Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}

$${a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}$$ $$ \Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1$$ $$ \Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2$$ For $$\matrix{ {n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr {n = 2} & {{a_4}{a_3} - {a_3}{a_2} = 2} \cr {n = 3} & {{a_5}{a_4} - {a_4}{a_3} = 2} \cr {} & {\matrix{ . \cr . \cr . \cr . \cr } } \cr {n = n} & {{{{a_{n + 2}}{a_{n + 1}} - {a_n}{a_{n + 1}} = 2} \over {{a_{n + 2}}{a_{n + 1}} = 2n + {a_1}{a_2}}}} \cr } $$ N

Sequences and Series — Page 6 | JEE Mathematics

Q51

Consider the sequence

a_{1}, a_{2}, a_{3}, \ldots

such that

a_{1}=1, a_{2}=2

and

a_{n+2}=\dfrac{2}{a_{n+1}}+a_{n}

for

\mathrm{n}=1,2,3, \ldots .

If

\left(\dfrac{\mathrm{a}_{1}+\dfrac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\dfrac{\mathrm{a}_{2}+\dfrac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\dfrac{\mathrm{a}_{3}+\dfrac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\dfrac{\mathrm{a}_{30}+\dfrac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)

, then

\alpha

is equal to :

A

-

30

B

-

31

C

-

60

D

-

61

Correct Answer

Option C

Solution

{a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}

\Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1

\Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2

For

\begin{array}{ll}{n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \\ {n = 2} & {{a_4}{a_3} - {a_3}{a_2} = 2} \\ {n = 3} & {{a_5}{a_4} - {a_4}{a_3} = 2} \\ {} & \begin{array}{ll}. \\ . \\ . \\ . \end{array} \\ {n = n} & {{{{a_{n + 2}}{a_{n + 1}} - {a_n}{a_{n + 1}} = 2} \over {{a_{n + 2}}{a_{n + 1}} = 2n + {a_1}{a_2}}}} \end{array}

Now,

{{({a_1}{a_2} + 1)} \over {{a_2}{a_3}}}\,.\,{{({a_2}{a_3} + 1)} \over {{a_3}{a_4}}}\,.\,{{({a_3}{a_4} + 1)} \over {{a_4}{a_5}}}\,.\,.....\,.\,{{({a_{30}}{a_{31}} + 1)} \over {{a_{31}}{a_{32}}}}

= {3 \over 4}\,.\,{5 \over 6}\,.\,{7 \over 8}\,.\,....\,.\,{{61} \over {62}}

= {2^{ - 60}}\left( {{}^{61}{C_{31}}} \right)

Q52

Let

a_1, a_2, a_3, \ldots

be an A.P. If

a_7=3

, the product

a_1 a_4

is minimum and the sum of its first

n

terms is zero, then

n !-4 a_{n(n+2)}

is equal to :

A 24

B

\dfrac{381}{4}

C 9

D

\dfrac{33}{4}

Correct Answer

Option A

Solution

$a_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d$

\begin{aligned} & a_{1} \cdot a_{4}=a(a+3 d) \\\\ & \Rightarrow(3-6 d)(3-6 d+3 d) \\\\ & \Rightarrow 3(1-2 d) 3(1-d) \\\\ & \Rightarrow 9\left(2 d^{2}-3 d+1\right) \end{aligned}

Let $f(d)=2 d^{2}-3 d+1$ $f^{\prime}(d)=4 d-3 \Rightarrow d=\dfrac{3}{4}$ $\therefore a=3-6 \cdot \dfrac{3}{4}=3-\dfrac{9}{2}=-\dfrac{3}{2}$ $S_{n}=0$ $\dfrac{n}{2}(29+(n-1) d)=0$ $\Rightarrow 2 \cdot\left(-\dfrac{3}{2}\right)+(n-1)\left(\dfrac{3}{4}\right)=0$ $\Rightarrow \quad 3=\dfrac{3}{4}(n-1)$ $\Rightarrow n=5$ Now, $n !-4 \cdot a_{n(n+2)}$

\begin{aligned} & =5 !-4 \cdot a_{35} \\\\ & =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\\\ & =120-(-6+102) \\\\ & =120-(96) \\\\ & =24 \end{aligned}

Q53

The sum of 10 terms of the series

{1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....

is

A

{{58} \over {111}}

B

{{56} \over {111}}

C

{{55} \over {111}}

D

{{59} \over {111}}

Correct Answer

Option C

Solution

\begin{aligned} & T_n=\frac{n}{1+n^2+n^4} \\\\ & =\frac{n}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)} \\\\ & =\frac{1}{2}\left[\frac{\left(\mathrm{n}^2+\mathrm{n}+1\right)-\left(\mathrm{n}^2-\mathrm{n}+1\right)}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\\\ & \Rightarrow \mathrm{T}_{\mathrm{n}}=\frac{1}{2}\left[\frac{1}{\left(\mathrm{n}^2-\mathrm{n}+1\right)}-\frac{1}{\left(\mathrm{n}^2+\mathrm{n}+1\right)}\right] \\\\ & \mathrm{S}_{\mathrm{n}}=\sum_{n=1}^{10} T_n \\\\ & =\frac{1}{2} \sum\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right) \\\\ & =\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)\right. \\\\ & \left.\dots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\left(\frac{1}{91}-\frac{1}{111}\right)\right] \end{aligned}

$\therefore$ $S=\dfrac{1}{2}\left(1-\dfrac{1}{111}\right)=\dfrac{55}{111}$

Q54

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is

A 7

B 14

C 3

D

\dfrac{9}{2}

Correct Answer

Option A

Solution

$\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \mathrm{ar}^{3}(\mathrm{a}, \mathrm{r}>0)$ $a^{4} r^{6}=1296$ $a^{2} r^{3}=36$ $a=\dfrac{6}{r^{3 / 2}}$ $a+a r+a r^{2}+a r^{3}=126$ $\dfrac{1}{\mathrm{r}^{3 / 2}}+\dfrac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\dfrac{\mathrm{r}^{2}}{\mathrm{r}^{3 / 2}}+\dfrac{\mathrm{r}^{3}}{\mathrm{r}^{3 / 2}}=\dfrac{126}{6}=21$ $\left(r^{-3 / 2}+r^{3 / 2}\right)+\left(r^{1 / 2}+r^{-1 / 2}\right)=21$ $\mathrm{r}^{1 / 2}+\mathrm{r}^{-1 / 2}=\mathrm{A}$ $\mathrm{r}^{-3 / 2}+\mathrm{r}^{3 / 2}+3 \mathrm{~A}=\mathrm{A}^{3}$ $\mathrm{A}^{3}-3 \mathrm{~A}+\mathrm{A}=21$ $\mathrm{A}^{3}-2 \mathrm{~A}=21$ $A=3$ $\sqrt{\mathrm{r}}+\dfrac{1}{\sqrt{\mathrm{r}}}=3$ $\mathrm{r}+1=3 \sqrt{\mathrm{r}}$ $r^{2}+2 r+1=9 r$ $r^{2}-7 r+1=0$ $\Rightarrow r_{1}+r_{2}=7$

Q55

Let

a, b, c>1, a^3, b^3

and

c^3

be in A.P., and

\log _a b, \log _c a

and

\log _b c

be in G.P. If the sum of first 20 terms of an A.P., whose first term is

\dfrac{a+4 b+c}{3}

and the common difference is

\dfrac{a-8 b+c}{10}

is

-444

, then

a b c

is equal to :

A 343

B 216

C

\dfrac{343}{8}

D

\dfrac{125}{8}

Correct Answer

Option B

Solution

2{b^3} = {a^3} + {c^3}

{\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)

\Rightarrow {(\log a)^3} = {(\log c)^3}

\Rightarrow \log a = \log c

\Rightarrow a = c

\Rightarrow a = b = c

{T_1} = 2a,d = - {{3a} \over 5}

{S_{20}} = - 444

\Rightarrow {{20} \over 2}\left( {2(2a) + (19)\left( { - {{3a} \over 5}} \right)} \right) = - 444

\Rightarrow 10{{(20a - 57a)} \over 5} = - 444

\Rightarrow 37a = 222

\Rightarrow a = 6

\Rightarrow abc = {(6)^3} = 216

Q56

Let

A_{1}

and

A_{2}

be two arithmetic means and

G_{1}, G_{2}, G_{3}

be three geometric means of two distinct positive numbers. Then

G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}

is equal to :

A

\left(A_{1}+A_{2}\right)^{2} G_{1} G_{3}

B

\left(A_{1}+A_{2}\right) G_{1}^{2} G_{3}^{2}

C

2\left(A_{1}+A_{2}\right) G_{1}^{2} G_{3}^{2}

D

2\left(A_{1}+A_{2}\right) G_{1} G_{3}

Correct Answer

Option A

Solution

Now, we have the following relations : Arithmetic progression : Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$ , we can say that $a$ , $A_1$ , $A_2$ , and $b$ are in an arithmetic progression.

This means there are three equal intervals between $a$ and $b$ , which are represented by the common difference $d$ .

To find the value of $d$ , we can use the following equation :

b - a = 3d

From this equation, we can find the value of $d$ :

d = \frac{b - a}{3}

A_1 = a + \frac{b - a}{3} = \frac{2a + b}{3}

A_2 = \frac{a + 2b}{3}

A_1 + A_2 = a + b

Geometric progression :

a, G_1, G_2, G_3, b \text{ are in G.P. }

r = \left(\frac{b}{a}\right)^{\frac{1}{4}}

G_1 = \left(a^3b\right)^{\frac{1}{4}}

G_2 = \left(a^2b^2\right)^{\frac{1}{4}}

G_3 = \left(ab^3\right)^{\frac{1}{4}}

We have the expression :

G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + \left(a^3b\right)^{\frac{1}{2}}\cdot\left(ab^3\right)^{\frac{1}{2}}

Simplify the expression :

a^3b + a^2b^2 + ab^3 + ab(a^2b^2)

Factor out $ab$ :

ab(a^2 + ab + b^2 + a^2b^2)

Combine the terms :

ab(a^2 + 2ab + b^2)

Rewrite the expression using the sum of squares :

ab(a + b)^2

Now, recall that $A_1 + A_2 = a + b$ . Substitute this into the expression :

G_1 \cdot G_3 \cdot (A_1 + A_2)^2

Q57

Let a

_1

, a

_2

, a

_3

, .... be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be 2 and the product of its 3rd and 5th terms be

\dfrac{1}{9}

. Then

6(a_2+a_4)(a_4+a_6)

is equal to

A 2

\sqrt2

B 2

C 3

\sqrt3

D 3

Correct Answer

Option D

Solution

Given the conditions : $a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$ $a_3 \cdot a_5 = \dfrac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \dfrac{1}{9} \Rightarrow a r^3 = \dfrac{1}{3}$ From this, we can form the equation $\dfrac{r^2}{3} + \dfrac{r^4}{3} = 2$ , which simplifies to $r^4 + r^2 = 6$ .

This can be factored to give $\left(r^2 - 2\right)\left(r^2 + 3\right) = 0$ , yielding $r^2 = 2$ (since $r^2$ cannot be $-3$ for real $r$ ).

So, we have $r = \sqrt{2}$ .

Substituting $r = \sqrt{2}$ into the equation $a r = \dfrac{1}{6}$ , we get $a = \dfrac{1}{6\sqrt{2}}$ .

Now, we find the value of $6(a_2+a_4)(a_4+a_6)$ : $6(a_2+a_4)(a_4+a_6) = 6\left(a r + a r^3\right)\left(a r^3 + a r^5\right)$ $= 6\left(\dfrac{1}{6\sqrt{2}} + \dfrac{1}{3\sqrt{2}}\right)\left(\dfrac{1}{3\sqrt{2}} + \dfrac{2}{3\sqrt{2}}\right)$ $= 6 \cdot \dfrac{1}{2} \cdot 1 = 3$ .

Q58

Let

s_{1}, s_{2}, s_{3}, \ldots, s_{10}

respectively be the sum to 12 terms of 10 A.P. s whose first terms are

1,2,3, \ldots .10

and the common differences are

1,3,5, \ldots \ldots, 19

respectively. Then

\sum\limits_{i=1}^{10} s_{i}

is equal to :

A 7360

B 7220

C 7260

D 7380

Correct Answer

Option C

Solution

We have 10 arithmetic progressions (A.P.s) with the first terms

a_i

and the common differences

d_i

, where

i = 1, 2, \ldots, 10

. The first terms are

a_i = i

and the common differences are

d_i = 2i - 1

.

Now, we need to find the sum of the first 12 terms for each A.P.

The formula for the sum of the first n terms of an A.P. is:

S_n = n\left(\frac{2a + (n - 1)d}{2}\right)

In this case, we need to find the sum of the first 12 terms for each A.P., so we have:

S_{12} = 12\left(\frac{2a + 11d}{2}\right)

Now, we can compute the sum

s_i

for each A.P.:

s_i = 12\left(\frac{2i + 11(2i - 1)}{2}\right) = 6(2i + 22i - 11) = 6(24i - 11)

Finally, we need to find the sum of all

s_i

for

i = 1, 2, \ldots, 10

:

\sum\limits_{i=1}^{10} s_i = 6\sum\limits_{i=1}^{10} (24i - 11) = 6\left(24\sum\limits_{i=1}^{10} i - 11\sum\limits_{i=1}^{10} 1\right)

The sum of the first 10 integers is

\sum\limits_{i=1}^{10} i = \frac{10(10 + 1)}{2} = 55

, so we have:

\sum\limits_{i=1}^{10} s_i = 6\left(24\cdot55 - 11\cdot10\right) = 6(1320 - 110) = 6\cdot1210 = 7260

Thus, the sum

\sum\limits_{i=1}^{10} s_i

is equal to 7260.

Q59

Let

be a sequence such that

a_{1}+a_{2}+\ldots+a_{n}=\dfrac{n^{2}+3 n}{(n+1)(n+2)}

. If

28 \sum\limits_{k=1}^{10} \dfrac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}

, where

\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}

are the first

\mathrm{m}

prime numbers, then

\mathrm{m}

is equal to

A 5

B 7

C 6

D 8

Correct Answer

Option C

Solution

Given the sum of the first n terms, $S_n = \dfrac{n^2+3n}{(n+1)(n+2)}$ , we can find the nth term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms : So,

a_n = S_n - S_{n-1}

Solving, we get :

a_n = \frac{n^2+3n}{(n+1)(n+2)} - \frac{(n-1)^2+3(n-1)}{n(n+1)}

Simplifying further, we find :

a_n = \frac{4}{n(n+1)(n+2)}

Then, we find the reciprocal of $a_n$ :

\frac{1}{a_n} = \frac{n(n+1)(n+2)}{4}

Now, we sum this over the first 10 terms :

\sum\limits_{k=1}^{10} \frac{1}{a_k} = \sum\limits_{k=1}^{10} \frac{k(k+1)(k+2)}{4}

Evaluating the sum :

\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} \left[\sum\limits_{k=1}^{10} k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)\right]

This can be rewritten as the sum of differences :

\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} [(1 \cdot 2 \cdot 3 \cdot 4 - 0) + (2 \cdot 3 \cdot 4 \cdot 5 - 1 \cdot 2 \cdot 3 \cdot 4) + \cdots + (10 \cdot 11 \cdot 12 \cdot 13 - 9 \cdot 10 \cdot 11 \cdot 12)]

\sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} (10 \cdot 11 \cdot 12 \cdot 13 - 0) = \frac{1}{16} \cdot 17160

Now, given the condition that :

28 \sum\limits_{k=1}^{10} \frac{1}{a_k} = p_1p_2p_3...p_m

Substituting the sum we've calculated:

28 \cdot \frac{1}{16} \cdot 17160 = p_1p_2p_3...p_m

This simplifies to :

30030 = p_1p_2p_3...p_m

The prime factorization of 30030 is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ , which consists of 6 primes.

Therefore, m is equal to 6.

Q60

Let

a, b, c

and

d

be positive real numbers such that

a+b+c+d=11

. If the maximum value of

a^{5} b^{3} c^{2} d

is

3750 \beta

, then the value of

\beta

is

A 110

B 108

C 90

D 55

Correct Answer

Option C

Solution

Given that

a+b+c+d=11

and the maximum value of

a^5 b^3 c^2 d

is

3750\beta

, you assumed the numbers to be

\frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d

. Applying the AM-GM inequality:

\frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d}{11} \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}

Since

a+b+c+d=11

, we have:

1 \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}

Now, raising both sides to the power of 11:

1^{11} \geq \frac{a^5 b^3 c^2 d}{5^5 3^3 2^2 1}

From the given information, we know that

a^5 b^3 c^2 d \geq 3750\beta

:

5^5 3^3 2^2 \geq 3750\beta

Now, we can solve for $\beta$ :

\beta \leq \frac{1}{3750} \cdot 5^5 3^3 2^2

Since we are looking for the maximum value of $\beta$ , we take the equality case:

\beta = \frac{1}{3750} \cdot 5^5 3^3 2^2

Calculating the value, we find that:

\beta = 90

So, the value of $\beta$ is 90.