Q1. Find the \(20^{th}\) and \(n^{th}\) terms of the G.P \(\dfrac{5}{2},\;\dfrac{5}{4},\;\dfrac{5}{8},\;\ldots\)

Solution

The given geometric progression (G.P.) is \(\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots\).

The first term \(a\) is \(\dfrac{5}{2}\).

The common ratio \(r\) is calculated as \(\dfrac{5/4}{5/2} = \dfrac{5}{4} \times \dfrac{2}{5} = \dfrac{1}{2}\).

The \(n\)th term of a G.P. is given by \(a_n = a r^{n-1}\).

Thus, the 20th term \(a_{20} = \dfrac{5}{2} \left( \dfrac{1}{2} \right)^{19} = \dfrac{5}{2} \times \dfrac{1}{2^{19}} = \dfrac{5}{2^{20}}\).

Similarly, the \(n\)th term is \(a_n = \dfrac{5}{2} \left( \dfrac{1}{2} \right)^{n-1} = \dfrac{5}{2} \times \dfrac{1}{2^{n-1}} = \dfrac{5}{2^n}\).

Q2. Find the \(12^{th}\) term of a G.P. whose \(8^{th}\) term is 192 and the common ratio is 2.

Solution

The 8th term of the geometric progression (G.P.) is \(a_8 = 192\), and the common ratio is \(r = 2\).

The formula for the nth term of a G.P. is \(a_n = a r^{n-1}\), where \(a\) is the first term.

For the 8th term, \(a_8 = a \cdot 2^{7} = 192\).

Solving for \(a\), \(a = \dfrac{192}{2^7} = \dfrac{192}{128} = \dfrac{3}{2}\).

Now, the 12th term is \(a_{12} = a \cdot 2^{11} = \dfrac{3}{2} \cdot 2^{11} = 3 \cdot 2^{10} = 3 \cdot 1024 = 3072\).

Q3. The \(5^{th}\), \(8^{th}\) and \(11^{th}\) terms of a G.P. are \(p\), \(q\) and \(s\), respectively. Show that \(q^2 = ps\).

Solution

The 5th, 8th, and 11th terms of the geometric progression (G.P.) are given as \(a_5 = p\), \(a_8 = q\), and \(a_{11} = s\), respectively.

In a G.P., the nth term is \(a_n = a r^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.

Thus, \(a_5 = a r^4 = p\), \(a_8 = a r^7 = q\), and \(a_{11} = a r^{10} = s\).

To show \(q^2 = p s\), compute \(q^2 = (a r^7)^2 = a^2 r^{14}\).

The right side is \(p s = (a r^4)(a r^{10}) = a^2 r^{14}\).

Therefore, \(q^2 = a^2 r^{14} = p s\).

Q4. The \(4^{th}\) term of a G.P. is square of its second term, and the first term is – 3. Determine its \(7^{th}\) term.

Solution

The first term of the geometric progression (G.P.) is \(a = -3\).

The second term is \(a_2 = a r = -3r\), and the fourth term is \(a_4 = a r^3 = -3 r^3\).

Given that the fourth term is the square of the second term, \(-3 r^3 = (-3r)^2\).

This simplifies to \(-3 r^3 = 9 r^2\).

Assuming \(r \neq 0\), divide both sides by \(r^2\): \(-3 r = 9\), so \(r = -3\).

The seventh term is \(a_7 = a r^6 = -3 \cdot (-3)^6\).

Since \((-3)^6 = 729\), \(a_7 = -3 \cdot 729 = -2187\).

Q5. Which term of the following sequences:
(a) \(2,\;2\sqrt{2},\;4,\ldots\) is 128?
(b) \(\sqrt{3},\;3\;3\sqrt{3}\;\ldots\) is 729?
(c) \(\dfrac{1}{3},\;\dfrac{1}{9},\;\dfrac{1}{27},\;\ldots\) is \(\dfrac{1}{19683}\)

Solution

(a) For the G.P. \(2, 2\sqrt{2}, 4, \ldots\), the first term \(a = 2\).

The common ratio \(r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}\).

The nth term is \(a_n = 2 (\sqrt{2})^{n-1}\).

Set \(a_n = 128\): \(2 (\sqrt{2})^{n-1} = 128\), so \((\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12}\).

Thus, \(n-1 = 12\), and \(n = 13\).

(b) For the G.P. \(\sqrt{3}, 3, 3\sqrt{3}, \ldots\), the first term \(a = \sqrt{3}\).

The common ratio \(r = \dfrac{3}{\sqrt{3}} = \sqrt{3}\).

The nth term is \(a_n = \sqrt{3} \cdot (\sqrt{3})^{n-1} = (\sqrt{3})^n\).

Set \(a_n = 729 = 3^6 = (3^{1/2})^12 = (\sqrt{3})^{12}\), so \(n = 12\).

(c) For the G.P. \(\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots\), the first term \(a = \dfrac{1}{3}\).

The common ratio \(r = \dfrac{1/9}{1/3} = \dfrac{1}{3}\).

The nth term is \(a_n = \dfrac{1}{3} \cdot \left( \dfrac{1}{3} \right)^{n-1} = \left( \dfrac{1}{3} \right)^n\).

Set \(a_n = \dfrac{1}{19683}\). Note \(19683 = 3^9\), so \(\left( \dfrac{1}{3} \right)^9 = \dfrac{1}{19683}\), hence \(n = 9\).

Q6. For what values of \(x\), the numbers \(-\dfrac{2}{7},\;x\;-\dfrac{7}{2}) are in G.P.?

Solution

For the numbers \(-\dfrac{2}{7}, x, -\dfrac{7}{2}\) to be in geometric progression (G.P.), the square of the middle term must equal the product of the first and third terms.

This condition gives \(x^2 = \left(-\dfrac{2}{7}\right) \left(-\dfrac{7}{2}\right)\).

Simplifying the right side: \(\dfrac{2}{7} \cdot \dfrac{7}{2} = 1\).

Thus, \(x^2 = 1\), so \(x = \pm 1\).

Alternatively, using common ratio: \(\dfrac{x}{-\frac{2}{7}} = \dfrac{-\frac{7}{2}}{x}\), which leads to \(\dfrac{7x}{2} = -\dfrac{2}{7} \cdot x\), confirming \(x^2 = 1\) and \(x = \pm 1\).

Q7. 0.15, 0.015, 0.0015, ... 20 terms.

Solution

The geometric progression (G.P.) is \(0.15, 0.015, 0.0015, \ldots\) up to 20 terms.

The first term \(a = 0.15\), and the common ratio \(r = \dfrac{0.015}{0.15} = 0.1 = \dfrac{1}{10}\).

The sum of the first \(n\) terms of a G.P. is \(S_n = \dfrac{a (1 - r^n)}{1 - r}\).

For \(n = 20\), \[\begin{aligned}S_{20} &= \dfrac{0.15 \left(1 - \left(\dfrac{1}{10}\right)^{20}\right)}{1 - \dfrac{1}{10}} \\\\&= \dfrac{0.15 \left(1 - 10^{-20}\right)}{\dfrac{9}{10}}\\\\ &=\dfrac{0.15 \times \dfrac{10}{9} \left(1 - 10^{-20}\right)}{1} \\\\&= \dfrac{1.5}{9} \left(1 - 10^{-20}\right) \\&= \dfrac{1}{6} \left(1 - 10^{-20}\right) \\&\approx 0.166\ldots \\&= \dfrac{1}{6}\end{aligned}\] since \(10^{-20}\) is negligible.

Q8. \(\sqrt{7},\;\sqrt{21},\;3\sqrt{7},\;\ldots n\) terms

Solution

The given terms form a geometric progression since the ratio of any term to its preceding term is constant. The first term is obtained directly from the sequence as \(a=\sqrt{7}\).

The common ratio is calculated by dividing the second term by the first term. Thus, \[ r=\dfrac{\sqrt{21}}{\sqrt{7}}=\sqrt{3} \] which confirms that the progression is geometric.

For a geometric progression with first term \(a\), common ratio \(r\neq 1\), and \(n\) terms, the sum of the first \(n\) terms is given by \[ S_n=\dfrac{a\left(r^{n}-1\right)}{r-1} \] This formula is applicable here since \(r=\sqrt{3}\neq 1\).

Substituting \(a=\sqrt{7}\) and \(r=\sqrt{3}\), we obtain \[ S_n=\dfrac{\sqrt{7}\left(\left(\sqrt{3}\right)^n-1\right)}{\sqrt{3}-1} \]

To rationalize the denominator, the numerator and denominator are multiplied by \(\sqrt{3}+1\). This simplifies the expression without altering its value. Hence, \[ S_n=\dfrac{\sqrt{7}\left(\sqrt{3}+1\right)\left(\left(\sqrt{3}\right)^n-1\right)}{2} \] which gives the required sum of the first \(n\) terms of the given geometric progression.

Q9. \(1,\; – a,\; a^2,\; – a^3,\; \ldots\; n\) terms (if \(a \ne – 1\)).

Solution

The given terms exhibit a constant ratio between successive terms, which confirms that the sequence is a geometric progression. The first term of the progression is clearly \(1\).

The common ratio is obtained by dividing the second term by the first term. Hence, \[ r=\dfrac{-a}{1}=-a \] which remains constant for all consecutive terms.

For a geometric progression with first term \(a_1\), common ratio \(r\neq 1\), and \(n\) terms, the sum of the first \(n\) terms is given by \[ S_n=\dfrac{a_1\left(1-r^n\right)}{1-r} \] This formula is applicable here since \(a\neq -1\), ensuring that the denominator is nonzero.

Substituting \(a_1=1\) and \(r=-a\), we get \[ S_n=\dfrac{1\left(1-\left(-a\right)^n\right)}{1-(-a)} \]

Simplifying the denominator yields the final expression for the required sum as \[ S_n=\dfrac{1-\left(-a\right)^n}{1+a} \] which represents the sum of the first \(n\) terms of the given geometric progression.

Q10. \(x^3,\; x^5,\; x^7,\; \ldots\; n\) terms (if \(x \ne \pm 1\)).

Solution

The given terms follow a fixed multiplicative pattern, so the sequence is a geometric progression. The first term of the progression is clearly identified as \(a=x^3\).

The common ratio is obtained by dividing the second term by the first term. Thus, \[ r=\dfrac{x^5}{x^3}=x^2 \] which remains constant throughout the progression. The condition \(x\ne \pm 1\) ensures that the common ratio is not equal to \(1\).

For a geometric progression with first term \(a\), common ratio \(r\neq 1\), and \(n\) terms, the sum of the first \(n\) terms is given by \[ S_n=\dfrac{a\left(1-r^n\right)}{1-r} \] This standard result is applicable in the present case.

Substituting \(a=x^3\) and \(r=x^2\), we obtain \[ S_n=\dfrac{x^3\left(1-\left(x^2\right)^n\right)}{1-x^2} \]

This expression gives the sum of the first \(n\) terms of the geometric progression \(x^3, x^5, x^7, \ldots\) and is valid for all values of \(x\) such that \(x\ne \pm 1\).

Q11. Evaluate \(\sum\limits^{11}_{k=1}(2+3^k)\)

Solution

The given summation can be rewritten by separating the constant term and the exponential term in each summand. Thus, \[ \sum_{k=1}^{11}(2+3^k)=\sum_{k=1}^{11}2+\sum_{k=1}^{11}3^k \] This separation is valid because summation is linear.

The first sum consists of the constant term \(2\) repeated for \(11\) terms. Hence, \[ \sum_{k=1}^{11}2=2+2+2+\cdots+2=2\times 11=22 \]

The second sum forms a geometric progression with first term \(a=3\) and common ratio \(r=3\). Therefore, the sum of the first \(11\) terms is given by \[ S_{11}=\dfrac{a(1-r^{11})}{1-r} \]

Substituting the values of \(a\) and \(r\), we obtain \[ S_{11}=\dfrac{3(1-3^{11})}{1-3}=\dfrac{3(1-3^{11})}{-2} \]

Combining both parts of the original sum, we get \[ 22+\dfrac{3(1-3^{11})}{-2} \] which simplifies to \[ 22+\dfrac{3^{12}-3}{2} \]

Evaluating this expression gives the final value of the given summation as \[ 265741 \]

Q12. The sum of first three terms of a G.P. is \(\dfrac{39}{10}\) and their product is 1. Find the common ratio and the terms.

Solution

The three given numbers are \(\dfrac{a}{r}, a, ar\). From the condition on their sum, we have \[ \dfrac{a}{r}+a+ar=\dfrac{39}{10} \] and from the condition on their product, \[ \dfrac{a}{r}\cdot a\cdot ar=1 \]

Simplifying the product, we observe that the factors \(r\) cancel out, giving \[ a^3=1 \] which implies \[ a=1 \]

Substituting \(a=1\) in the equation for the sum, we obtain \[ \dfrac{1}{r}+1+r=\dfrac{39}{10} \]

Multiplying throughout by \(r\) to clear the denominator gives \[ 1+r+r^2=\dfrac{39r}{10} \]

Multiplying both sides by \(10\), we get \[ 10r^2+10r+10-39r=0 \]

Combining like terms leads to \[ 10r^2-29r+10=0 \]

Factoring the quadratic expression, we write \[ 10r^2-25r-4r+10=0 \] which gives \[ 5r(2r-5)-2(2r-5)=0 \]

Taking the common factor, we obtain \[ (2r-5)(5r-2)=0 \]

Hence, \[ 2r=5 \quad \text{or} \quad 5r=2 \] which yields \[ r=\dfrac{5}{2} \quad \text{or} \quad r=\dfrac{2}{5} \]

When \(r=\dfrac{5}{2}\), the numbers are \[ \dfrac{2}{5},\;1,\;\dfrac{5}{2} \] and when \(r=\dfrac{2}{5}\), the numbers are \[ \dfrac{5}{2},\;1,\;\dfrac{2}{5} \]

Q13. How many terms of G.P. \(3,\; 3^2,\; 3^3,\; \ldots \) are needed to give the sum 120?

Solution

The given series \(3, 3^2, 3^3, \ldots\) is a geometric progression since each term is obtained by multiplying the preceding term by \(3\). Hence, the first term is \(a=3\) and the common ratio is \(r=3\).

The sum of the first \(n\) terms of a geometric progression, where \(r\neq 1\), is given by \[ \begin{aligned} S_n &= \dfrac{a(r^n-1)}{r-1} \end{aligned} \]

Substituting \(a=3\), \(r=3\), and the given sum \(S_n=120\), we obtain \[ \begin{aligned} 120 &= \dfrac{3(3^n-1)}{3-1} \\ 120 &= \dfrac{3(3^n-1)}{2} \end{aligned} \]

Multiplying both sides by \(2\), we get \[ \begin{aligned} 240 &= 3(3^n-1) \end{aligned} \]

Dividing both sides by \(3\), we have \[ \begin{aligned} 80 &= 3^n-1 \end{aligned} \]

Adding \(1\) to both sides gives \[ \begin{aligned} 3^n &= 81 \end{aligned} \]

Since \(81=3^4\), it follows that \[ \begin{aligned} n &= 4 \end{aligned} \]

Therefore, \(4\) terms of the given geometric progression are required to obtain the sum \(120\).

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.

Solution

Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). The sum of the first three terms is given as \(16\), so \[ \begin{aligned} S_3=\dfrac{a(1-r^3)}{1-r}=16 \end{aligned} \]

The sum of the next three terms is \(128\), which means the sum of the first six terms is \[ \begin{aligned} S_6=16+128=144 \end{aligned} \] Using the formula for the sum of the first six terms of a geometric progression, we have \[ \begin{aligned} S_6=\dfrac{a(1-r^6)}{1-r}=144 \end{aligned} \]

Dividing the expression for \(S_3\) by that for \(S_6\), the common factor \(\dfrac{a}{1-r}\) cancels out, giving \[ \begin{aligned} \dfrac{1-r^3}{1-r^6}=\dfrac{16}{144} \end{aligned} \]

Let \(r^3=x\). Then the equation becomes \[ \begin{aligned} \dfrac{1-x}{1-x^2}=\dfrac{16}{144} \end{aligned} \]

Cross-multiplying, we obtain \[ \begin{aligned} 144(1-x)&=16(1-x^2)\\ 144-144x&=16-16x^2\\ 16x^2-144x+128&=0 \end{aligned} \]

Dividing throughout by \(16\), we get \[ \begin{aligned} x^2-9x+8=0 \end{aligned} \]

Factoring, \[ \begin{aligned} (x-8)(x-1)=0 \end{aligned} \] which gives \(x=8\) or \(x=1\).

Since \(x=r^3\), the value \(x=1\) would give \(r=1\), which is not possible because the sums of successive groups of terms are different. Hence, \[ \begin{aligned} r^3=8\\ r=2 \end{aligned} \]

Substituting \(r=2\) in the expression for \(S_3\), \[ \begin{aligned} 16=\dfrac{a(1-2^3)}{1-2} \end{aligned} \] which simplifies to \[ \begin{aligned} 16=\dfrac{a(1-8)}{-1}=7a \end{aligned} \] Hence, \[ \begin{aligned} a=\dfrac{16}{7} \end{aligned} \]

Therefore, the sum of the first \(n\) terms of the geometric progression is \[ \begin{aligned} S_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{16}{7}\left(2^n-1\right) \end{aligned} \]

Q15. Given a G.P. with \(a = 729\) and \(7^{th}\) term 64, determine \(S_7\).

Solution

The first term of the given geometric progression is \(a=729\) and the seventh term is \(64\). For a geometric progression, the \(n^{\text{th}}\) term is given by \(a_n=ar^{n-1}\). Hence, \[ \begin{aligned} a_7 &= ar^6 \end{aligned} \]

Substituting the given values, we obtain \[ \begin{aligned} 64 &= 729r^6 \end{aligned} \] which gives \[ \begin{aligned} r^6 &= \dfrac{64}{729}=\left(\dfrac{2}{3}\right)^6 \end{aligned} \]

Since the terms of the progression are positive, the common ratio is taken as \[ \begin{aligned} r=\dfrac{2}{3} \end{aligned} \]

The sum of the first \(n\) terms of a geometric progression is given by \[ \begin{aligned} S_n=\dfrac{a(r^n-1)}{r-1} \end{aligned} \] This formula is applicable here as \(r\neq 1\).

Substituting \(a=729\), \(r=\dfrac{2}{3}\), and \(n=7\), we get \[ \begin{aligned} S_7 &= \dfrac{729\left[\left(\dfrac{2}{3}\right)^7-1\right]}{\dfrac{2}{3}-1} \end{aligned} \]

Simplifying the denominator and rearranging the terms, \[ \begin{aligned} S_7 &= 729\times 3\left[1-\left(\dfrac{2}{3}\right)^7\right] \end{aligned} \]

Evaluating the expression gives \[ \begin{aligned} S_7 &= 2059 \end{aligned} \] which is the required sum of the first seven terms of the given geometric progression.

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.

Solution

Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). The sum of the first two terms is given as \(-4\). Using the formula for the sum of the first two terms of a geometric progression, we have \[ \begin{aligned} S_2=\dfrac{a(1-r^2)}{1-r}=-4 \end{aligned} \]

The third and fifth terms of the geometric progression are given by \[ \begin{aligned} a_3 &= ar^2\\ a_5 &= ar^4 \end{aligned} \] According to the given condition, the fifth term is four times the third term. Hence, \[ \begin{aligned} ar^4 &= 4ar^2 \end{aligned} \]

Since \(a\neq 0\), dividing both sides by \(a r^2\) gives \[ \begin{aligned} r^2=4 \end{aligned} \] which implies \[ \begin{aligned} r=\pm 2 \end{aligned} \]

First, taking \(r=2\), and substituting this value in the expression for \(S_2\), we get \[ \begin{aligned} -4 &= \dfrac{a(1-2^2)}{1-2} \end{aligned} \]

Simplifying, \[ \begin{aligned} -4 &= \dfrac{a(1-4)}{-1}=3a \end{aligned} \] which gives \[ \begin{aligned} a=-\dfrac{4}{3} \end{aligned} \]

Hence, one such geometric progression is \[ \begin{aligned} -\dfrac{4}{3},\; -\dfrac{8}{3},\; -\dfrac{16}{3},\;\ldots \end{aligned} \]

Next, taking \(r=-2\), and substituting in the sum formula, we have \[ \begin{aligned} -4 &= \dfrac{a(1-(-2)^2)}{1-(-2)} \end{aligned} \]

Simplifying, \[ \begin{aligned} -4 &= \dfrac{a(1-4)}{3}=-a \end{aligned} \] which yields \[ \begin{aligned} a=4 \end{aligned} \]

Therefore, another geometric progression satisfying the given conditions is \[ \begin{aligned} 4,\;-8,\;16,\;-32,\;\ldots \end{aligned} \]

Q17. If the \(4^{th}\), \(10^{th}\) and \(16^{th}\) terms of a G.P. are \(x,\; y\)and \(z\), respectively. Prove that \(x,\; y,\; z\) are in G.P.

Solution

Let the first term of the given geometric progression be \(a\) and the common ratio be \(r\). The fourth, tenth, and sixteenth terms of the progression are given respectively by \[ \begin{aligned} a_4 &= ar^3 = x\\ a_{10} &= ar^9 = y\\ a_{16} &= ar^{15} = z \end{aligned} \]

To prove that \(x\), \(y\), and \(z\) are in geometric progression, it is sufficient to show that the ratio of the third term to the second term is equal to the ratio of the second term to the first term.

Consider the ratios \[ \begin{aligned} \dfrac{z}{y} &= \dfrac{ar^{15}}{ar^9}\\ \dfrac{y}{x} &= \dfrac{ar^9}{ar^3} \end{aligned} \]

Simplifying both ratios, we obtain \[ \begin{aligned} \dfrac{z}{y} &= r^{15-9}=r^6\\ \dfrac{y}{x} &= r^{9-3}=r^6 \end{aligned} \]

Since \[ \begin{aligned} \dfrac{z}{y}=\dfrac{y}{x} \end{aligned} \] the common ratio between successive terms is the same.

Hence, \(x\), \(y\), and \(z\) are in geometric progression.

Q18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Solution

The given sequence consists of numbers formed by repeating the digit \(8\). Each term can be expressed in terms of powers of \(10\). Writing the sum of the first \(n\) terms, we have \[ \begin{aligned} 8+88+888+\ldots \end{aligned} \]

Factoring out \(8\) from each term, the sum becomes \[ \begin{aligned} 8\left(1+11+111+\ldots\right) \end{aligned} \]

Each number of the form \(1, 11, 111, \ldots\) can be written as \(\dfrac{10^k-1}{9}\). Hence, \[ \begin{aligned} 8\left(1+11+111+\ldots\right) = \dfrac{8}{9}\left(9+99+999+\ldots\right) \end{aligned} \]

Now, \[ \begin{aligned} 9+99+999+\ldots = (10-1)+(10^2-1)+(10^3-1)+\ldots \end{aligned} \]

Separating the sums, we get \[ \begin{aligned} \dfrac{8}{9}\left[(10+10^2+10^3+\ldots+10^n)-(1+1+1+\ldots+n)\right] \end{aligned} \]

The sum \(1+1+1+\ldots\) taken \(n\) times is equal to \(n\). The series \(10+10^2+10^3+\ldots+10^n\) is a geometric progression with first term \(a=10\) and common ratio \(r=10\). Hence, \[ \begin{aligned} 10+10^2+10^3+\ldots+10^n = \dfrac{10(10^n-1)}{9} \end{aligned} \]

Substituting these results, the required sum becomes \[ \begin{aligned} \dfrac{8}{9}\left[\dfrac{10(10^n-1)}{9}-n\right] \end{aligned} \]

Simplifying, we obtain \[ \begin{aligned} \dfrac{80}{81}(10^n-1)-\dfrac{8}{9}n \end{aligned} \] which is the sum of the first \(n\) terms of the sequence \(8, 88, 888, 8888, \ldots\).

Q19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, \(\mathrm{\dfrac{1}{2}}\)

Solution

The two given sequences have the same number of terms, so the products of their corresponding terms can be formed term by term. Writing the sequences together, we have \[ \begin{aligned} 2,\;4,\;8,\;16,\;32\\ 128,\;32,\;8,\;2,\;\dfrac{1}{2} \end{aligned} \]

Multiplying the corresponding terms of the two sequences, we obtain \[ \begin{aligned} (2\times128)+(4\times32)+(8\times8)+(16\times2)+(32\times\dfrac{1}{2}) \end{aligned} \]

Evaluating each product gives \[ \begin{aligned} 256+128+64+32+16 \end{aligned} \]

Adding these values, the required sum of the products of the corresponding terms is \[ \begin{aligned} 496 \end{aligned} \]

Q20. Show that the products of the corresponding terms of the sequences \(a,\; ar,\; ar^2,\;\ldots\;ar^{n – 1}\) and \(A,\; AR,\; AR^2,\; \ldots\; AR^{n – 1}\) form a G.P, and find the common ratio.

Solution

The given sequences are \[ \begin{aligned} a,\; ar,\; ar^2,\;\ldots,\; ar^{n-1} \end{aligned} \] and \[ \begin{aligned} A,\; AR,\; AR^2,\;\ldots,\; AR^{n-1} \end{aligned} \] Both are geometric progressions.

Forming the products of the corresponding terms of the two sequences, we obtain \[ \begin{aligned} aA,\; (ar)(AR),\; (ar^2)(AR^2),\;\ldots,\; (ar^{n-1})(AR^{n-1}) \end{aligned} \]

Simplifying each product, the resulting sequence becomes \[ \begin{aligned} aA,\; aArR,\; aAr^2R^2,\;\ldots,\; aA(rR)^{\,n-1} \end{aligned} \]

To examine whether this sequence is a geometric progression, we find the ratio of any term to its preceding term. Thus, \[ \begin{aligned} \text{Common ratio} &= \dfrac{aAr^2R^2}{aArR} \end{aligned} \]

Simplifying, we get \[ \begin{aligned} \text{Common ratio} = rR \end{aligned} \]

Since the ratio of successive terms is constant, the products of the corresponding terms form a geometric progression with common ratio \(rR\).

Q21. 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 \(4^{th}\) by 18.

Solution

Let the four numbers forming the geometric progression be \(a, ar, ar^2, ar^3\), where \(a\) is the first term and \(r\) is the common ratio.

The third term is greater than the first term by \(9\). Hence, \[ \begin{aligned} a_3 &= ar^2\\ ar^2 &= a+9\\ ar^2-a &= 9\\ a(r^2-1) &= 9 \end{aligned} \]

The second term is greater than the fourth term by \(18\). Therefore, \[ \begin{aligned} a_2 &= a_4+18\\ ar &= ar^3+18\\ ar-ar^3 &= 18 \end{aligned} \]

Factoring the left-hand side, we get \[ \begin{aligned} ar(1-r^2) &= 18\\ -ar(r^2-1) &= 18 \end{aligned} \]

Using the earlier result \(a(r^2-1)=9\), substitution gives \[ \begin{aligned} -9r &= 18 \end{aligned} \] which yields \[ \begin{aligned} r=-2 \end{aligned} \]

Substituting \(r=-2\) in the relation \(a(r^2-1)=9\), we have \[ \begin{aligned} a(4-1) &= 9\\ 3a &= 9\\ a &= 3 \end{aligned} \]

Hence, the required geometric progression is \[ \begin{aligned} 3,\;-6,\;12,\;-24 \end{aligned} \]

Q22. If the \(p^{th}\), \(q^{th}\) and \(r^{th}\) terms of a G.P. are \(a,\; b\) and \(c\), respectively. Prove that \(a^{q – r} b^{r – p}c^{P – q} = 1\).

Solution

Let the first term of the geometric progression be \(A\) and the common ratio be \(r\). Then the \(p^{\text{th}}\), \(q^{\text{th}}\), and \(r^{\text{th}}\) terms are given respectively by \[ \begin{aligned} a_p &= Ar^{p-1}=a\\ a_q &= Ar^{q-1}=b\\ a_r &= Ar^{r-1}=c \end{aligned} \]

Raising each term to the required powers, we obtain \[ \begin{aligned} a^{\,q-r} &= \left(Ar^{p-1}\right)^{q-r}\\ b^{\,r-p} &= \left(Ar^{q-1}\right)^{r-p}\\ c^{\,p-q} &= \left(Ar^{r-1}\right)^{p-q} \end{aligned} \]

Multiplying these three expressions, we get \[ \begin{aligned} a^{q-r} b^{r-p} c^{p-q} &= A^{(q-r)+(r-p)+(p-q)} \\ &\quad \times r^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \end{aligned} \]

Now, \[ \begin{aligned} (q-r)+(r-p)+(p-q)=0 \end{aligned} \] and \[ \begin{aligned} (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)=0 \end{aligned} \]

Hence, the above expression reduces to \[ \begin{aligned} A^{0}r^{0}=1 \end{aligned} \]

Therefore, \[ \begin{aligned} a^{q-r} b^{r-p} c^{p-q}=1 \end{aligned} \] which proves the required result.

Q23. If the first and the \(n^{th}\) 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

Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). Then the \(n^{\text{th}}\) term is given by \[ \begin{aligned} a_n=ar^{n-1}=b \end{aligned} \]

The product \(P\) of the first \(n\) terms of the geometric progression is \[ \begin{aligned} P &= a\cdot ar\cdot ar^2\cdot \ldots \cdot ar^{n-1} \end{aligned} \]

Collecting like factors, this product can be written as \[ \begin{aligned} P &= a^n r^{1+2+3+\ldots+(n-1)} \end{aligned} \]

The sum of the first \((n-1)\) natural numbers is \[ \begin{aligned} 1+2+3+\ldots+(n-1)=\dfrac{n(n-1)}{2} \end{aligned} \]

Hence, \[ \begin{aligned} P=a^n r^{\frac{n(n-1)}{2}} \end{aligned} \]

Squaring both sides, we obtain \[ \begin{aligned} P^2=a^{2n}r^{n(n-1)} \end{aligned} \]

From the relation \(ar^{n-1}=b\), we have \[ \begin{aligned} ab=a^2 r^{n-1} \end{aligned} \]

Raising both sides to the power \(n\), we get \[ \begin{aligned} (ab)^n=\left(a^2 r^{n-1}\right)^n=a^{2n}r^{n(n-1)} \end{aligned} \]

Comparing this with the expression obtained for \(P^2\), we conclude that \[ \begin{aligned} P^2=(ab)^n \end{aligned} \] which proves the required result.

Q24. 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 \(\dfrac{1}{r^n}\).

Solution

Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). The sum of the first \(n\) terms of a geometric progression is given by \[ \begin{aligned} S_n=\dfrac{a(1-r^n)}{1-r} \end{aligned} \]

The \((n+1)^{\text{th}}\) term of the geometric progression is \[ \begin{aligned} a_{n+1}=ar^n \end{aligned} \]

The sum of the terms from the \((n+1)^{\text{th}}\) term to the \((2n)^{\text{th}}\) term is equal to the sum of the first \(n\) terms of a geometric progression whose first term is \(ar^n\) and common ratio is \(r\). Hence, \[ \begin{aligned} S_{2n-n}= \dfrac{ar^n(1-r^n)}{1-r} \end{aligned} \]

Now, taking the ratio of the sum of the first \(n\) terms to the sum of the next \(n\) terms, we obtain \[ \begin{aligned} \dfrac{S_n}{S_{2n-n}} &= \dfrac{\dfrac{a(1-r^n)}{1-r}}{\dfrac{ar^n(1-r^n)}{1-r}} \end{aligned} \]

Canceling the common factors, this simplifies to \[ \begin{aligned} \dfrac{S_n}{S_{2n-n}}=\dfrac{1}{r^n} \end{aligned} \]

Hence, the ratio of the sum of the first \(n\) terms to the sum of the terms from the \((n+1)^{\text{th}}\) to the \((2n)^{\text{th}}\) term is \(\dfrac{1}{r^n}\), as required.

Q25. If \(a,\; b,\; c\) and d are in G.P. show that \((a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2\).

Solution

Since \(a, b, c,\) and \(d\) are in geometric progression, let the first term be \(a\) and the common ratio be \(r\). Then the terms can be written as \[ \begin{aligned} a,\quad ar,\quad ar^2,\quad ar^3 \end{aligned} \]

Now consider the left-hand side of the given expression. We have \[ \begin{aligned} a^2+b^2+c^2 &= a^2+(ar)^2+(ar^2)^2 \\ &= a^2\left(1+r^2+r^4\right) \end{aligned} \] and \[ \begin{aligned} b^2+c^2+d^2 &= (ar)^2+(ar^2)^2+(ar^3)^2 \\ &= a^2r^2\left(1+r^2+r^4\right) \end{aligned} \]

Multiplying these two expressions, we get \[ \begin{aligned} (a^2+b^2+c^2)(b^2+c^2+d^2) &= a^2\left(1+r^2+r^4\right)\cdot a^2r^2\left(1+r^2+r^4\right) \\ &= a^4r^2\left(1+r^2+r^4\right)^2 \end{aligned} \]

Now consider the right-hand side. We have \[ \begin{aligned} ab+bc+cd &= a(ar)+(ar)(ar^2)+(ar^2)(ar^3) \\ &= a^2r\left(1+r^2+r^4\right) \end{aligned} \]

Squaring this expression gives \[ \begin{aligned} (ab+bc+cd)^2 &= a^4r^2\left(1+r^2+r^4\right)^2 \end{aligned} \]

Since both sides reduce to the same expression, it follows that \[ \begin{aligned} (a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2 \end{aligned} \] which proves the required result.

Q26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Solution

Let the four terms of the required geometric progression be \(3, ar, ar^2, 81\). Since the first term is \(3\), we have \(a=3\).

The fourth term of a geometric progression is given by \[ \begin{aligned} a_4 &= ar^3 \end{aligned} \] Substituting the known values, we obtain \[ \begin{aligned} 3r^3 &= 81 \end{aligned} \]

Dividing both sides by \(3\), we get \[ \begin{aligned} r^3 &= 27 \end{aligned} \] which gives \[ \begin{aligned} r &= \sqrt[3]{27}=3 \end{aligned} \]

Now, the second and third terms are \[ \begin{aligned} ar &= 3\times 3=9\\ ar^2 &= 3\times 3^2=27 \end{aligned} \]

Hence, the required geometric progression is \[ \begin{aligned} 3,\;9,\;27,\;81 \end{aligned} \]

Q27. Find the value of \(n\) so that \(\dfrac{a^{n+1}+ b^{n+1}}{a^n+b^n}\) may be the geometric mean between \(a\) and \(b\).

Solution

For the given expression to be the geometric mean between \(a\) and \(b\), it must be equal to \(\sqrt{ab}\). Hence, we write \[ \begin{aligned} \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab} \end{aligned} \]

Multiplying both sides by \((a^n+b^n)\), we obtain \[ \begin{aligned} a^{n+1}+b^{n+1}=\sqrt{ab}\,(a^n+b^n) \end{aligned} \]

Writing \(\sqrt{ab}=a^{1/2}b^{1/2}\), the right-hand side becomes \[ \begin{aligned} a^{1/2}b^{1/2}a^n+a^{1/2}b^{1/2}b^n = a^{n+\frac12}b^{\frac12}+a^{\frac12}b^{n+\frac12} \end{aligned} \]

Thus, we have \[ \begin{aligned} a^{n+1}+b^{n+1} = a^{n+\frac12}b^{\frac12}+a^{\frac12}b^{n+\frac12} \end{aligned} \]

Rearranging terms, \[ \begin{aligned} a^{n+\frac12}\left(a^{\frac12}-b^{\frac12}\right) + b^{n+\frac12}\left(b^{\frac12}-a^{\frac12}\right)=0 \end{aligned} \]

Factoring further, \[ \begin{aligned} \left(a^{\frac12}-b^{\frac12}\right) \left(a^{n+\frac12}-b^{n+\frac12}\right)=0 \end{aligned} \]

Since \(a\neq b\), it follows that \[ \begin{aligned} a^{n+\frac12}=b^{n+\frac12} \end{aligned} \] which is possible only when \[ \begin{aligned} n+\frac12=0 \end{aligned} \]

Hence, \[ \begin{aligned} n=-\frac12 \end{aligned} \] which is the required value of \(n\).

Q28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio \((3+2\sqrt{2}):(3-2\sqrt{2})\).

Solution

Let the two numbers be \(a\) and \(b\). It is given that their sum is six times their geometric mean. Hence, \[ \begin{aligned} a+b=6\sqrt{ab} \end{aligned} \]

Squaring both sides, we obtain \[ \begin{aligned} (a+b)^2=36ab \end{aligned} \]

Using the identity \((a-b)^2=(a+b)^2-4ab\), we have \[ \begin{aligned} (a-b)^2 &= 36ab-4ab\\ (a-b)^2 &= 32ab \end{aligned} \]

Taking the positive square root, since the ratio sought is positive, \[ \begin{aligned} a-b=\sqrt{32ab}=4\sqrt{2ab} \end{aligned} \]

Now we have the two relations \[ \begin{aligned} a+b=6\sqrt{ab}\\ a-b=4\sqrt{2ab} \end{aligned} \]

Adding these equations, we get \[ \begin{aligned} 2a=6\sqrt{ab}+4\sqrt{2ab} \end{aligned} \]

Dividing by \(2\), \[ \begin{aligned} a=3\sqrt{ab}+2\sqrt{2ab} \end{aligned} \]

Taking \(\sqrt{ab}\) as common, this gives \[ \begin{aligned} a=\sqrt{ab}(3+2\sqrt{2}) \end{aligned} \]

From \(a+b=6\sqrt{ab}\), we find \[ \begin{aligned} b &= 6\sqrt{ab}-a\\ &= 6\sqrt{ab}-\sqrt{ab}(3+2\sqrt{2})\\ &= \sqrt{ab}(3-2\sqrt{2}) \end{aligned} \]

Therefore, the ratio of the two numbers is \[ \begin{aligned} \dfrac{a}{b} = \dfrac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}} = \dfrac{3+2\sqrt{2}}{3-2\sqrt{2}} \end{aligned} \]

Hence, the given numbers are in the ratio \((3+2\sqrt{2}):(3-2\sqrt{2})\), as required.

Q29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \(A\pm \sqrt{(A+G)(A-G)}\).

Solution

Let the two positive numbers be \(x\) and \(y\). Since \(A\) is the arithmetic mean and \(G\) is the geometric mean between them, we have \[ \begin{aligned} A &= \dfrac{x+y}{2}\\ G &= \sqrt{xy} \end{aligned} \]

From the definition of arithmetic mean, we get \[ \begin{aligned} x+y = 2A \end{aligned} \] and from the definition of geometric mean, \[ \begin{aligned} xy = G^2 \end{aligned} \]

Now consider the identity \[ \begin{aligned} (x-y)^2 = (x+y)^2 - 4xy \end{aligned} \] Substituting the obtained values, we get \[ \begin{aligned} (x-y)^2 &= (2A)^2 - 4G^2\\ &= 4(A^2 - G^2) \end{aligned} \]

Taking square roots, \[ \begin{aligned} x-y = 2\sqrt{A^2 - G^2} \end{aligned} \] Since \[ \begin{aligned} A^2 - G^2 = (A+G)(A-G) \end{aligned} \] we obtain \[ \begin{aligned} x-y = 2\sqrt{(A+G)(A-G)} \end{aligned} \]

Now we have the two equations \[ \begin{aligned} x+y &= 2A\\ x-y &= 2\sqrt{(A+G)(A-G)} \end{aligned} \]

Adding these equations, we get \[ \begin{aligned} 2x = 2A + 2\sqrt{(A+G)(A-G)} \end{aligned} \] which gives \[ \begin{aligned} x = A + \sqrt{(A+G)(A-G)} \end{aligned} \]

Subtracting the equations, we obtain \[ \begin{aligned} 2y = 2A - 2\sqrt{(A+G)(A-G)} \end{aligned} \] hence, \[ \begin{aligned} y = A - \sqrt{(A+G)(A-G)} \end{aligned} \]

Therefore, the two positive numbers are \[ \begin{aligned} A \pm \sqrt{(A+G)(A-G)} \end{aligned} \] which proves the required result.

Q30. 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 \(2^{nd}\) hour, \(4^{th}\) hour and \(n^{th}\) hour ?

Solution

The number of bacteria doubles every hour, so the growth of bacteria follows a geometric progression. The number of bacteria present initially is \(30\), which is the first term of the progression.

Since the bacteria double every hour, each term is obtained by multiplying the previous term by \(2\). Hence, the common ratio is \[ \begin{aligned} r=2 \end{aligned} \]

The number of bacteria at the end of the second hour is \[ \begin{aligned} 30\times 2^2 = 120 \end{aligned} \] and at the end of the fourth hour is \[ \begin{aligned} 30\times 2^4 = 480 \end{aligned} \]

In general, the number of bacteria present at the end of the \(n^{\text{th}}\) hour is given by the \((n+1)^{\text{th}}\) term of the geometric progression. Therefore, \[ \begin{aligned} a_{n+1} = a r^n = 30\times 2^n \end{aligned} \]

Thus, the required numbers of bacteria at the end of the second hour, fourth hour, and \(n^{\text{th}}\) hour are \(120\), \(480\), and \(30\times 2^n\), respectively.

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?

Solution

The amount obtained on a principal when interest is compounded annually is given by the formula \[ \begin{aligned} A=P\left(1+\dfrac{r}{100}\right)^n \end{aligned} \] where \(P\) is the principal, \(r\) is the annual rate of interest, and \(n\) is the time in years.

Here, the principal deposited is \(P=500\), the rate of interest is \(r=10\%\) per annum, and the time period is \(n=10\) years. Substituting these values, we get \[ \begin{aligned} A=500\left(1+\dfrac{10}{100}\right)^{10} \end{aligned} \]

Simplifying the expression, \[ \begin{aligned} A=500\left(\dfrac{11}{10}\right)^{10} \end{aligned} \]

Evaluating this gives \[ \begin{aligned} A \approx 1296.87 \end{aligned} \]

Hence, Rs \(500\) will amount to approximately Rs \(1296.87\) after \(10\) years when invested at an annual interest rate of \(10\%\) compounded annually.

Q32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Solution

Let the roots of the required quadratic equation be \(a\) and \(b\). It is given that the arithmetic mean of the roots is \(8\) and the geometric mean is \(5\).

From the definition of arithmetic mean, we have \[ \begin{aligned} \dfrac{a+b}{2}=8 \end{aligned} \] which gives \[ \begin{aligned} a+b=16 \end{aligned} \]

From the definition of geometric mean, we have \[ \begin{aligned} \sqrt{ab}=5 \end{aligned} \] and hence \[ \begin{aligned} ab=25 \end{aligned} \]

A quadratic equation whose roots are \(a\) and \(b\) is given by \[ \begin{aligned} x^2-(a+b)x+ab=0 \end{aligned} \]

Substituting the values of \(a+b\) and \(ab\), we obtain \[ \begin{aligned} x^2-16x+25=0 \end{aligned} \]

This is the required quadratic equation.