Exercise 8.2 NCERT Math Class 11 Best Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16 Q.17 Q.18 Q.19 Q.20 Q.21 Q.22 Q.23 Q.24 Q.25 Q.26 Q.27 Q.28 Q.29 Q.30 Q.31 Q.32

Exercise 8.2 NCERT Math Class 11 contains 32 questions. Each and every question has been fully solved, describing all steps. You can understand the solutions just by reading it.

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Exercise 8.2 NCERT Math Class 11 Solutions

1. Find the 20^th and n^th terms of the G.P.
$\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, ...$

Solution

The given G.P. is $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, ...$
We know that
n^th term of a G.P. = $a r^{n - 1}$
From the given G.P., we have,
The first term $(a)$ = $\frac{5}{2}$
and the common ratio $(r)$ = $\frac{2^{n d} t e r m}{1^{s t} t e r m}$ = $\frac{\frac{5}{4}}{\frac{5}{2}}$ = $\frac{5}{4} \times \frac{2}{5} = \frac{1}{2}$
$∴$ 20^th term = $\frac{5}{2} \times {(\frac{1}{2})}^{20 - 1} = \frac{5}{2} \times {(\frac{1}{2})}^{19} = \frac{5}{2} \times \frac{1}{2^{19}} = \frac{5}{2^{20}}$
$n^{t h}$ term = $\frac{5}{2} \times {((\frac{1}{2}))}^{n - 1} = \frac{5}{2 \times 2^{n - 1}}$ $= \frac{5}{2^{n}}$

2. Find the 20^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

Solution

It is given that 8^th term = 192 and common ratio (r) = 2.
$∵$ 8^th term = 192
$\Rightarrow a r^{8 - 1} = 192$
$\Rightarrow a \times 2^{7} = 192$
$\Rightarrow a \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 192 \Rightarrow a \times 128 = 192$
$\Rightarrow a = \frac{192}{128} =$ $\frac{3}{2}$
Now, 20^th term = $\frac{3}{2} \times 2^{20 - 1} = \frac{3}{2} \times 2^{19} = 3 \times 2^{18}$ = 3072

3. The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

How to show: Method

We will take LHS first and find its value. After that, we will take RHS and find its value. If done correctly, values obtained in RHS and LHS will match. And that will be our proof.

Solution

It is given that
$5^{th}$ term = p,
$8^{th}$ term = q,
$11^{th}$ term = s
We have to show that $q^{2} = p s$ .
Taking LHS, we get,
LHS = $q^{2}$
= (8^th term)² ...(since q = 8^th term)
= ( $a r^{8 - 1}$ )²
= ( $a r^{7}$ )²
= $a^{2} r^{14}$
$∴$ LHS = $a^{2} r^{14}$
Now, taking RHS, we get,
RHS = ps
= (5^th term)(11^th term)
= ( $a r^{5 - 1}$ )( $a r^{11 - 1}$ )
= $a r^{4} \times$ $a r^{10}$
= $a^{(1 + 1)} < \times r^{(10 + 4)}$
= $a^{2} r^{14}$
$∴$ RHS = $a^{2} r^{14}$
$∵$ LHS = RHS, hence proved.

4. 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.

How to Solve

To determine the 7^th term, we need the first term and the common ratio. First term is given, you have to find only the common ratio using the equation formed by the condition given in the first sentence.

Solution

Let $a$ and $r$ be the first term and the common ratio of the G.P. respectively.
According to the question,
$a = - 3$ and
4^th term = (2^nd term)²
$\Rightarrow a r^{4 - 1}$ = ( $a r^{2 - 1}$ )²
$\Rightarrow a r^{3}$ = $(a r^{1})$ ²
$\Rightarrow a r^{3}$ = $a^{2} r^{2}$
Dividing by $a r^{2}$ on both sides, we get,
$r$ = $a$
$\Rightarrow r$ = $- 3$ ...(Given $a$ = -3)
Now, 7^th term = $a r^{7 - 1}$ = (-3) $\times$ (-3)⁶ = -2187

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

How to Solve

Since in each of the questions, the first term and the common ratio can be found easily. We will suppose the given term as n^th term and solve the resulting equation. We just need to find the value of n. And that will be our answer.

Solution

5. (a) The given G.P. is 2, $2 \sqrt{2}$ , 4, ...
Here, $a$ = 2, and r = $\frac{2 \sqrt{2}}{2}$ = $\sqrt{2}$
Let 128 is the n^th term.
$\Rightarrow$ n^th term = 128
$\Rightarrow a r^{n - 1}$ = 128
$\Rightarrow 2 \times$ ( $\sqrt{2}$ )^{n - 1} = 128
$\Rightarrow 2 \times$ ( $2^{\frac{1}{2}}$ )^{n - 1} = 128 ...( $∵ \sqrt{2}$ = 2^{$\frac{1}{2}$})
$\Rightarrow 2 \times$ ( $2^{\frac{n - 1}{2}}$ ) = 128 ... $∵$ ( $a^{m}$ )ⁿ = $a^{m \times n}$
$\Rightarrow$ $2^{\frac{n - 1}{2} + 1}$ = 128 ... $∵$ $a^{n} \times a^{m} = a^{m + n}$
$\Rightarrow 2^{\frac{n - 1 + 2}{2}}$ = 128
$\Rightarrow 2^{\frac{n + 1}{2}}$ = $2^{7}$ ... $∵$ 128 = 2⁷
In the above equation, since the bases are equal, the exponents or the powers must be same.
So, $\frac{n + 1}{2} = 7$
$\Rightarrow n + 1 = 14$
$\Rightarrow n = 14 - 1 = 13$
Hence, 128 is the 13^th term
(b) The given G.P. is $\sqrt{3}$ , 3, $3 \sqrt{3}$ , ...
Here, $a = \sqrt{3}$ and $r = \frac{3}{\sqrt{3}} = \sqrt{3}$
Let 729 is the n^th term.
$\Rightarrow$ n^th term = 729
$\Rightarrow a r^{n - 1}$ = 729
$\Rightarrow \sqrt{3} \times$ ( $\sqrt{3}$ )^{n - 1} = 729
$\Rightarrow 3^{\frac{1}{2}} \times 3^{\frac{n - 1}{2}}$ = 729
$\Rightarrow 3^{\frac{1}{2} + \frac{n - 1}{2}}$ = 729
$\Rightarrow 3^{\frac{1 + n - 1}{2}}$ = 3⁶ ... $∵$ 729 = 3⁶
$\Rightarrow 3^{\frac{n}{2}}$ = 3⁶
Since bases are same in the equation, the powers or exponents must be same.
So, $\frac{n}{2} = 6 \Rightarrow n = 12$
Hence, 729 is the 12^th term.
(c) The given G.P. is $\frac{1}{3}$ , $\frac{1}{9}$ , $\frac{1}{27}$ , ...
Here, $a = \frac{1}{3}$ and $r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}$
Let $\frac{1}{19683}$ is the n^th term.
$\Rightarrow$ n^th term = $\frac{1}{19683}$
$\Rightarrow a r^{n - 1} = \frac{1}{19683}$
$\Rightarrow \frac{1}{3} \times$ ( $\frac{1}{3}$ )^n-1 $= \frac{1}{19683}$
$\Rightarrow$ ( $\frac{1}{3}$ )^{1 + n - 1} $= \frac{1}{19683}$
$\Rightarrow$ ( $\frac{1}{3}$ )ⁿ $=$ $\frac{1}{3^{9}}$
$\Rightarrow \frac{1}{3^{n}} = \frac{1}{3^{9}}$ Taking Reciprocals on both sides, we get,
$3^{n} = 3^{9}$
Since bases are equal, powers must be same.
So, n = 9.
Hence, $\frac{1}{19683}$ is the 9^th term.

6. For what values of x, the numbers $- \frac{2}{7}, x, - \frac{7}{2}$ are in G.P.?

How to Solve

Since the given terms are in G.P. , their common ratio between its consecutive terms must be same. We can find the value of $x$ by equating the ratio obtained by dividing the second term by the first term and the third term by the second term. Solving the resulting equation will give us the values of $x$ .

Solution

Since the numbers $- \frac{2}{7}, x, - \frac{7}{2}$ are in G.P., the common ratio should be same between every consecutive numbers.
So, $\frac{Second Term}{First Term} = \frac{Third Term}{Second Term}$
$\Rightarrow \frac{x}{- \frac{2}{7}} = \frac{- \frac{7}{2}}{x}$
$\Rightarrow x^{2} = \frac{- 7}{2} \times \frac{2}{- 7}$
$\Rightarrow x^{2} = 1$
$\Rightarrow \sqrt{x^{2}} = \sqrt{1}$ ...(taking square roots)
$\Rightarrow |x| = 1$
$\Rightarrow x = \pm 1$
Hence, the value of x is 1 or −1

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

7. 0.15, 0.015, 0.0015, ... 20 terms.

Solution

Here, $a = 0.15$ and $r = \frac{0.015}{0.15} = 0.1 We know that S_{n} = \frac{a (1 - r^{n})}{1 - r} ... for 0 < r < 1$
$\Rightarrow S_{20} = \frac{0.15 (1 - {0.1}^{20})}{1 - 0.1}$
$= \frac{0.15}{0.9} (1 - {0.1}^{20})$
$= \frac{15}{90} (1 - {0.1}^{20})$
$= \frac{1}{6} (1 - {0.1}^{20})$

8. $\sqrt{7}, \sqrt{21}, 3 \sqrt{7},$ ... n terms.

Solution

Here, $a = \sqrt{7}$ and $r = \frac{\sqrt{21}}{\sqrt{7}} = \frac{\sqrt{7} \times \sqrt{3}}{\sqrt{7}} = \sqrt{3}$
We know that
$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$
$S_{n} = \frac{\sqrt{7} [({\sqrt{3)}}^{n} - 1]}{\sqrt{3} - 1}$
Rationalising the denominator, we get,
$S_{n} = \frac{\sqrt{7} [({\sqrt{3)}}^{n} - 1]}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$
$S_{n} = \frac{\sqrt{7} (\sqrt{3} + 1) [({\sqrt{3)}}^{n} - 1]}{2}$
$∴ S_{n} = \frac{\sqrt{7}}{2} (\sqrt{3} - 1) (3^{\frac{n}{2}} - 1)$

9. 1, $- a, a^{2}, - a^{3}, ... n$ terms. (if $a \neq - 1$ )

Solution

Here, $a = 1$ and $r = \frac{- a}{1} = - a$
We know that
$S_{n} = \frac{a (1 - r^{n})}{1 - r}$
$∴ S_{n} = \frac{1 [1 - (- a)^{n}]}{1 - (- a)} =$ $\frac{[1 - (- a)^{n}]}{1 + a}$

10. $x^{3}, x^{5}, x^{7},$ ... n terms (if $x \neq \pm 1$ )

Solution

Here, $a = x^{3}$ and $r = \frac{x^{5}}{x^{3}} = x^{2}$
We know that
$S_{n} = \frac{a (1 - r^{n})}{1 - r}$
$∴ S_{n} = \frac{x^{3} [1 - (x^{2})^{n}]}{1 - x^{2}}$ $= \frac{x^{3} (1 - x^{2 n})}{1 - x^{2}}$

11. Evaluate $\sum_{k = 1}^{11} (2 + 3^{k})$

How to solve

The sigma $\sum_{k = 1}^{11}$ (2+3^k) means sum of all terms formed by putting the value of k in the expression till k = 11. And after that we regroup the terms which can make a useful sequence.

Solution

We have,
$\sum_{k = 1}^{11} (2 + 3^{k})$ = (2+3¹) + (2+3²) + (2+3³) + ... 11 times.
= (2+2+2+ ... 11 times) + (3+3²+3³+ ... 11 times)

The second bracket contains
 a G.P., where  $a = 3$  and 
 $r = \frac{3^{2}}{3} = 3$

= 2

\times

11 +

\frac{a (r^{n} - 1)}{r - 1}

= 22 +

\frac{3 (3^{11} - 1)}{3 - 1}

= 22 +

\frac{3}{2} (3^{11} - 1)

12. 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.

Concept

Whenever you are given the sum and product of three or four terms in the question, you must always suppose the terms as supposed in the solution.

Solution

Let the first three terms of the G.P. be $\frac{a}{r}, a, a r$ .
According to the given question,
$\frac{a}{r} + a + a r = \frac{39}{10}$ ... (1)
$\frac{a}{r} \times a \times a r = 1$ ... (2)
From equ⁽²⁾, we get,
$\frac{a}{r} \times a \times a r = 1$ , cancelling r, we get
$a \times a \times a = 1$
$\Rightarrow a^{3} = 1$
$\Rightarrow a^{3} - 1 = 0$
$\Rightarrow (a - 1) (a^{2} + a + 1) = 0$
$\Rightarrow (a - 1) = 0 o r (a^{2} + a + 1) = 0$
But, $(a^{2} + a + 1) \neq 0$
As B²-4AC = 1²-4×1×1 = -3<0,
That means roots are not real.
So, $a - 1 = 0$
$\Rightarrow a = 1$
Putting the value of $a$ in equ⁽¹⁾,
$\frac{1}{r} + 1 + r = \frac{39}{10}$
$\Rightarrow \frac{1 + r + r^{2}}{r} = \frac{39}{10}$
$\Rightarrow 10 \times (1 + r + r^{2}) = 39 r$
$\Rightarrow 10 + 10 r + 10 r^{2} = 39 r$
$\Rightarrow 10 r^{2} + 10 r - 39 r + 10 = 0$
$\Rightarrow 10 r^{2} - 29 r + 10 = 0$
$\Rightarrow 10 r^{2} - 25 r - 4 r + 10 = 0$
$\Rightarrow 5 r (2 r - 5) - 2 (2 r - 5) = 0$
$\Rightarrow (2 r - 5) (5 r - 2) = 0$
$\Rightarrow (2 r - 5) = 0 o r (5 r - 2) = 0$
$\Rightarrow r = \frac{5}{2} o r r = \frac{2}{5}$
There are two common ratios, which means there will be two sets of associated terms.
So, when $a = 1$ and $r = \frac{5}{2}$
The terms are $\frac{a}{r} = \frac{1}{\frac{5}{2}} = \frac{2}{5},$
$a = 1,$
$a r = 1 \times \frac{5}{2} = \frac{5}{2}$
Ans: ( $\frac{2}{5}, 1, \frac{5}{2}$ )
Similarly, when $a = 1$ and $r = \frac{2}{5}$
Ans: ( $\frac{5}{2}, 1, \frac{2}{5}$ )

13. How many terms of G.P. 3, 3², 3³, ... are needed to give the sum 120?

Solution

Let n terms of the given G.P. are needed to give the sum 120. So, we can write,
$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$
Here, $S_{n} = 120$ , $a = 3$ , and $r = {\frac{3}{3}}^{2} = 3$
$\Rightarrow \frac{3 (3^{n} - 1)}{3 - 1} = 120$
$\Rightarrow \frac{3}{2} (3^{n} - 1) = 120$
$\Rightarrow 3^{n} - 1 = 120 \times \frac{2}{3}$
$\Rightarrow 3^{n} - 1 = 80$
$\Rightarrow 3^{n} = 80 + 1$
$\Rightarrow 3^{n} = 81$
$\Rightarrow 3^{n} = 3^{4}$
Since bases are equal on both sides of the equation, the exponents must also be equal.
$\Rightarrow n = 4$
Hence, the sum of 4 terms are needed to give the sum 120.

14. 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 three terms of the G.P. be $a, a r$ and $a r^{2}$ .
Also, let the next three terms of the G.P. be $a r^{3}, a r^{4},$ and $a r^{5}$ , where $a$ is the first term and r is the common ratio of the G.P.
According to the given question, we can write,
$a + a r + a r^{2} = 16$ ... (1)
$a r^{3} + a r^{4} + a r^{5} = 128$ ... (2)
On dividing (1) by (2), we get,
$\frac{a + a r + a r^{2}}{a r^{3} + a r^{4} + a r^{5}} = \frac{16}{128}$ ,
$\Rightarrow \frac{a (1 + r + r^{2})}{a r^{3} (1 + r + r^{2})} = \frac{16}{128}$
$\Rightarrow \frac{1}{r^{3}} = \frac{1}{8}$ ( $∵ \frac{16}{128} = \frac{16}{16 \times 8} = \frac{1}{8}$ )
$\Rightarrow r^{3} = 8$
$\Rightarrow r^{3} = 2^{3}$
$\Rightarrow r = 2$ ( $∵$ Exponents are same, the bases must be same)
Putting the value of r in (1), we get,
$a + 2 a + 4 a = 16$
$\Rightarrow 7 a = 16$
$\Rightarrow a = \frac{16}{7}$
Now, $S_{n} = \frac{a (r^{n} - 1)}{r - 1} = \frac{\frac{16}{7} (2^{n} - 1)}{2 - 1} = \frac{16}{7} (2^{n} - 1)$
Hence, the first term = $\frac{16}{7}$ ,
Common Ratio = 2, and
Sum to n terms = $S_{n} = \frac{16}{7} (2^{n} - 1)$

15. Given a G.P. with $a = 729$ and $7^{th}$ term = 64, determine $S_{7}$

Solution

According to the given question,
first term ( $a$ ) = 729, and
7^th term = 64
$\Rightarrow a_{7} = 64$
$\Rightarrow a r^{7 - 1} = 64$
$\Rightarrow 729 \times r^{6} = 64$ .... ( $∵ a = 729$ )
$\Rightarrow r^{6} = \frac{64}{729}$
$\Rightarrow r^{6} = \frac{2^{6}}{3^{6}}$
$\Rightarrow r^{6} = {(\frac{2}{3})}^{6}$
$\Rightarrow r^{6} - {((\frac{2}{3}))}^{6} = 0$
$\Rightarrow {((r^{2}))}^{3} - {(({((\frac{2}{3}))}^{2}))}^{3} = 0$
$\Rightarrow [[r^{2} - {((\frac{2}{3}))}^{2}]] [[r^{4} + \frac{2}{3} r^{2} + {((\frac{2}{3}))}^{4}]] = 0$ $\Rightarrow [[r^{2} - {((\frac{2}{3}))}^{2}]] = 0 o r$
$[[r^{4} + \frac{2}{3} r^{2} + {((\frac{2}{3}))}^{4}]] = 0$
As the equation $[[r^{4} + \frac{2}{3} r^{2} + {((\frac{2}{3}))}^{4}]] = 0$ does not have real roots, we can neglect it.
So, $[[r^{2} - {((\frac{2}{3}))}^{2}]] = 0$
$\Rightarrow r^{2} - {((\frac{2}{3}))}^{2} = 0$
$\Rightarrow ((r - \frac{2}{3})) ((r + \frac{2}{3})) = 0$
$\Rightarrow ((r - \frac{2}{3})) = 0 o r ((r + \frac{2}{3})) = 0$
$\Rightarrow r = \frac{2}{3} o r r = - \frac{2}{3}$
Since the first term and common ratios are known, we can find S₇.
$∵ S_{n} = \frac{a (1 - r^{n})}{1 - r}$
When $r = \frac{2}{3}$
$∴ S_{7} = \frac{729 [1 - {(\frac{2}{3})}^{7}]}{(1 - \frac{2}{3})}$
$\Rightarrow S_{7} = \frac{729 \times [1 - \frac{2^{7}}{3^{7}}]}{\frac{1}{3}} =$ $729 \times 3 \times [1 - \frac{128}{2187}]$
$\Rightarrow S_{7} = 729 \times 3 \times \frac{2059}{2187}$ $= 2187 \times \frac{2059}{2187}$
$\Rightarrow S_{7} = 2059$
When $r = - \frac{2}{3}$
$S_{7} = \frac{729 [[1 - {((\frac{- 2}{3}))}^{7}]]}{1 - ((\frac{- 2}{3}))}$ $= \frac{729 [[1 - {((\frac{- 128}{2187}))}^{7}]]}{1 + \frac{2}{3}}$
$S_{7} = \frac{729 [[1 + \frac{128}{2187}]]}{\frac{5}{3}}$ $= 729 \times \frac{3}{5} ((\frac{2315}{2187}))$ $= \frac{2187}{5} \times \frac{2315}{2187}$ $= 463$
Hence, $S_{7} = 2059 o r 463$

16. Find a G.P. for which sum of the first two terms is $- 4$ and fifth term is 4 times the third term.

Solution

Let the given G.P. be
$a_{1}, a_{2}, a_{3}, a_{4}, a_{5},$ ...
According to the question, we have,
$a_{1} + a_{2} = - 4$ ... (1)
$a_{5} = 4 \times a_{3}$ ... (2)
Let $a$ and $r$ be the first term and common ratio respectively.
From (1), we get,
$a_{1} + a_{2} = - 4$
$\Rightarrow a + a r = - 4$
$\Rightarrow a (1 + r) = - 4$ ...(3)
From (2), we get,
$a_{5} = 4 \times a_{3}$
$\Rightarrow a r^{5 - 1} = 4 \times a r^{3 - 1}$
$\Rightarrow a r^{4} = 4 \times a r^{2}$
Dividing by $a r^{2}$ on both sides,
$\Rightarrow \frac{a r^{4}}{a r^{2}} = \frac{4 \times a r^{2}}{a r^{2}}$
$\Rightarrow r^{2} = 4$
$\Rightarrow r^{2} - 4 = 0$
$\Rightarrow r^{2} - 2^{2} = 0$
$\Rightarrow (r - 2) (r + 2) = 0$
$∵ (a^{2} - b^{2}) = (a - b) (a + b)$
$\Rightarrow (r - 2) = 0 o r (r + 2) = 0$
$\Rightarrow r = 2 o r r = - 2$
Putting the value of r in equ⁽³⁾,
When r = 2,
$a (1 + 2) = - 4$
$\Rightarrow a \times 3 = - 4$
$\Rightarrow a = \frac{- 4}{3}$
In this case, G.P. is given by
$a, a r, a r^{2}, a r^{3}, a r^{4}, . . .$
$\frac{- 4}{3}, \frac{- 4}{3} \times 2, \frac{- 4}{3} \times 2^{2}, \frac{- 4}{3} \times 2^{3}, \frac{- 4}{3} \times 2^{4}, . . .$
$\frac{- 4}{3}, \frac{- 8}{3}, \frac{- 16}{3}, \frac{- 32}{3}, \frac{- 64}{3}, . . .$
When r = -2,
$a (1 - 3) = - 4$
$\Rightarrow a \times (- 2) = - 4$
$\Rightarrow a = \frac{- 4}{- 2} = 2$
In this case, G.P. is given by
$a, a r, a r^{2}, a r^{3}, a r^{4}, . . .$
$2, 2 \times (- 2), 2 \times {((- 2))}^{2},$ $2 \times {((- 2))}^{3}, 2 \times {((- 2))}^{4}, . . .$
$2, - 4, 8, - 16, 32, . . .$

17. If the $4^{th}, 10^{th}$ and $16^{th}$ terms of a G.P. are $x, y$ and $z$ respectively. Prove that $x, y,$ and $z are in G.P.$

Solution

Let the first term and common ratio of the G.P. be

a

and

r

respectively.
According to the given question, we have,

4^{t h} t e r m = a_{4} = x

10^{t h} t e r m = a_{10} = y

16^{t h} t e r m = a_{16} = z

To prove that

x, y, a n d z

are in G.P., we need to show that

\frac{y}{x} = \frac{z}{y}

i.e. common ratios are equal.

\frac{y}{x} =

\frac{10^{t h} t e r m}{4^{t h} t e r m} = \frac{a_{10}}{a_{4}} = \frac{a r^{10 - 1}}{a r^{4 - 1}} = \frac{a r^{9}}{a r^{3}} = r^{6}

\frac{z}{y} = \frac{16^{t h} t e r m}{10^{t h} t e r m} = \frac{a_{16}}{a_{10}} = \frac{a r^{16 - 1}}{a r^{10 - 1}} = \frac{a r^{15}}{a r^{9}} = r^{6}

Since

\frac{y}{x} = \frac{z}{y}

x, y, a n d z

are in G.P. Hence, Proved.

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

Solution

The given sequence is
$8, 88, 888, 8888, . . .$
Sum to n terms
$S_{n} = 8 + 88 + 888 + 8888 + . . .$ to n terms
$= 8 (1 + 11 + 111 + 1111 + . . . t o n t e r m s)$
$= \frac{8}{9} \times 9 (1 + 11 + 111 + 1111 + . . . t o n t e r m s)$
$= \frac{8}{9} \times (9 + 99 + 999 + 9999 + . . . t o n t e r m s)$
$= \frac{8}{9} \times [(10 - 1) + (100 - 1) + (1000 - 1) + (10000 - 1) + . . . t o n t e r m s]$
$= \frac{8}{9} \times [(10 + 100 + 1000 + 10000 + . . . t o n t e r m s) - (1 + 1 + 1 + 1 + . . . t o n t e r m s)]$
$= \frac{8}{9} \times [(10 + 10^{2} + 10^{3} + 10^{4} + . . . t o n t e r m s) - n]$
Here, $10 + 10^{2} + 10^{3} + 10^{4} + . . . t o n t e r m s$ is a G.P.
with $a = 10$ and $r = 10$ . So, using the formula to calculate its sum, we get.
$= \frac{8}{9} \times [[\{{\frac{10 (10^{n} - 1)}{10 - 1}}\} - n]]$
$= \frac{8}{9} \times [[\{{\frac{10}{9} (10^{n} - 1)}\} - n]]$
$= \frac{8}{9} \times \{{\frac{10}{9} (10^{n} - 1)}\} - \frac{8 n}{9}$
$= \frac{80}{81} (10^{n} - 1) - \frac{8 n}{9}$ .

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, $\frac{1}{2}$

Solution

The given sequences are
$2, 4, 8, 16, 32$
$128, 32, 8, 2, \frac{1}{2}$
Sum of the products of the corresponding terms of the sequences
$= 2 \times < 128 + 4 \times 32 + 8 \times 8 + 16 \times 2 + 32 \times \frac{1}{2}$
$= 256 + 128 + 64 + 32 + 16$
Clearly, it is a G.P. with $a = 256$ and $r = \frac{128}{256} = \frac{1}{2}$ .
$&#x2235$ Total number of terms in the given series is 5.
So, $S_{5} = \frac{256 \times [[1 - {((\frac{1}{2}))}^{5}]]}{((1 - \frac{1}{2}))}$ $= \frac{256 \times [[1 - \frac{1}{32}]]}{\frac{1}{2}}$ = $256 \times 2 \times [[\frac{32 - 1}{32}]]$ $= 16 \times 31$ = 496
Hence, the sum of the products of the corresponding terms of the sequences is 496.

20. Show that the products of the corresponding terms of the sequences $a, a r, a r^{2}, . . ., a r^{n - 1}$ and $A, A R, A R^{2}, . . . A R^{n-1}$ form a G.P., and find the common ratio.

Solution

The given sequences are
$a, a r, a r^{2}, . . ., a r^{n - 1}$
$A, A R, A R^{2}, . . . A R^{n-1}$
The sequence obtained by multiplying the corresponding terms of the above sequences is given by
$a A, a A r R, a A r^{2} R^{2}, . . ., a A r^{n-1} R^{n-1}$

To show that the above sequence forms a G.P., we only need to show that common ratio is same between every consecutive terms of the above sequence.

r_{1} = \frac{\sec o n d t e r m}{f i r s t t e r m} = \frac{a A r R}{a A} = r R

and

r_{2} = \frac{\sec o n d t e r m}{f i r s t t e r m} = \frac{a A r^{2} R^{2}}{a A r R} = r R

∵ r_{1} = r_{2}

Hence, the sequence obtained by multiplying the corresponding terms of the given sequences forms a G.P. as the common ratios are same and equal to

r R

21. 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 in G.P. be
$a, a r, a r^{2}, a r^{3}$
Where $a$ is the first term and $r$ is the common ratio.
According to the given question,
$a r^{2} - a = 9$
$\Rightarrow a (r^{2} - 1) = 9$ ... (1)
$a r - a r^{3} = 18$
$\Rightarrow a r (1 - r^{2}) = 18$ ... (2)
Dividing equation (1) by equation (2), we get,
$\frac{a (r^{2} - 1)}{a r (1 - r^{2})} = \frac{9}{18}$
$\Rightarrow \frac{- a (1 - r^{2})}{a r (1 - r^{2})} = \frac{9}{18}$
$\Rightarrow \frac{- 1}{r} = \frac{1}{2}$
$\Rightarrow r = - 2$
Putting the value of 'r' in equation (1), we get,
$\Rightarrow a [{(-2)}^{2} - 1] = 9$
$\Rightarrow a \times 3 = 9$
$\Rightarrow a = \frac{9}{3} = 3$
Hence, the four numbers are
$a, a r, a r^{2}, a r^{3}$
$= 3, 3 \times (- 2), 3 \times {((- 2))}^{2}, 3 \times {((- 2))}^{3}$
$= 3, - 6, 12, - 24$

22. 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 and common ratio of the G.P. be $A$ and $R$ respectively.
According to the given question, we have,
$p^{t h} t e r m = a$
$\Rightarrow A R^{p - 1} = a$ ... (1)
$q^{t h} t e r m = b$
$\Rightarrow A R^{q - 1} = b$ ... (2)
$r^{t h} t e r m = c$
$\Rightarrow A R^{r - 1} = c$ ... (3)
To Prove: $a^{q - r} b^{r - p} c^{p - q} = 1$
Taking LHS, we have,
$a^{q - r} b^{r - p} c^{p - q}$
$= {((A R^{p - 1}))}^{q - r} {((A R^{q - 1}))}^{r - p} {((A R^{r - 1}))}^{p - q}$
$= A^{q - r} {((R^{p - 1}))}^{q - r} \times A^{r - p} {((R^{q - 1}))}^{r - p} \times A^{p - q} {((R^{r - 1}))}^{p - q}$
$= A^{q - r} \times A^{r - p} \times A^{p - q} \times R^{(p - 1) (q - r)} \times R^{(q - 1) (r - p)} \times R^{(r - 1) (p - q)}$
$= A^{q - r + r - p + p - q} \times R^{(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)}$
$= A^{0} \times R^{p q - p r - q + r + q r - q p - r + p + r p - r q - p + q}$
$= A^{0} \times R^{0} = 1 \times 1$ = 1 = RHS
Hence proved!

23. 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 $P^{2} = {(a b)}^{n}$ .

Solution

It is given that first term and $n^{th}$ term of the G.P. are $a$ and $b$ respectively.
Let $r$ be the common ratio of the G.P. . Then, the n terms of the G.P. can be written as
$a, a r, a r^{2}, a r^{3}, . . ., a r^{n - 1}$
$∵ P =$ product of n terms.
$= a \times a r \times a r^{2} \times a r^{3} \times . . . \times a r^{n - 1}$
$= (a \times a \times a \times a \times . . . n t i m e s) \times r^{1 + 2 + 3 + . . . + (n - 1)}$
Here, exponent of r is an arithmetic series. So,
$P = a^{n} \times r^{[[\frac{n - 1}{2} \{{1 + (n - 1)}\}]]}$
$= a^{n} \times r^{[\frac{n - 1}{2} \times n]}$
$= a^{n} \times r^{[[\frac{n (n - 1)}{2}]]}$
To prove: $P^{2} = {(a b)}^{n}$
Taking LHS, we get,
$P^{2} = {[[a^{n} \times r^{[[\frac{n (n - 1)}{2}]]}]]}^{2}$
$= {((a^{n}))}^{2} \times {[[r^{[[\frac{n (n - 1)}{2}]]}]]}^{2}$
$= a^{2 n} \times r^{\frac{n (n - 1)}{2} \times 2}$
$= a^{2 n} \times r^{n (n - 1)}$
$= {((a^{2} \times r^{n - 1}))}^{n}$
$= {((a \times a r^{n - 1}))}^{n}$
$= {((a b))}^{n}$ ... As b is the $n^{th}$ term.
= RHS
Hence proved!

24. 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 $\frac{1}{r^{n}}$ .

Solution

Let the first term and common ratio of the G.P. be
$a$ and $r$ respectively.
The G.P., up to 2n number of terms, is as follows:
$a, a r, a r^{2}, . . ., a r^{n - 1}, a r^{n}, a r^{n + 1}, a r^{n + 2}, . . ., a r^{2 n - 1}$
Here, the number of terms from $' a'$ to $' a r^{n - 1}'$ is n
and the number of terms from $' a r^{n}'$ to $' a r^{2 n - 1}'$ is
also n .
To show: $\frac{Sum of first n terms}{Sum of terms from (n+1)th to (2n)th terms}$ = $\frac{1}{r^{n}}$
Taking LHS, we get,
$\frac{Sum of first n terms}{Sum of terms from (n+1)th to (2n)th terms}$
= $\frac{a + a r + a r^{2} + a r^{3} + . . . + a r^{n - 1}}{a r^{n} + a r^{n + 1} + a r^{n + 2} + . . . + a r^{2 n - 1}}$
= $\frac{\frac{a (r^{n} - 1)}{r - 1}}{\frac{a r^{n} (r^{n} - 1)}{r - 1}}$ = $\frac{a (r^{n} - 1)}{r - 1} \times \frac{r - 1}{a r^{n} (r^{n} - 1)}$ = $\frac{1}{r^{n}}$ = RHS
Hence proved!

25. If $a, b, c$ and $d$ are in G.P.. Show that
$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})$ $= {((a b + b c + c d))}^{2}$

Solution

Given that $a, b, c$ and $d$ are in G.P.
$\Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = Common Ratio (r)$
Equating in pairs, we get,
$\frac{b}{a} = \frac{c}{b}$ $\Rightarrow b^{2} = a c$ ... (1)
$\frac{c}{b} = \frac{d}{c}$ $\Rightarrow c^{2} = b d$ ... (2)
$\frac{b}{a} = \frac{d}{c}$ $\Rightarrow b c = a d$ ... (3)
To show: $(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})$ = ${((a b + b c + c d))}^{2}$
Taking LHS, we get,
$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})$
$= a^{2} b^{2} + a^{2} c^{2} + a^{2} d^{2} + b^{4} + b^{2} c^{2} + b^{2} d^{2} + c^{2} b^{2} + c^{4} + c^{2} d^{2}$
$= a^{2} b^{2} + b^{2} c^{2} + c^{2} d^{2} + (b^{4} + a^{2} c^{2}) + (c^{4} + b^{2} d^{2}) + (a^{2} d^{2} + c^{2} b^{2})$
$= a^{2} b^{2} + b^{2} c^{2} + c^{2} d^{2} + (b^{2} \times b^{2} + a c \times a c) + (c^{2} \times c^{2} + b d \times b d) + (a d \times a d + c b \times c b)$
$= a^{2} b^{2} + b^{2} c^{2} + c^{2} d^{2} + (b^{2} \times a c + b^{2} \times a c) + (c^{2} \times b d + c^{2} \times b d) + (a d \times c b + a d \times c b)$
Using equations (1), (2) and (3) appropriately,
so that we could get the RHS.
$= a^{2} b^{2} + b^{2} c^{2} + c^{2} d^{2} + (a b^{2} c + a b^{2} c) + (b c^{2} d + b c^{2} d) + (a b c d + a b c d)$
$= a^{2} b^{2} + b^{2} c^{2} + c^{2} d^{2} + 2 a b^{2} c + 2 b c^{2} d + 2 a b c d$
= ${((a b + b c + c d))}^{2}$
= RHS, Hence Proved!

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

Solution

Let the two numbers between 3 and 81 be p and q such that the resulting sequence is G.P.
So, the resulting G.P. can be written as
$3, p, q, 81$
Here, $a = 3$
$4^{th}$ term = 81
$\Rightarrow a r^{4 - 1} = 81$
$\Rightarrow 3 r^{3} = 81$
$\Rightarrow r^{3} = \frac{81}{3}$
$\Rightarrow r^{3} = 27$
Taking cube roots on both sides, we get,
$r = \sqrt[3]{27}$
$\Rightarrow r = \sqrt[3]{3 \times 3 \times 3}$
$\Rightarrow r = 3$
So, $p = a r = 3 r = 3 \times 3 = 9$
$q = a r^{2} = 3 \times {(3)}^{2} = 3 \times 9 = 27$
Hence, the required two numbers between 3 and 81 are $9$ and $27$ .

27. 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 $b$ .

Solution

$∵ \frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the geometric mean between $a$ and $b$ .
So, according to the definition of geometric mean
between $a$ and $b$ , we can write,
$\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}} = \sqrt{a b}$
$\Rightarrow a^{n + 1} + b^{n + 1} = \sqrt{a b} \times (a^{n} + b^{n})$
&nbsp &nbsp squaring on both sides
$\Rightarrow {((a^{n + 1} + b^{n + 1}))}^{2} = {[[\sqrt{a b} \times (a^{n} + b^{n})]]}^{2}$
$\Rightarrow a^{2 (n + 1)} + b^{2 (n + 1)} + 2 \times a^{n + 1} \times b^{n + 1}$ $= a b {((a^{n} + b^{n}))}^{2}$
$\Rightarrow a^{2 n + 2} + b^{2 n + 2} + 2 \times a^{n + 1} \times b^{n + 1} = a b (a^{2 n} + b^{2 n} + 2 a^{n} b^{n})$
$\Rightarrow a^{2 n + 2} + b^{2 n + 2} +$ $2 \times a^{n + 1} \times b^{n + 1}$ $= a^{2 n + 1} b + a b^{2 n + 1} +$ $2 \times a^{n + 1} \times b^{n + 1}$
$\Rightarrow a^{2 n + 2} + b^{2 n + 2} = a^{2 n + 1} b + a b^{2 n + 1}$
$\Rightarrow a^{2 n + 2} - a^{2 n + 1} b = a b^{2 n + 1} - b^{2 n + 2}$
$\Rightarrow a^{2 n + 1} (a - b) = b^{2 n + 1} (a - b)$
Dividing by ( $a - b$ ) on both sides, we get,
$\Rightarrow a^{2 n + 1} = b^{2 n + 1}$
$\Rightarrow \frac{a^{2 n + 1}}{b^{2 n + 1}} = 1$
$\Rightarrow {((\frac{a}{b}))}^{2 n + 1} = {((\frac{a}{b}))}^{0}$ ... $∵ 1 =$ ${((\frac{a}{b}))}^{0}$
$\Rightarrow 2 n + 1 = 0$
$\Rightarrow n = \frac{- 1}{2}$
Hence, the value of n so that $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ will be the
geometric mean between $a$ and $b$ is $\frac{- 1}{2}$ .

28. 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 are $a$ and $b$ .
According to the given question, we have,
$a + b = 6 \sqrt{a b}$ ... (1)
$∵ {((a - b))}^{2} = {((a + b))}^{2} - 4 a b$
$\Rightarrow a - b = \sqrt{{((a + b))}^{2} - 4 a b}$ $= \sqrt{{((6 \sqrt{a b}))}^{2} - 4 a b}$
$= \sqrt{36 a b - 4 a b}$ $= \sqrt{32 a b}$ $= 4 \sqrt{2 a b}$
$∴ a - b$ $= 4 \sqrt{2 a b}$ ... (2)

Adding equations (1) and (2), we get,
$a + b = 6 \sqrt{a b}$
$a - b$ $= 4 \sqrt{2 a b}$

$2 a = 6 \sqrt{a b} + 4 \sqrt{2 a b}$
$\Rightarrow a = 3 \sqrt{a b} + 2 \sqrt{2 a b}$
$\Rightarrow a = (3 + 2 \sqrt{2}) \sqrt{a b}$
Putting the value of $a$ in equation (1), we get,
$a + b = 6 \sqrt{a b}$
$\Rightarrow (3 + 2 \sqrt{2}) \sqrt{a b} + b = 6 \sqrt{a b}$
$\Rightarrow b = 6 \sqrt{a b} - (3 + 2 \sqrt{2}) \sqrt{a b}$
$\Rightarrow b = 6 \sqrt{a b} - 3 \sqrt{a b} - 2 \sqrt{2} \sqrt{a b}$
$\Rightarrow b = 3 \sqrt{a b} - 2 \sqrt{2} \sqrt{a b}$
$\Rightarrow b = (3 - 2 \sqrt{2}) \sqrt{a b}$
Now, $\frac{a}{b} = \frac{(3 + 2 \sqrt{2}) \sqrt{a b}}{(3 - 2 \sqrt{2}) \sqrt{a b}} = \frac{(3 + 2 \sqrt{2})}{(3 - 2 \sqrt{2})}$
Hence, proved!

29. 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 are $a$ and $b$
$∵ A$ is Arithmetic Mean of $a$ and $b$ .
$\Rightarrow \frac{(a + b)}{2} = A$
$\Rightarrow a + b = 2 A$ ... (1)
$∵ G$ is Geometric Mean of $a$ and $b$ .
$\Rightarrow \sqrt{a b} = G$
$\Rightarrow a b = G^{2}$ ... (2)
We know that
${((a - b))}^{2} = {((a + b))}^{2} - 4 a b$
$\Rightarrow a - b = \sqrt{{((a + b))}^{2} - 4 a b}$ $= \sqrt{{((2 A))}^{2} - 4 G^{2}}$
$\Rightarrow a - b = \sqrt{4 A^{2} - 4 G^{2}}$
$\Rightarrow a - b = 2 \sqrt{A^{2} - G^{2}}$ ... (3)
Adding equations (1) and (3), we get,
$a + b = 2 A$
$a - b = 2 \sqrt{A^{2} - G^{2}}$

2 a = 2 A + 2 \sqrt{A^{2} - G^{2}}

\Rightarrow a = A + \sqrt{A^{2} - G^{2}}

...Dividing by 2 on both sides

\Rightarrow a = A + \sqrt{(A - G) (A + G)}

Putting the value of

a

in equation (1), we get,

a + b = 2 A

\Rightarrow A + \sqrt{A^{2} - G^{2}} + b = 2 A

\Rightarrow b = 2 A - (A + \sqrt{A^{2} - G^{2}})

\Rightarrow b = A - \sqrt{A^{2} - G^{2}}

\Rightarrow b = A - \sqrt{(A - G) (A + G)}

Hence, the two positive numbers are

A \pm \sqrt{(A - G) (A + G)}

30. 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

From the given question, we can conclude the following:
Number of bacteria at the start ( $a$ ) = 30
Number of bacteria at the end of 1 hour ( $a r$ ) = $2 \times 30 = 60$
Number of bacteria at the end of 2 hours ( $a r^{2}$ ) = $2 \times 60 = 120$
Number of bacteria at the end of 3 hours ( $a r^{3}$ ) = $2 \times 120 = 240$
..................................................
Number of bacteria at the end of n years ( $a r^{n}$ )

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Exercise 8.2 NCERT Math Class 11 Solutions

1. Find the 20th and nth terms of the G.P. 52,54,58,...

Solution

2. Find the 20th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Solution

3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

How to show: Method

Solution

4. The 4th term of a G.P. is square of its second term, and the first term is -3. Determine its 7th term.

How to Solve

Solution

5. Which term of the following sequences:(a) 2, 22, 4, ... is 128 ? (b) 3, 3, 33, ... is 729?(c) 13, 19, 127, ... is 119683?

How to Solve

Solution

6. For what values of x, the numbers −27,x,−72 are in G.P.?

How to Solve

Solution

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

7. 0.15, 0.015, 0.0015, ... 20 terms.

Solution

8. 7,21,37, ... n terms.

Solution

9. 1, −a,a2,−a3, ...n terms. (if a≠−1)

Solution

10. x3,x5,x7, ... n terms (if x≠±1)

Solution

11. Evaluate ∑k=111(2+3k)

How to solve

Solution

12. The sum of first three terms of a G.P. is 3910 and their product is 1. Find the common ratio and the terms.

Concept

Solution

13. How many terms of G.P. 3, 32, 33, ... are needed to give the sum 120?

Solution

14. 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

15. Given a G.P. with a=729 and 7th term = 64, determine S7

Solution

16. Find a G.P. for which sum of the first two terms is -4 and fifth term is 4 times the third term.

Solution

17. If the 4th,10th and 16th terms of a G.P. are x,y and z respectively. Prove that x, y, and z are in G.P.

Solution

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

Solution

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 12

Solution

20. Show that the products of the corresponding terms of the sequences a,ar,ar2,...,arn-1 and A,AR,AR2,...ARn-1 form a G.P., and find the common ratio.

Solution

21. 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.

Solution

22. If the pth,qth and rth terms of a G.P. are a,b and c, respectively. Prove that aq-r br-p cp-q=1

Solution

23. 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 P2=(ab)n.

Solution

24. 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 1rn.

Solution

25. If a,b,c and d are in G.P.. Show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2

Solution

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

Solution

27. Find the value of n so that an+1+bn+1an+bn may be the geometric mean between a and b.

Solution

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+22)∶(3-22).

Solution

29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A±(A+G)(A-G)

Solution

30. 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?

Solution

See other exercises

All Exercises NCERT Math Class 11 Solutions

1. Find the 20^th and n^th terms of the G.P.
$\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, ...$

2. Find the 20^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

3. The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

4. 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.

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

6. For what values of x, the numbers $- \frac{2}{7}, x, - \frac{7}{2}$ are in G.P.?

8. $\sqrt{7}, \sqrt{21}, 3 \sqrt{7},$ ... n terms.

9. 1, $- a, a^{2}, - a^{3}, ... n$ terms. (if $a \neq - 1$ )

10. $x^{3}, x^{5}, x^{7},$ ... n terms (if $x \neq \pm 1$ )

11. Evaluate $\sum_{k = 1}^{11} (2 + 3^{k})$

12. 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.

13. How many terms of G.P. 3, 3², 3³, ... are needed to give the sum 120?

15. Given a G.P. with $a = 729$ and $7^{th}$ term = 64, determine $S_{7}$

16. Find a G.P. for which sum of the first two terms is $- 4$ and fifth term is 4 times the third term.

17. If the $4^{th}, 10^{th}$ and $16^{th}$ terms of a G.P. are $x, y$ and $z$ respectively. Prove that $x, y,$ and $z are in G.P.$

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, $\frac{1}{2}$

20. Show that the products of the corresponding terms of the sequences $a, a r, a r^{2}, . . ., a r^{n - 1}$ and $A, A R, A R^{2}, . . . A R^{n-1}$ form a G.P., and find the common ratio.

21. 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.

22. If the $p^{th}, q^{th}$ and $r^{th}$ terms of a G.P. are $a, b$ and $c,$ respectively. Prove that
$a^{q - r} b^{r - p} c^{p - q} = 1$

23. 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 $P^{2} = {(a b)}^{n}$ .

24. 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 $\frac{1}{r^{n}}$ .

25. If $a, b, c$ and $d$ are in G.P.. Show that
$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})$ $= {((a b + b c + c d))}^{2}$

27. 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 $b$ .

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

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

30. 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?