Solution: Let m be the first term of the AP and d be the common difference Let I be the GP’s first term and s be the common ration The Ap’s nth term is given as $t_{n}=a+{(n-1)}d$ in which the first...
Solution: An AP’s sum of n terms is given by ${S_{n}}={\frac{n}{2}}(2a+(n-1)d)$ Where the first term is 'a' and the common difference is 'd'. It is given that $S_{p}=q$ and $S_{q}=p$ $\Rightarrow...
Solution: It is given that $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in the form of A.P., and $d$ is the common difference, We now need to prove that $\operatorname{Sec}...
Solution: It is given that $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in the form of AP in which $a_{i}\lt{0}$ for all i. To prove that: $\begin{array}{l}...
Solution: As per the question $\left(3^{3}-2^{3}\right)+\left(5^{3}-4^{3}\right)+\left(7^{3}-6^{3}\right)+\ldots$ Let's assume the series be...
Solution: The $n^{\text {th }}$ term of Geometric Progression is given by $t_{n}=a r^{n-1}$ in which the first term is $a$ and the common difference is $r$ $\mathrm{p}^{\text {th }}$ term is given...
Solution: It is given that the first term is ' $a$ ' and the sum of first '$p$' terms is $S_{p}=0$ We now need to find the sum of the next 'q' terms So, the total terms are $p+q$ As a result, Sum of...
Solution: Option (C) 45 is correct. Explanation: It is given that the third term of G.P, ${{T}_{3}}~=\text{ }4$ We need to find the product of first 5 terms It is known that, ${{T}_{n}}~=\text{...
Solution: Let the sum got by the lead position group be a Rs and d be distinction in sum As the thing that matters is same subsequently the runner up will get a – d and the third spot a – 2d, etc...
Solution: Given at start he needs to run 24m to get the main potato then 28 m as the following potato is 4m away from first, and so on Thus the succession of his running will be 24, 28, 32 … There...
Solution: Say ABC is a triangle with $AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }20$ cm Let’s say that D, E and F are respectively the midpoints of AC, CB and AB that are joined to...
Solution: It is given that the sum of interior angles of a polygon having ‘n’ sides is denoted by $(n-2)\times {{180}^{\circ }}$ The sum of angles having 3 sides i.e n $=\text{ }3\text{ }is\text{...
Solution: It is given that on the day first he made five frames then 2 frames more than the previous i.e. 7 and then 9 and so on Therefore the making of frames each day forms a sequence of 5, 7, 9…...
Solution: It is given to us that in first month the man’s salary is Rs.5200 and then it increases by Rs.320 every month Therefore, 5200, 5200 + 320, 5200 + 640… will be the sequence so formed of his...
Solution: It is given that the total amount saved is 66000 in 20 years, therefore ${{S}_{20}}~=\text{ }66000$ Let ‘a’ be the amount saved in first year. Then every year if he increases Rs. 200 every...
How about we expect x to be the quantity of days in which 150 laborers finish the work. Then, at that point, from the inquiry, we have \[\mathbf{150x}\text{ }=\text{ }\mathbf{150}\text{ }+\text{...
Given, the expense of machine \[=\text{ }\mathbf{Rs}\text{ }\mathbf{15625}\] Additionally, considering that the machine deteriorates by 20% consistently. Consequently, its worth after consistently...
Given, the man saved Rs 10000 in a bank at the pace of 5% basic premium yearly. Subsequently, the interest in first year \[=\text{ }\left( \mathbf{5}/\mathbf{100} \right)\text{ }\mathbf{x}\text{...
It's seen that, The quantities of letters sent structures a \[\mathbf{G}.\mathbf{P}.:\text{ }\mathbf{4},\text{ }\mathbf{42},\text{ }\ldots \text{ }\mathbf{48}\] Here, \[\mathbf{initial}\text{...
Given, Shamshad Ali purchases a bike for Rs 22000 and pays Rs 4000 in real money. Thus, the neglected sum \[=\text{ }\mathbf{Rs}\text{ }\mathbf{22000}\text{ }\text{ }\mathbf{Rs}\text{...
Given, the rancher pays Rs 6000 in real money. In this way, the neglected sum \[=\text{ }\mathbf{Rs}\text{ }\mathbf{12000}\text{ }\text{ }\mathbf{Rs}\text{ }\mathbf{6000}\text{ }=\text{...
nth term of the numerator \[=~n{{\left( n~+\text{ }1 \right)}^{2}}~\] \[=~{{n}^{3}}~+\text{ }2{{n}^{2}}~+\text{ }n\] and the nth term of the denominator \[=~{{n}^{2}}\left( n~+\text{ }1 \right)\]...
solution:
As the question says, we have In this way, from (1) and (2), we have \[\mathbf{9S22}\text{ }=\text{ }\mathbf{S3}\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{8S1} \right).\]
The given series is \[\mathbf{3}\text{ }+\text{ }\mathbf{7}\text{ }+\text{ }\mathbf{13}\text{ }+\text{ }\mathbf{21}\text{ }+\text{ }\mathbf{31}\text{ }+\text{ }\ldots \] \[\mathbf{S}\text{ }=\text{...
Given series is \[\mathbf{2}\text{ }\times \text{ }\mathbf{4}\text{ }+\text{ }\mathbf{4}\text{ }\times \text{ }\mathbf{6}\text{ }+\text{ }\mathbf{6}\text{ }\times \text{ }\mathbf{8}\text{ }+\text{...
(i) Given, \[\mathbf{5}\text{ }+\text{ }\mathbf{55}\text{ }+\text{ }\mathbf{555}\text{ }+\text{ }\ldots \] Let \[\mathbf{Sn}\text{ }=\text{ }\mathbf{5}\text{ }+\text{ }\mathbf{55}\text{ }+\text{...
Given a, b, c are in A.P. Thus, \[\mathbf{b}\text{ }\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{c}\text{ }\text{ }\mathbf{b}\text{ }\ldots \text{ }\left( \mathbf{1} \right)\] What's more,...
Leave the two numbers alone an and b. \[A.M\text{ }=\text{ }\left( a\text{ }+\text{ }b \right)/2\text{ }and\text{ }G.M.\text{ }=\text{ }\surd ab\] From the inquiry, we have
Given, an and b are the foundations of \[\mathbf{x2}\text{ }\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{p}\text{ }=\text{ }\mathbf{0}\] Along these lines, we have \[\mathbf{a}\text{ }+\text{...
Given, a, b, c,and d are in G.P. Along these lines, we have \[\therefore \mathbf{b2}\text{ }=\text{ }\mathbf{ac}\text{ }\ldots \text{ }\left( \mathbf{I} \right)\] \[\mathbf{c2}\text{ }=\text{...
If are in A.P
We should expect t and d to be the initial term and the normal distinction of the A.P. individually. Then, at that point, the nth term of the A.P. is given by, \[\mathbf{a}\text{ }=\text{...
Give the terms access G.P. be \[\mathbf{a},\text{ }\mathbf{ar},\text{ }\mathbf{ar2},\text{ }\mathbf{ar3},\text{ }\ldots \text{ }\mathbf{arn}\text{ }\text{ }\mathbf{1}\ldots \] Structure the inquiry,...
Given, On cross duplicating, we have Likewise, given On cross duplicating, we have From (1) and (2), we get \[\mathbf{b}/\mathbf{a}\text{ }=\text{ }\mathbf{c}/\mathbf{b}\text{ }=\text{...
How about we think about the terms in A.P. to be \[\mathbf{a},\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{d},\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{2d},\text{ }\mathbf{a}\text{ }+\text{...
We should consider the terms in the G.P.to be \[\mathbf{T1},\text{ }\mathbf{T2},\text{ }\mathbf{T3},\text{ }\mathbf{T4},\text{ }\ldots \text{ }\mathbf{T2n}.\] The quantity of terms \[=\text{...
How about we think about the three numbers in G.P. to be as \[\mathbf{a},\text{ }\mathbf{ar},\text{ }\mathbf{and}\text{ }\mathbf{ar2}.\] Then, at that point, from the inquiry, we have...
How about we consider an and r to be the initial term and the normal proportion of the G.P. separately. Given, \[\mathbf{a}\text{ }=\text{ }\mathbf{1}\] \[\mathbf{a3}\text{ }=\text{...
Considering that the amount of certain terms in a G.P is \[\mathbf{315}.\] Leave the quantity of terms alone n. We realize that, amount of terms is Considering that the initial term an is...
Considering that, \[\mathbf{f}\text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{y} \right)\text{ }=\text{ }\mathbf{f}\text{ }\left( \mathbf{x} \right)\text{ }\times \text{ }\mathbf{f}\text{ }\left(...
It's perceived from the inquiry that, the points of the polygon will frame an A.P. with normal contrast \[d\text{ }=\text{ }5{}^\circ \text{ }and\text{ }initial\text{ }term\text{ }a\text{ }=\text{...
Given, The primary portion of the advance is \[Rs\text{ }100.\] The second portion of the credit is \[Rs\text{ }105,\]etc as the portion increments by \[Rs\text{ }5\]consistently. In this way, the...
How about we consider \[{{a}_{1}},\text{ }{{a}_{2}},\text{ }\ldots \text{ }{{a}_{m}}~be~m\]numbers to such an extent that \[1,\text{ }{{a}_{1}},\text{ }{{a}_{2}},\text{ }\ldots \text{...
Solution:- The A.M among \[a\text{ }and\text{ }b\]is given by, \[\left( a\text{ }+\text{ }b \right)/2\] Then, at that point, as indicated by the inquiry, Along these lines, the worth of \[n\text{...
We need to initially track down the two-digit numbers, which when partitioned by 4, yield 1 as remaining portion. They are: \[\mathbf{13},\text{ }\mathbf{17},\text{ }\ldots \text{...
We should expect\[{{A}_{1}},\text{ }{{A}_{2}},\text{ }{{A}_{3}},\text{ }{{A}_{4}},\text{ }and\text{ }{{A}_{5}}\] to be five numbers between \[8\text{ }and\text{ }26~\]with the end goal that...
How about we take \[a\text{ }and\text{ }d\]to be the initial term and the normal contrast of the A.P. separately. \[{{a}_{m}}~=~a~+\text{ }(m~\text{ }1)d~=\text{ }164\text{ }\ldots \text{ }\left( 1...
How about we take \[a\text{ }and\text{ }d\]to be the initial term and the normal contrast of the A.P. separately. Then, at that point, it given that Hence, the given result is proved.
Solution:- How about we take \[a\text{ }and\text{ }d\]to be the initial term and the normal contrast of the A.P. separately. Then, at that point, it given that
How about we take \[a\text{ }and\text{ }d\]to be the initial term and the normal contrast of the A.P. separately. Then, at that point, it given that Along these lines, the amount of \[\left( p\text{...
Let \[a1,\text{ }a2,\text{ }and\text{ }d1,\text{ }d2\]be the initial terms and the normal distinction of the first and second math movement individually. Then, at that point, from the inquiry we...
First how about we discover the whole numbers from\[\mathbf{1}\text{ }\mathbf{to}\text{ }\mathbf{100}\] , which are distinguishable by 2. What's more, they are\[\mathbf{2},\text{ }\mathbf{4},\text{...
We realize that, \[{{S}_{n}}~=\text{ }n/2\text{ }\left[ 2a\text{ }+\text{ }\left( n-1 \right)d \right]\] From the inquiry we have, On looking at the coefficients of\[{{n}^{2}}\] on the two sides, we...
Given, the \[{{k}^{th}}~\]term of the A.P. is . \[5k~+\text{ }1.\] \[{{k}^{th}}~term\text{ }=~{{a}_{k}}~=~a~+\text{ }(k~\text{ }1)d\] And, \[\begin{array}{*{35}{l}} a~+\text{ }\left( k~\text{ }1...
Given A.P., \[25,\text{ }22,\text{ }19,\text{ }\ldots \] Here, Initial term, \[a\text{ }=\text{ }25\]and Normal distinction, \[d\text{ }=\text{ }22\text{ }\text{ }25\text{ }=\text{ }-\text{ }3\]...
Solution:-
First how about we discover the numbers somewhere in the range of \[\mathbf{200}\text{ }\mathbf{and}\text{ }\mathbf{400}\] which are detachable by \[\mathbf{7}.\] The numbers are:...
How about we consider the amount of n terms of the given A.P. as \[\text{ }25.\] We realized that, \[{{S}_{n}}~=\text{ }n/2\text{ }\left[ 2a\text{ }+\text{ }\left( n-1 \right)d \right]\] where...
Given, The initial term \[\left( a \right)\]of an \[A.P\text{ }=\text{ }2\] How about we accept d be the normal contrast of the A.P. In this way, the A.P. will be \[2,\text{ }2\text{ }+\text{...
We should take an and d to be the initial term and the normal distinction of the A.P. individually. Thus, we have
The regular numbers lying somewhere in the range of \[100\text{ }and\text{ }1000,\]which are products of \[5,\text{ }are\text{ }105,\text{ }110,\text{ }\ldots \text{ }995.\] It plainly frames an...
The odd numbers from \[1\text{ }to\text{ }2001\]are\[~1,\text{ }3,\text{ }5,\text{ }\ldots \text{ }1999,\text{ }2001.\] It obviously shapes an arrangement in \[A.P.\] Where, the initial term,...
Given, \[\begin{array}{*{35}{l}} 1\text{ }=\text{ }{{a}_{1}}~=\text{ }{{a}_{2}} \\ {{a}_{n}}~=\text{ }{{a}_{n\text{ }\text{ }1~}}+\text{ }{{a}_{n\text{ }\text{ }2}},\text{ }n\text{ }>\text{ }2 ...
We should think about the three numbers in A.P. as \[\mathbf{a}\text{ }\text{ }\mathbf{d},\text{ }\mathbf{a},\text{ }\mathbf{and}\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{d}.\] Then, at that point,...
Given, \[{{a}_{1}}~=\text{ }{{a}_{2}},\text{ }{{a}_{n}}~=\text{ }{{a}_{n-1}}~\text{ }1\] Then, at that point, \[\begin{array}{*{35}{l}} {{a}_{3}}~=\text{ }{{a}_{2}}~\text{ }1\text{ }=\text{ }2\text{...
How about we take an and d to be the initial term and the normal distinction of the A.P. individually. We realize that, the kth term of A. P. is given by \[\mathbf{ak}\text{ }=\text{...
Given, \[{{a}_{n}}~=\text{ }{{a}_{n-1}}/n\text{ }and\text{ }{{a}_{1}}~=\text{ }-1\] Then, at that point, \[\begin{array}{*{35}{l}} {{a}_{2}}~=\text{ }{{a}_{1}}/2\text{ }=\text{ }-1/2 \\...
Given, nth term of the series as: \[\mathbf{a}\text{ }=\text{ }\left( \mathbf{2n}\text{ }\text{ }\mathbf{1} \right)\mathbf{2}\text{ }=\text{ }\mathbf{4n2}\text{ }\text{ }\mathbf{4n}\text{ }+\text{...
Given, \[{{a}_{n}}~=\text{ }3{{a}_{n-1}}~+\text{ }2\text{ }and\text{ }{{a}_{1}}~=\text{ }3\] Then, at that point, \[\begin{array}{*{35}{l}} {{a}_{2}}~=\text{ }3{{a}_{1}}~+\text{ }2\text{ }=\text{...
Solution:- On substituting\[n~=\text{ 20}\], we get
Given, nth term of the series as: \[\mathbf{a}\text{ }=\text{ }\mathbf{n2}\text{ }+\text{ }\mathbf{2n}\] Then, at that point, the amount of n terms of the series can be communicated as
Given, \[{{n}^{th}}~\]term of the grouping is \[{{a}_{n}}~=\text{ }{{\left( -1 \right)}^{n-1}}~{{n}^{3}}\] On substituting\[n~=\text{ }9\], we get \[{{a}_{9}}~=\text{ }{{\left( -1...
Given, \[\mathbf{a}\text{ }=\text{ }\left( \mathbf{n}\text{ }+\text{ }\mathbf{1} \right)\text{ }\left( \mathbf{n}\text{ }+\text{ }\mathbf{4} \right)\text{ }=\text{ }\mathbf{n}\left(...
Given series is \[\mathbf{12}\text{ }+\text{ }\left( \mathbf{12}\text{ }+\text{ }\mathbf{22} \right)\text{ }+\text{ }\left( \mathbf{12}\text{ }+\text{ }\mathbf{22}\text{ }+\text{ }\mathbf{32}\text{...
Given, \[{{n}^{th}}\]term of the grouping is \[{{a}_{n}}~=\text{ }{{n}^{2}}/{{2}^{n}}\] Presently, on substituting\[~n~=\text{ }7\], we get \[{{a}_{7}}~=\text{ }{{7}^{2}}/{{2}^{7}}~=\text{...
Given series is \[\mathbf{3}\text{ }\times \text{ }\mathbf{8}\text{ }+\text{ }\mathbf{6}\text{ }\times \text{ }\mathbf{11}\text{ }+\text{ }\mathbf{9}\text{ }\times \text{ }\mathbf{14}\text{ }+\text{...
Given series is \[\mathbf{52}\text{ }+\text{ }\mathbf{62}\text{ }+\text{ }\mathbf{72}\text{ }+\text{ }\ldots \text{ }+\text{ }\mathbf{202}\] It's seen that, nth term, \[\mathbf{a}\text{ }=\text{...
Given series is \[\mathbf{3}\text{ }\times \mathbf{12}\text{ }+\text{ }\mathbf{5}\text{ }\times \text{ }\mathbf{22}\text{ }+\text{ }\mathbf{7}\text{ }\times \text{ }\mathbf{32}\text{ }+\text{...
Given series is \[\mathbf{1}\text{ }\times \text{ }\mathbf{2}\text{ }\times \text{ }\mathbf{3}\text{ }+\text{ }\mathbf{2}\text{ }\times \text{ }\mathbf{3}\text{ }\times \text{ }\mathbf{4}\text{...
Given series is \[1\text{ }\times \text{ }2\text{ }+\text{ }2\text{ }\times \text{ }3\text{ }+\text{ }3\text{ }\times \text{ }4\text{ }+\text{ }4\text{ }\times \text{ }5\text{ }+\text{ }\ldots \]...
Given, the \[{{k}^{th}}\]term of the A.P. is \[5k~+\text{ }1.\] \[{{k}^{th}}~term\text{ }=~{{a}_{k}}~=~a~+\text{ }(k~\text{ }1)d\] Also, \[\begin{array}{*{35}{l}} a~+\text{ }\left( k~\text{ }1...
Given A.P., \[25,\text{ }22,\text{ }19,\text{ }\ldots \] Here, Initial term, \[a\text{ }=\text{ }25\]and Normal distinction, \[d\text{ }=\text{ }22\text{ }\text{ }25\text{ }=\text{ }-\text{ }3\]...
Solution:-
How about we consider the amount of \[n\]terms of the given A.P. as \[\text{ }25.\] We realized that, \[{{S}_{n}}~=\text{ }n/2\text{ }\left[ 2a\text{ }+\text{ }\left( n-1 \right)d \right]\]...
Given, The initial term \[\left( a \right)\]of an A.P \[=\text{ }2\] How about we accept d be the normal distinction of the A.P. In this way, the A.P. will be \[2,\text{ }2\text{ }+\text{ }d,\text{...
The normal numbers lying somewhere in the range of\[100\text{ }and\text{ }1000\], which are products of \[5,\text{ }are\text{ }105,\text{ }110,\text{ }\ldots \text{ }995.\] It unmistakably frames a...
The odd whole numbers from \[1\text{ }to\text{ }2001\text{ }are\text{ }1,\text{ }3,\text{ }5,\text{ }\ldots \text{ }1999,\text{ }2001.\] It unmistakably frames a succession in A.P. Where, the...
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}} ...
Given G.P. is \[5/2,\text{ }5/4,\text{ }5/8,\ldots \ldots \] Here, \[a\text{ }=\text{ }First\text{ }term\text{ }=\text{ }5/2\] \[r\text{ }=\text{ }Common\text{ }proportion\text{ }=\text{ }\left( 5/4...