad.png)
Also, it is known that,
![S subscript n space equals space fraction numerator a space open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction
therefore space S subscript 7 space equals space fraction numerator 729 open square brackets 1 minus open parentheses begin display style 2 over 3 end style close parentheses to the power of 7 close square brackets over denominator 1 minus begin display style 2 over 3 end style end fraction
space space space space space space space space space space space space equals space 3 space cross times space 729 space open square brackets 1 minus open parentheses 2 over 3 close parentheses to the power of 7 close square brackets
space space space space space space space space space space space space equals open parentheses 3 close parentheses to the power of 7 space open square brackets fraction numerator open parentheses 3 close parentheses to the power of 7 space minus space open parentheses 2 close parentheses to the power of 7 over denominator open parentheses 3 close parentheses to the power of 7 end fraction close square brackets
space space space space space space space space space space space space equals space open parentheses 3 close parentheses to the power of 7 space minus open parentheses 2 close parentheses to the power of 7
space space space space space space space space space space space space space equals space 2187 space minus space 128
space space space space space space space space space space space space space equals space 2059](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=62a294d9f446b4e4b71fb443340ca94f.png)
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Q16 |
Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term. |
Ans: |
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
![S subscript 2 space equals space minus 4 space equals space fraction numerator a open parentheses 1 minus r squared close parentheses over denominator 1 minus r end fraction space space space space space space space space space... left parenthesis 1 right parenthesis
a subscript 5 space equals space 4 space cross times space a subscript 3
a r to the power of 4 space equals space 4 a r squared
rightwards double arrow r squared space equals space 4
therefore space r space equals space plus-or-minus 2](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=6443ab7bc30f73851196327d50c8626b.png)
From (1), we obtain
.....
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Q17 |
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. |
Ans: |
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a4 = a r3 = x … (1)
a10 = a r9 = y … (2)
a16 = a r15 = z … (3)
Dividing (2) by (1), we obtain
![y over x space equals space fraction numerator a r to the power of 9 over denominator a r cubed end fraction space rightwards double arrow y over x space equals space r to the power of 6](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=eb75b6c46b74b876ebe95c32d8a3035d.png)
Dividing (3) by (2), we obtain
![z over y space equals space fraction numerator a r to the power of 15 over denominator a r to the power of 9 end fraction rightwards double arrow z over y equals r to the power of 6](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=e805918ae67c3db225b64d6e54375a95.png)
∴ ![y over x space equals space z over y](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=a866de7a2afc67af89e0a42dcad2b67e.png)
Thus, x, y, z are in G. P.
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Q18 |
Find the sum to n terms of the sequence, 8, 88, 888, 8888… |
Ans: |
The given sequence is 8, 88, 888, 8888…
This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
![equals 8 over 9 open square brackets 9 plus 99 plus 999 plus 9999 plus.......... t o space n space t e r m s close square brackets
equals 8 over 9 open square brackets open parentheses 10 minus 1 close parentheses plus open parentheses 10 squared minus 1 close parentheses plus open parentheses 10 cubed minus 1 close parentheses plus open parentheses 10 to the power of 4 minus 1 close parentheses plus........ t o to the power of space n space t e r m s close square brackets
equals 8 over 9 open square brackets open parentheses 10 plus 10 squared space plus space....... n space t e r m s close parentheses space minus space open parentheses 1 plus 1 plus 1 plus... n space t e r m s close parentheses close square brackets
equals 8 over 9 open square brackets fraction numerator 10 open parentheses 10 to the power of n minus 1 close parentheses over denominator 10 minus 1 end fraction space minus space n close square brackets
equals 8 over 9 open square brackets fraction numerator 10 open parentheses 10 to the power of n minus 1 close parentheses over denominator 9 end fraction space minus space n close square brackets
equals 80 over 81 open parentheses 10 to the power of n minus 1 close parentheses space minus space 8 over 9 n](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=e8b29d2b02b90b86bde85f422fd59ba0.png)
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Q19 |
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, . |
Ans: |
Required sum = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 ×![1 half](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=41891f5e8ace60c819a1c76af3792a55.png)
![equals space 64 space open square brackets 4 space plus space 2 space plus space 1 space plus 1 half space plus space 1 over 2 squared close square brackets](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=b327cbaaa6c21348f5f8bb8986328cb8.png)
Here, 4, 2, 1, is a G.P.
First term, a = 4
Common ratio, r =![1 half](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=41891f5e8ace60c819a1c76af3792a55.png)
It is known that,
![S subscript n space equals space fraction numerator a space open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction
therefore space S subscript 5 space equals space fraction numerator 4 open square brackets 1 minus open parentheses begin display style 1 half end style close parentheses to the power of 5 close square brackets over denominator 1 minus begin display style 1 half end style end fraction equals fraction numerator 4 open square brackets 1 minus begin display style 1 over 32 end style close square brackets over denominator begin display style 1 half end style end fraction equals fraction numerator 8 open parentheses 32 minus 1 close parentheses over denominator 32 end fraction equals 31 over 4
therefore space R e q u i r e d space s u m space equals space 64 space open parentheses 31 over 4 close parentheses space equals space open parentheses 16 close parentheses open parentheses 31 close parentheses space equals 496](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=5d26dff8a86485ca4731c0aa0927509c.png)
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Q20 |
Show that the products of the corresponding terms of the sequences a,ar,ar2, ...arn-1 and A, AR, AR2, ,,,ARn-1 form a G.P, and find the common ratio. |
Ans: |
It has to be proved that the sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P.
![fraction numerator S e c o n d space t e r m over denominator F i r s t space t e r m end fraction space equals space fraction numerator a r A R over denominator a A end fraction equals r R
fraction numerator T h i r d space t e r m over denominator S e c o n d space t e r m end fraction space equals space fraction numerator a r squared A R squared over denominator a r A R end fraction equals r R](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=81cb688125e7ddec80bf2224504ef914.png)
Thus, the above sequence forms a G.P. and the common ratio is rR.
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Q21 |
Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18. |
Ans: |
Let a be the first term and r be the common ratio of the G.P.
a1 = a, a2 = ar, a3 = ar2, a4 = ar3
By the given condition,
a3 = a1 + 9
⇒ ar2 = a + 9 … (1)
a2 = a4 + 18
⇒ ar = ar3 + 18 … (2)
From (1) and (2), we obtain
a(r2 – 1) = 9 … (3)
ar (1– r2) = 18 … (4)
Dividing (4) by (3), we obtain
![fraction numerator a r open parentheses 1 minus r squared close parentheses over denominator a open parentheses r squared minus 1 close parentheses end fraction equals 18 over 9
rightwards double arrow minus r space equals space 2
rightwards double arrow r space equals space minus 2](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=f43f302afb3506f4b32993cfbe058349.png)
Substituting the value of r in (1), we obtain
4a = a + 9
⇒ 3a = 9
∴ a = 3
Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 i.e., 3¸–6, 12, and –24.
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Q22 |
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-rbr-pcp-q=1 |
Ans: |
Let A be the first term and R be the common ratio of the G.P.
According to the given information,
ARp–1 = a
ARq–1 = b
ARr–1 = c
aq–r br–p cp–q
= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r-p) × Ap–q × R(r –1)(p–q)
= Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)
= A0 × R0
= 1
Thus, the given result is proved.
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Q23 |
If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. |
Ans: |
The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.
b = arn–1 … (1)
P = Product of n terms
= (a) (ar) (ar2) … (arn–1)
= (a × a ×…a) (r × r2 × …rn–1)
= an r 1 + 2 +…(n–1) … (2)
Here, 1, 2, …(n – 1) is an A.P.
∴1 + 2 + ……….+ (n – 1)
![equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 plus left parenthesis n minus 1 minus 1 right parenthesis cross times 1 close square brackets equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 plus n minus 2 close square brackets equals fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction
P space equals space a to the power of n r to the power of fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction end exponent
therefore space P squared space equals space a to the power of 2 n end exponent space r to the power of n open parentheses n minus 1 close parentheses end exponent
space space space space space space space space space space space space equals space open square brackets a squared space r to the power of left parenthesis n minus 1 right parenthesis end exponent close square brackets to the power of n
space space space space space space space space space space space space equals open square brackets a cross times a r to the power of n minus 1 end exponent close square brackets to the power of n
space space space space space space space space space space space space equals open parentheses a b close parentheses to the power of n space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=887054ac43407f4faabce5755dc76a28.png)
Thus, the given result is proved.
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Q24 |
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
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Ans: |
Let a be the first term and r be the common ratio of the G.P.
![S u m space o f space f i r s t space n space t e r m s space equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=a4911f9583980a63e2fddd105de03e47.png)
Since there are n terms from (n +1)th to (2n)th term,
Sum of terms from(n + 1)th to (2n)th term =![fraction numerator a subscript n plus 1 end subscript open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=4b727592dccf9bf72a3c098ed93b0cfe.png)
a n +1 = ar n + 1 – 1 = arn
Thus, required ratio = ![fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator open parentheses 1 minus r close parentheses end fraction cross times fraction numerator open parentheses 1 minus r close parentheses over denominator a r to the power of n open parentheses 1 minus r to the power of n close parentheses end fraction equals 1 over r to the power of n](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=a571618e87c0d0d9d660c7c35a5b3671.png)
Thus, 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 .
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Q25 |
If a, b, c and d are in G.P. show that
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Ans: |
a, b, c, d are in G.P.
Therefore,
bc = ad … (1)
b2 = ac … (2)
c2 = bd … (3)
It has to be proved that,
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
R.H.S.
= (ab + bc + cd)2
= (ab + ad + cd)2 [Using (1)]
= [ab + d (a + c)]2
= a2b2 + 2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
[Using (2) and (3) and rearranging terms]
= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2)
= L.H.S.
∴ L.H.S. = R.H.S.
∴ ![open parentheses a squared plus b squared plus c squared close parentheses open parentheses b squared plus c squared plus d squared close parentheses equals open parentheses a b plus b c plus c d close parentheses squared](https://www.saralstudy.com/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=2264805241d83a565ba83837ce598842.png)
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Q26 |
Insert two numbers between 3 and 81 so that the resulting sequence is G.P. |
Ans: |
Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1, G2, 81, forms a G.P.
Let a be the first term and r be the common ratio of the G.P.
∴81 = (3) (r)3
⇒ r3 = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G1 = ar = (3) (3) = 9
G2 = ar2 = (3) (3)2 = 27
Thus, the required two numbers are 9 and 27.
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Q27 |
Find the value of n so that may be the geometric mean between a and b. |
Ans: |
G. M. of a and b is .
By the given condition, ![fraction numerator a to the power of n plus 1 end exponent space plus space b to the power of n plus 1 end exponent over denominator a to the power of n space plus space b to the power of n end fraction equals square root of a b end root](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=026402be541bed3eb0d088e635df0972.png)
Squaring both sides, we obtain
![open parentheses a to the power of n plus 1 end exponent space plus space b to the power of n plus 1 end exponent close parentheses squared over open parentheses a to the power of n space plus space b to the power of n close parentheses squared equals a b
rightwards double arrow a to the power of 2 n plus 2 end exponent space plus space 2 a to the power of n plus 1 end exponent b to the power of n plus 1 end exponent space plus space b to the power of 2 n plus 2 end exponent space equals space open parentheses a b close parentheses open parentheses a to the power of 2 n end exponent space plus space 2 a to the power of n b to the power of n space plus space b to the power of 2 n end exponent close parentheses
rightwards double arrow a to the power of 2 n plus 2 end exponent space plus space 2 a to the power of n plus 1 end exponent b to the power of n plus 1 end exponent space plus space b to the power of 2 n plus 2 end exponent space equals space a to the power of 2 n plus 1 end exponent b space plus space 2 a to the power of n plus 1 end exponent b to the power of n plus 1 end exponent space plus space a b to the power of 2 n plus 1 end exponent
rightwards double arrow a to the power of 2 n plus 2 end exponent space plus space b to the power of 2 n plus 2 end exponent space equals space a to the power of 2 n plus 1 end exponent b space plus space a b to the power of 2 n plus 1 end exponent
rightwards double arrow a to the power of 2 n plus 2 end exponent space minus a to the power of 2 n plus 1 end exponent b space equals space a b to the power of 2 n plus 1 end exponent space minus space b to the power of 2 n plus 2 end exponent
rightwards double arrow space a to the power of 2 n plus 1 end exponent open parentheses a minus b close parentheses equals b to the power of 2 n plus 1 end exponent open parentheses a minus b close parentheses
rightwards double arrow open parentheses a over b close parentheses to the power of 2 n plus 1 end exponent space equals space 1 space equals space open parentheses a over b close parentheses to the power of 0
rightwards double arrow 2 n space plus 1 space equals space 0
rightwards double arrow n space equals space fraction numerator minus 1 over denominator 2 end fraction](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=18d5ea0635ce7374ce080cc9829c30ee.png)
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Q28 |
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio |
Ans: |
Let the two numbers be a and b.
G.M. = ![square root of a b end root](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=00ff5a829b5bcebf0fb35081ac8c10a1.png)
According to the given condition,
![a space plus space b space equals space 6 square root of a b end root space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
rightwards double arrow open parentheses a space plus space b close parentheses squared space equals space 36 open parentheses a b close parentheses](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=134adcda9b488db0bf29f202a19d4290.png)
Also,
![open parentheses a minus b close parentheses squared space equals space open parentheses a plus b close parentheses squared space minus space 4 a b space equals space 36 a b space minus 4 a b space equals space 32 a b
rightwards double arrow a minus b space equals space square root of 32 space square root of a b end root space
equals space 4 space square root of 2 space square root of a b end root space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=1d4eab354c2a514fd483bcc498ea5b50.png)
Adding (1) and (2), we obtain
![2 a space equals space open parentheses 6 space plus space 4 square root of 2 close parentheses square root of a b end root
rightwards double arrow a space equals space open parentheses 3 space plus space 2 square root of 2 close parentheses square root of a b end root](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=dd2498c74fb1f3b703f19a7790058df8.png)
Substituting the value of a in (1), we obtain
![b space equals space 6 square root of a b end root space minus space open parentheses 3 plus 2 square root of 2 close parentheses square root of a b end root
rightwards double arrow b space equals space open parentheses 3 space minus space 2 square root of 2 close parentheses space square root of a b end root
a over b space equals space fraction numerator open parentheses 3 space plus 2 square root of 2 close parentheses square root of a b end root over denominator open parentheses 3 minus 2 square root of 2 close parentheses square root of a b end root end fraction space equals fraction numerator 3 space plus space 2 square root of 2 over denominator 3 minus 2 square root of 2 end fraction space](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=ebdf60544dba2288e20b14fa242b7a88.png)
Thus, the required ratio is .
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Q29 |
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . |
Ans: |
It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.
∴ ![A M space equals space A space equals space fraction numerator a space plus space b over denominator 2 end fraction space space space space space space space space space space space space... left parenthesis 1 right parenthesis
G M space equals space G space equals space square root of a b end root space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=85f72245750cb828e42d58a02f2d4045.png)
From (1) and (2), we obtain
a + b = 2A … (3)
ab = G2 … (4)
Substituting the value of a and b from (3) and (4) in the identity (a – b)2 = (a + b)2 – 4ab, we obtain
(a – b)2 = 4A2 – 4G2 = 4 (A2–G2)
(a – b)2 = 4 (A + G) (A – G)
![open parentheses a minus b close parentheses space equals space 2 square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root space space space space space space space space space space... left parenthesis 5 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=21ffaec87b3349a6213b44d00a4f40d8.png)
From (3) and (5), we obtain
![2 a space equals space 2 A space plus space 2 square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root
rightwards double arrow a space equals space A space plus space square root of open parentheses A space plus space G close parentheses open parentheses A minus G close parentheses end root](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=a598c735a71aec23b94cdfcadac3b6a4.png)
Substituting the value of a in (3), we obtain
![b space equals space 2 A space minus A space minus space square root of open parentheses A space plus space G close parentheses open parentheses A minus G close parentheses end root space equals space A space minus space square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=8ec2c8a90c6bb991f60d087a627ef8e0.png)
Thus, the two numbers are .
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Q31 |
What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually? |
Ans: |
The amount deposited in the bank is Rs 500.
At the end of first year, amount = = Rs 500 (1.1)
At the end of 2nd year, amount = Rs 500 (1.1) (1.1)
At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on
∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)
= Rs 500(1.1)10 |
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Q32 |
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. |
Ans: |
Let the root of the quadratic equation be a and b.
According to the given condition,
![A. M. space equals fraction numerator a plus b over denominator 2 end fraction equals 8 rightwards double arrow a plus b space equals space 16 space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis
G. M. space equals space square root of a b end root space equals space 5 space rightwards double arrow a b space equals space 25 space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis](/saralkocadmin/ckeditor_4.4.5/ckeditor//plugins/ckeditor_wiris/integration/showimage.php?formula=2b28d87c9644c963ee7064be6fdf5f31.png)
The quadratic equation is given by,
x2– x (Sum of roots) + (Product of roots) = 0
x2 – x (a + b) + (ab) = 0
x2 – 16x + 25 = 0 [Using (1) and (2)]
Thus, the required quadratic equation is x2 – 16x + 25 = 0
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