We should consider the underlying foundations of the quadratic condition to be \[a~and~b.\] Then, at that point, we have We realize that, A quadratic condition can be framed as, \[\begin{align}...

Given, The sum saved in the bank is  \[Rs\text{ }500.\] Toward the finish of \[~first\text{ }year,\text{ }amount\]\[=\text{ }Rs\text{ }500\left( 1\text{ }+\text{ }1/10 \right)~=\text{ }Rs\text{...

Given, the quantity of bacteria copies each hour. Henceforth, the quantity of bacteria after consistently will frame a G.P. Here we have, \[a\text{ }=\text{ }30\text{ }and\text{ }r\text{ }=\text{...

Solution:- Considering that \[A\text{ }and\text{ }G\]are A.M. furthermore, G.M. between two positive numbers. Furthermore, let these two positive numbers be\[a\text{ }and\text{ }b\].

Solution:- Consider the two numbers be\[a\text{ }and\text{ }b\]. Then, at that point, \[G.M.\text{ }=\text{ }\surd ab.\] From the inquiry, we have

Solution:- We realize that, The G. M. of \[a\text{ }and\text{ }b\]is given by \[\surd ab.\] Then, at that point, from the inquiry, we have By squaring both sides, we get Performing cross increase in...

How about we accept \[{{G}_{1}}~and~{{G}_{2}}\]to be two numbers somewhere in the range of \[3\text{ }and\text{ }81\]with the end goal that the series \[3,~{{G}_{1}},~{{G}_{2}},\text{ }81~\]frames a...

Given, \[~a,~b,~c,~d~\]are in G.P. Thus, we have \[\begin{array}{*{35}{l}} bc~=~ad~\text{ }\ldots \text{ }\left( 1 \right)  \\ {{b}^{2}}~=~ac~~\ldots \text{ }\left( 2 \right)  \\...

SOLUTION:- Let \[a\]be the initial term and \[r\]be the normal proportion of the G.P. Since there are n terms from \[~{{(n~+1)}^{th}}~to\text{ }{{(2n)}^{th}}~term,\] Amount of terms from...

Given, the initial term of the G.P is an and the last term is \[b.\] Hence, The G.P. \[~a,~ar,~a{{r}^{2}},~a{{r}^{3}},\text{ }\ldots ~a{{r}^{n}}^{1},\]is the place where r is the normal proportion....

We should take \[A\]to be the initial term and \[R\]to be the normal proportion of the G.P. Then, at that point, as per the inquiry, we have \[\begin{array}{*{35}{l}} A{{R}^{p}}^{1~}=~a  \\...

Consider \[a\]to be the initial term and \[r\]to be the normal proportion of the G.P. Then, at that point, \[{{a}_{1}}~=~a,~{{a}_{2}}~=~ar,~{{a}_{3}}~=~a{{r}^{2}},~{{a}_{4}}~=~a{{r}^{3}}\] From the...

To be demonstrated: The grouping, \[aA,~arAR,~a{{r}^{2}}A{{R}^{2}},\text{ }\ldots a{{r}^{n}}^{1}A{{R}^{n}}^{1}\] shapes a G.P. Presently, we have In this way, the above arrangement frames a G.P....

The necessary aggregate \[=~2\text{ }x\text{ }128\text{ }+\text{ }4\text{ }x\text{ }32\text{ }+\text{ }8\text{ }x\text{ }8\text{ }+\text{ }16\text{ }x\text{ }2\text{ }+\text{ }32\text{ }x\text{...

Given arrangement: \[8,\text{ }88,\text{ }888,\text{ }8888\ldots \] This arrangement isn't a G.P. In any case, it tends to be changed to G.P. by composing the terms as \[{{S}_{n}}~=\text{ }8\text{...

Let \[a\]be the initial term and\[~r\] be the normal proportion of the G.P. As per the given condition, \[\begin{array}{*{35}{l}} {{a}_{4}}~=~a~{{r}^{3}}~=~x~\ldots \text{ }\left( 1 \right)  \\...

Consider \[a\]to be the initial term and r to be the normal proportion of the G.P. Given, \[{{S}_{2}}~=\text{ }-4\] Then, at that point, from the inquiry we have Also, \[\begin{array}{*{35}{l}}...

Given, \[a~=\text{ }729\text{ }and~{{a}_{7}}~=\text{ }64\] Leave \[r\]alone the normal proportion of the G.P. Then, at that point, we realize that, \[{{a}_{n}}~=~a\text{ }{{r}^{n}}^{1}\]...

We should accept the G.P. to be \[a,~ar,~a{{r}^{2}},~a{{r}^{3}},\text{ }\ldots \] Then, at that point, as per the inquiry, we have \[\begin{array}{*{35}{l}} a~+~ar~+~a{{r}^{2}}~=\text{ }16\text{...

Given G.P. is How about we consider that \[n\]terms of this G.P. be needed to acquire the amount of \[120.\] We realize that, Here, \[a\text{ }=\text{ }3\text{ }and\text{ }r\text{ }=\text{ }3\]...

Let \[a/r,\text{ }a,\text{ }ar\]be the initial three terms of the \[G.P.\] \[\begin{array}{*{35}{l}} a/r\text{ }+\text{ }a\text{ }+\text{ }ar\text{ }=\text{ }39/10\text{ }\ldots \ldots \text{...

SOLUTION:-

Given G.P. is \[{{x}^{3}},\text{ }{{x}^{5}},\text{ }{{x}^{7}},\text{ }\ldots \] Here, we have \[a~=~{{x}^{3}}~and~r~=~{{x}^{5}}/{{x}^{3}}~=\text{ }{{x}^{2}}\]

The given G.P. is \[1,\text{ }-\text{ }a,\text{ }a2,\text{ }-\text{ }a3\text{ }\ldots \text{ }.\] Here, the initial term \[=\text{ }a1\text{ }=\text{ }1\] Furthermore, the normal proportion...

The given G.P is \[\surd 7,\text{ }\surd 21,\text{ }3\surd 7,\text{ }\ldots \text{ }.\] Here, \[a\text{ }=\text{ }\surd 7\]and

Given \[G.P.,\text{ }0.15,\text{ }0.015,\text{ }0.00015,\text{ }\ldots \] Here, \[a\text{ }=\text{ }0.15\text{ }and\text{ }r\text{ }=\text{ }0.015/0.15\text{ }=\text{ }0.1\]

The given numbers are \[-\text{ }2/7,\text{ }x,\text{ }-\text{ }7/2.\] Normal proportion \[=\text{ }x/\left( -\text{ }2/7 \right)\text{ }=\text{ }-\text{ }7x/2\] Additionally, normal proportion...

Given arrangement, \[1/3,\text{ }1/9,\text{ }1/27,\text{ }\ldots \] \[a\text{ }=\text{ }1/3\text{ }and\text{ }r\text{ }=\text{ }\left( 1/9 \right)/\left( 1/3 \right)\text{ }=\text{ }1/3\] Taking the...

(i) The given arrangement, \[2,\text{ }2\surd 2,\text{ }4,\ldots \] We have, \[a\text{ }=\text{ }2\text{ }and\text{ }r\text{ }=\text{ }2\surd 2/2\text{ }=\text{ }\surd 2\] Taking the...

We should consider \[a\]to be the initial term and \[r\]to be the normal proportion of the \[G.P.\] Given, \[a\text{ }=\text{ }\text{ }3\] Also, we realize that, \[\begin{align} &...

How about we take \[a\]to be the initial term and \[r\]to be the normal proportion of the G.P. Then, at that point, as indicated by the inquiry, we have \[\begin{array}{*{35}{l}}...

Given, The normal proportion of the G.P., \[r\text{ }=\text{ }2\] What's more, let\[~a\] be the initial term of the G.P. Presently, \[\begin{array}{*{35}{l}} {{a}_{8}}~=~ar{{~}^{81}}~=~a{{r}^{7}} ...