According to the given ques, Hence, the common ratio is \[3/5.\]
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152. Find its common ratio.
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Geometric Progressions
How many terms of the series 2 + 6 + 18 + ….. must be taken to make the sum equal to 728?
. According to the given question, G.P: \[2\text{ }+\text{ }6\text{ }+\text{ }18\text{ }+\text{ }\ldots ..\] Here, \[a\text{ }=\text{ }2\] And \[r\text{ }=\text{ }6/2\text{ }=\text{ }3\] Also given,...
Find the sum of G.P.: 3, 6, 12, …., 1536.
According to the given question, G.P: \[3,\text{ }6,\text{ }12,\text{ }\ldots .,\text{ }1536\] So, \[a\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }1536\] and \[r\text{ }=\text{ }6/3\text{ }=\text{...
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.
According to the given question, For a G.P., \[r\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }486\] and \[{{S}_{n}}~=\text{ }728\] \[1458\text{ }-\text{ }a\text{ }=\text{ }728\text{ }x\text{ }2\text{...
The 4th term and the 7th term of a G.P. are 1/27 and 1/729 respectively. Find the sum of n terms of the G.P.
According to the given question, \[{{t}_{4}}~=\text{ }1/27\] and \[{{t}_{7~}}=\text{ }1/729\] We know that, \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4~}}=\text{...
A boy spends Rs.10 on first day, Rs.20 on second day, Rs.40 on third day and so on. Find how much, in all, will he spend in 12 days?
Amount spent on \[{{1}^{st}}~day\text{ }=\text{ }Rs\text{ }10\] Amount spent on \[{{2}^{nd}}~day\text{ }=\text{ }Rs\text{ }20\] Amount spent on \[{{3}^{rd}}~day\text{ }=\text{ }Rs\text{ }40\]...
The first term of a G.P. is 27 and its 8th term is 1/81. Find the sum of its first 10 terms.
\[First\text{ }term\text{ }\left( a \right)\text{ }of\text{ }a\text{ }G.P\text{ }=\text{ }27\] And, \[{{8}^{th}}~term\text{ }=\text{ }{{t}_{8}}~=\text{ }a{{r}^{8\text{ }-\text{ }1}}~=\text{ }1/81\]...
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?
According to the given question, G.P: \[1\text{ }+\text{ }4\text{ }+\text{ }16\text{ }+\text{ }64\text{ }+\text{ }\ldots \ldots ..\] Here, \[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{...
Find the sum of G.P.:
(i) (ii) Solution: (i) According to the given question Here, \[a\text{ }=\text{ }\left( x\text{ }+\text{ }y \right)/\text{ }\left( x\text{ }-\text{ }y \right)\] And \[~r\text{ }=\text{ }1/\left[...
Find the sum of G.P.: (i) 1 – 1/2 + 1/4 – 1/8 + …….. to 9 terms (ii) 1 – 1/3 + 1/32 – 1/33 + ……… to n terms
(i) According to the given question G.P: \[1\text{ }-\text{ }1/2\text{ }+\text{ }1/4\text{ }-\text{ }1/8\text{ }+\text{ }\ldots \ldots ..\text{ }to\text{ }9\text{ }terms\] Here, \[a\text{ }=\text{...
Find the sum of G.P.: (i) 1 + 3 + 9 + 27 + ………. to 12 terms (ii) 0.3 + 0.03 + 0.003 + 0.0003 +….. to 8 terms.
(i) According to the given question G.P: \[1\text{ }+\text{ }3\text{ }+\text{ }9\text{ }+\text{ }27\text{ }+\text{ }\ldots \ldots \ldots .\text{ }to\text{ }12\text{ }terms\] Here, \[a\text{ }=\text{...
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that: (q – r) log a + (r – p) log b + (p – q) log c = 0
The first term of the G.P. be A and its common ratio be R. Hence, \[{{p}^{th}}~term\text{ }=\text{ }a\text{ }\Rightarrow \text{ }A{{R}^{p\text{ }-\text{ }1}}~=\text{ }a\] \[{{q}^{th}}~term\text{...
Find the G.P. 1/27, 1/9, 1/3, ……, 81; find the product of fourth term from the beginning and the fourth term from the end.
According to the given question, G.P. \[1/27,\text{ }1/9,\text{ }1/3,\text{ }\ldots \ldots ,\text{ }81\] Here, \[a\text{ }=\text{ }1/27,\] \[common\text{ }ratio\text{ }\left( r \right)\text{...
Find the third term from the end of the G.P. 2/27, 2/9, 2/3, ……., 162
Given series: \[2/27,\text{ }2/9,\text{ }2/3,\text{ }\ldots \ldots .,\text{ }162\] Here, \[a\text{ }=\text{ }2/27\] \[r\text{ }=\text{ }\left( 2/9 \right)\text{ }/\text{ }\left( 2/27 \right)\]...
Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32
Given series: \[\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32\] Here, \[a\text{ }=\text{ }\surd 2\] \[r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2\]...
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term
According to the given question, Product of \[{{3}^{rd}}~and\text{ }{{8}^{th}}\] terms of a G.P. is \[243\] The general term of a G.P. First term \[a\] And Common ratio \[r\]is given by,...
If the first and the third terms of a G.P are 2 and 8 respectively, find its second term.
According to the given question, \[{{t}_{1}}~=\text{ }2\text{ }and\text{ }{{t}_{3}}~=\text{ }8\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{1}}~=\text{...
Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the G.P.
Given, \[{{t}_{4}}~=\text{ }1/18\text{ }and\text{ }{{t}_{7}}~=\text{ }-1/486\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4}}~=\text{ }a{{r}^{4\text{ }-\text{...
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
According to the given question, \[{{t}_{5}}~=\text{ }81\text{ }and\text{ }{{t}_{2}}~=\text{ }24\] We know that, General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So,...
Which term of the G.P. :
Solution: In the given G.P. First term, \[a\text{ }=\text{ }-10\] Common ratio, \[r\text{ }=\text{ }\left( 5/\surd 3 \right)/\text{ }\left( -10 \right)\text{ }=\text{ }1/\left( -2\surd 3 \right)\]...
Find the seventh term of the G.P: 1, √3, 3, 3 √3, …..
First term is \[\left( a \right)\text{ }=\text{ }1\] And, common ratio\[\left( r \right)\text{ }=\text{ }\surd 3/1\text{ }=\text{ }\surd 3\] We know that, the general term is \[{{t}_{n}}~=\text{...
Find the nth term of the series: 1, 2, 4, 8, ……..
It's seen that, the initial term is \[\left( a \right)\text{ }=\text{ }1\] What's more, typical ratio \[\left( r \right)\text{ }=\text{ }2/\text{ }1\text{ }=\text{ }2\] We realize that, the overall...
Find the 10th term of the G.P. :
Answer It can be written as \[12,\text{ }4,\text{ }4/3,\text{ }\ldots ..\] It's seen that, the initial term is \[\left( a \right)\text{ }=\text{ }12\] What's more, typical ratio \[\left( r...
Find the 8th term of the sequence:
Answer It's seen that, the initial term is \[\left( a \right)\text{ }=\text{ }1\] What's more, typical ratio \[\left( r \right)\text{ }=\text{ }\surd 3/1\text{ }=\text{ }\surd 3\] We realize that,...
Find the 9th term of the series: 1, 4, 16, 64, …..
It's seen that, the initial term is \[\left( a \right)\text{ }=\text{ }1\] What's more, typical ratio\[\left( r \right)\text{ }=\text{ }4/1\text{ }=\text{ }4\] We realize that, the overall term is...
Find which of the following sequence form a G.P.: 9, 12, 16, 24, ………
Given arrangement: \[9,\text{ }12,\text{ }16,\text{ }24,\text{ }\ldots ...\] Since, \[12/9\text{ }=\text{ }4/3;\text{ }16/12\text{ }=\text{ }4/3;\text{ }24/16\text{ }=\text{ }3/2\] \[12/9\text{...
Find which of the following sequence form a G.P.: (i) 8, 24, 72, 216, ……… (ii) 1/8, 1/24, 1/72, 1/216, ………
(i) Given arrangement:\[~8,\text{ }24,\text{ }72,\text{ }216,\text{ }\ldots \text{ }\ldots \] Since, \[24/8\text{ }=\text{ }3,\text{ }72/24\text{ }=\text{ }3,\text{ }216/72\text{ }=\text{ }3\]...