- Find the 20th and nth terms of the G.P. \(\frac{5}{2},\frac{5}{4},\frac{5}{8},...\)
Solution 👉 Click Here
Given that the GP is
\(\frac{5}{2},\frac{5}{4},\frac{5}{8},...\)
The first term is
\(t_1=a=\frac{5}{2}\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{\frac{5}{4}}{\frac{5}{2}} =\frac{1}{2}\)
Now, the 20th term is
\(t_{20}=ar^{19}=\frac{5}{2}*(\frac{1}{2})^{19}=\frac{5}{1048576}\)
Next, the nth term is
\(t_n=ar^{n-1}=\frac{5}{2}*(\frac{1}{2})^{n-1}=\frac{5}{2^n}\)
This completes the solution
- Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution 👉 Click Here
Given that the 8th term is
\(t_{8}=ar^7=192\)
The common ratio is
\(r=2\)
So, the using 8th term, we get
\(t_8=192\)
or\(ar^7=192\)
or \(a.2^7=192\)
or \(a=\frac{3}{2}\)
Now, the 12th term is
\(t_{12}=ar^{11}=\frac{3}{2}*(2^{11})=3072\)
This completes the solution
- The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that \(q^2 = ps\).
Solution 👉 Click Here
Given that
\(t_{5}=ar^4=p\)
\(t_{8}=ar^7=q\)
\(t_{11}=ar^{10}=s\)
By dividing, we get
\(r^3=\frac{t_8}{t_5}=\frac{q}{p}\)
\(r^3=\frac{t_{11}}{t_8}=\frac{s}{q}\)
So, the using both \(r^3\), we get
\(r^3=r^3\)
or\(\frac{q}{p}=\frac{s}{q}\)
or \(q^2 = ps\)
This completes the solution
- The 4th term of a G.P. is square of its second term, and the first term is – 3.
Determine its 7th term.
Solution 👉 Click Here
Given that
\(t_4=(t_2)^2\)
or\(ar^3=(ar)^2\)
or\(ar^3=a^2r^2\)
or\(a=r\)
Also, given that
\(a=-3\)
Now, 7th term is
\(t_7=ar^6\)
or\(t_7=r^7\)
or\(t_7=(-3)^7\)
or\(t_7=-2187\)
This completes the solution
- Which term of the following sequences:
- \(2,2\sqrt{2},4,... \) is 128 ?
Solution 👉 Click Here
Given that
\(2,2\sqrt{2},4,... \)
The first term is
\(t_1=a=2\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{2}{2\sqrt{2}} =\sqrt{2}\)
The given term is
\(t_n=128\)
or\(ar^{n-1}=128\)
or\(2*(\sqrt{2})^{n-1}=128\)
or\((\sqrt{2})^{n-1}=64\)
or\((\sqrt{2})^{n-1}=(\sqrt{2})^{12}\)
or\(n-1=12\)
or\(n=13\)
This completes the solution
- \(\sqrt{3},3,3\sqrt{3},... \) is 729 ?
Solution 👉 Click Here
Given that
\(\sqrt{3},3,3\sqrt{3},... \)
The first term is
\(t_1=a=\sqrt{3}\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{3}{\sqrt{3}} =\sqrt{3}\)
The given term is
\(t_n=729\)
or\(ar^{n-1}=729\)
or\(\sqrt{3}*(\sqrt{3})^{n-1}=729\)
or\((\sqrt{3})^n=(\sqrt{3})^{12}\)
or\(n=12\)
This completes the solution
- \(\frac{1}{3},\frac{1}{9},\frac{1}{27},...\) is \(\frac{1}{19683}\)
Solution 👉 Click Here
Given that
\(\frac{1}{3},\frac{1}{9},\frac{1}{27},...\) is \(\frac{1}{19683}\)
The first term is
\(t_1=a=\frac{1}{3}\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{\frac{1}{3}}{\frac{1}{9}} =\frac{1}{3}\)
The given term is
\(t_n=\frac{1}{19683}\)
or\(ar^{n-1}=\frac{1}{19683}\)
or\(\frac{1}{3}*(\frac{1}{3})^{n-1}=\frac{1}{19683}\)
or\((\frac{1}{3})^n=(\frac{1}{3})^9\)
or\(n=9\)
This completes the solution
- For what values of x, the numbers \( \frac{2}{7},x,\frac{7}{2}\) are in G.P.?
Solution 👉 Click Here
Given that
\( \frac{2}{7},x,\frac{7}{2}\)
We know that, three consecutive terms a,b,c are in GP if \(b^2=ac\)
So, for three numbers \( \frac{2}{7},x,\frac{7}{2}\), we can write that
\(x^2=\frac{2}{7}*\frac{7}{2}\)
or\(x=1\)
This completes the solution
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:
- 0.15, 0.015, 0.0015, ... 20 terms
Solution 👉 Click Here
Given that
0.15, 0.015, 0.0015, ...
The first term is
\(t_1=a=0.15\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{0.015}{0.15} =0.1\)
We have to cimpute the sum up to 20 terms
So,
n=20
Now, sum of 20terms is
\(S_n=\frac{a(r^n-1)}{r-1}\)
or \(S_n=\frac{0.15((0.1)^{20}-1)}{0.1-1}\)
or \(S_n=\frac{0.15(1-(0.1)^{20})}{0.9}\)
or\(x=\frac{1}{6} (1-(0.1)^{20})\)
This completes the solution
- \(\sqrt{7},\sqrt{21},3\sqrt{7},...\) n terms
Solution 👉 Click Here
Given that
\(\sqrt{7},\sqrt{21},3\sqrt{7},...\)
The first term is
\(t_1=a=\sqrt{7}\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{\sqrt{21}}{\sqrt{7}} =\sqrt{3}\)
We have to cimpute the sum up to n terms
Now, sum of nterms is
\(S_n=\frac{a(r^n-1)}{r-1}\)
or \(S_n=\frac{\sqrt{7}((\sqrt{3})^n-1)}{\sqrt{3}-1}\)
or \(S_n=[\sqrt{7}(\sqrt{3}+1)]\frac{\sqrt{3}^n-1}{2}\)
or \(S_n=[\sqrt{7}(\sqrt{3}+1)]\frac{3^{\frac{n}{2}}-1}{2}\)
This completes the solution
- \(1,-a,a^2,-a^3,...\) n terms (if a≠-1)
Solution 👉 Click Here
Given that
\(1,-a,a^2,-a^3,...\)
The first term is
\(t_1=a=1\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{-a}{1} =-a\)
We have to cimpute the sum up to n terms
Now, sum of nterms is
\(S_n=\frac{a(r^n-1)}{r-1}\)
or \(S_n=\frac{1((-a)^n-1)}{-a-1}\)
or \(S_n=\frac{1-(-a)^n}{a+1}\)
This completes the solution
- \(x^3,x^5,x^7,...\) n terms (if \(n \ne \pm\) 1)
Solution 👉 Click Here
Given that
\(x^3,x^5,x^7,...\)
The first term is
\(t_1=a=x^3\)
The common ratio is
\(r=\frac{t_2}{t_1}=\frac{x^5}{x^3} =x^2\)
We have to cimpute the sum up to n terms
Now, sum of nterms is
\(S_n=\frac{a(r^n-1)}{r-1}\)
or \(S_n=\frac{x^3((x^2)^n-1)}{x^2-1}\)
or \(S_n=\frac{x^3(x^{2n}-1)}{x^2-1}\)
This completes the solution
- Evaluate \(\displaystyle \sum_{k=1}^{11} (2+3^k)\)
Solution 👉 Click Here
Given that
\(\displaystyle \sum_{k=1}^{11} (2+3^k)\)
=\(\displaystyle \sum_{k=1}^{11} 2+\sum_{k=1}^{11} 3^k\)
=\(\displaystyle 11.2+\sum_{k=1}^{11} 3^k\)
=\( 22+ (3^1+3^2+3^3+...+3^{11})\)
=\( 22+ \frac{a(r^n-1)}{r-1} \) where a=3, r=3,n=11
=\( 22+ \frac{3(3^{11}-1)}{3-1} \)
=\( 22+ \frac{3}{2} (177146)\)
This completes the solution
- The sum of first three terms of a G.P. is \(\frac{39}{10}\) and their product is 1. Find the common ratio and the terms.
- How many terms of G.P. \(3, 3^2, 3^3\), … are needed to give the sum 120?
- 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.
- Given a G.P. with a = 729 and 7th term 64, determine \(S_7\).
- Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
- 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.
- Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
- Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and \(128, 32, 8, 2,\frac{1}{2}\)
- Show that the products of the corresponding terms of the sequences \(a, ar, ar^2,…ar^{n – 1}\) and \(A, AR, AR^2, … AR^{n – 1}\) form a G.P, and find the common ratio.
- Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
- If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that \(a^{q – r} b^{r – p}c^{p – q} = 1\).
Solution 👉 Click Here
Let A be the first term and R be the common ration , then
\(t_p=AR^{p-1}=a\)
\(t_q=AR^{q-1}=b\)
\(t_r=AR^{q-1}=c\)
Now,
\(a^{q – r} b^{r – p}c^{p – q}\)
or\( (AR^{p-1})^{q-r}.(AR^{q-1})^{r-p}. (AR^{q-1})^{p-q} \)
or\( A^0.R^0 \)
or1
This completes the solution
- If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that \(p^2 = (ab)^n\).
Solution 👉 Click Here
Let A be the first term and R be the common ration , then
\(t_1=A=a\)
\(t_n=AR^{n-1}=b\)
Now,
\(ab=A.AR^{n-1}\)
or\(ab= A^2R^{n-1}\)
Next
\(P=A*AR*AR^2*...*AR^{n-1}\)
or\(P=A^n* (R*R^2*...R^{n-1})\)
or\(P=A^n* R^{\frac{n(n-1)}{2}}\)
or\(P^2=A^{2n}* R^{n(n-1)}\)
or\(P^2=(A^2* R^{n-1})^n\)
or\(P^2=(ab)^n\)
This completes the solution
- 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 \(\frac{1}{r^n}\)
- If a, b, c and d are in G.P. show that \((a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2\) .
- Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
- Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n} \) may be the geometric mean between a and b.
- The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio \( \left ( 3+2\sqrt{2} \right ): \left ( 3-2\sqrt{2}\right )\)
- If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \( A \pm \sqrt{(A+G)(A-G)}\)
- The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?
- What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
- If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
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