Sequences and Series Test - 9

Given that \(S_n = 3n + 2n ^2 \)

\(S_1 = 3(1) + 2(1) ^2 = 5\)

\(S_2\) = 3(2) + 2(4) = 14

\(S_1 = a_1 = 5\)

\(S_2 – S_1 = a_2 = 14 – 5 = 9\)

\(\therefore\) Common difference d = \(a_2 - a_1\) 

= 9 – 5 = 4

Given that \(T_3 = 4\)

⇒ \(ar ^{3 – 1 }= 4 [ \therefore T_n = ar ^{n – 1} ] \)

⇒ \(ar ^2 = 4\)

Product of first 5 terms = a.ar. \(ar^2.ar^3.ar^4\)

\(= a ^5 r ^{10} = (ar^ 2 )^ 5 = (4)^ 5\)

\(T_n \)= a + (n – 1)d

\(\therefore T_9 \) = a + 8d

and \(T_{13}\) = a + 12d

As per the given condition

9[a+8d] = 13[a + 12d]

⇒ 9a + 72d = 13a + 156d ⇒ - 4a = 84d

⇒ a = - 21d ….(i)

Now \(T_{22}\) = a + 21d = - 21d + 21d = 0 [from eq. (i)]

Since x, 2y, 3z are in A.P.

\(\therefore\) 2y – x = 3z – 2y

⇒ 4y = x + 3z …(i)

Now x, y, z are in G.P.

\(\therefore \) Common ratio \(r = \frac{y}{x} = \frac{z}{y} \dots(ii)\)

\(\therefore \) \(y^2 = xz \dots(ii)\)

Putting the value of x from eq. (i), we get

\(y ^2 = (4y – 3z)z \implies y ^2 = 4yz – 3z ^2\)

\(\implies 3z ^2 – 4yz + y ^2 = 0\)

\(\implies 3z^ 2 – 3yz – yz + y ^2 = 0\)

⇒ 3z(z – y) – y(z – y) = 0

⇒ (3z – y) (z – y) = 0

⇒ 3z – y = 0 and z – y = 0

⇒ 3z = y and z ≠ y [\(\because\) z and y are distinct numbers]

⇒ \(\frac{z}{y} = \frac{1}{3} \implies r = \frac{1}{3}\) (from eq. (ii))

\(S_n = \frac{n}{2}[2a + (n - 1)d]\)

\(\therefore S_{2n} = \frac{2n}{2}[2a +(2n - 1)d]\)

\(S_{3n} = \frac{3n}{2}[2a +(3n - 1)d]\)

As per the condition of the question, we have

\(S_{2n} = 3. S_n\)

⇒ \(\frac{2n}{2}[2a +(2n -1)d] \) 

= \(3.\frac{n}{2}[2a +(n-1)d]\)

⇒ 2[2a + (2n – 1)d] = 3[2a + (n – 1)d]

⇒ 4a + (4n – 2)d = 6a + (3n – 3)d

⇒ 6a + (3n – 3)d – 4a – (4n – 2)d = 0

⇒ 2a + (3n – 3 – 4n + 2)d = 0

⇒ 2a + (- n – 1)d = 0

⇒ 2a – (n + 1)d = 0

⇒ 2a = (n + 1)d (i)

Now, 

\(S_{3n} : S_n = \frac{3n}{2}[2a +(3n - 1)d]: \frac{n}{2}[2a + (n - 1)d]\)

\(= \frac{\frac{3n}{2}[2a +(3n - 1)d}{\frac{n}{2}[2a +(n-1)d]}\)

\( \\ =\frac{3[2a + (3n - 1)d]}{2a + (n - 1)d}\)

\(= \frac{3[(n + 1)d + (3n - 1)d]}{(n+1)d + (n - 1)d}\)

\(= \frac{3d[n+1 + 3n - 1]}{d(n+1 + n -1)} = \frac{3[4n]}{2n}\)

= 6

We know that AM ≥ GM

\(\therefore \frac{4^x + 4^{1-x}}{2} \geq \sqrt{4^x .4^{1-x}}\) 

\(\implies 4^x + 4^{1-x} \geq 2 \sqrt{4^{x+1-x}}\)

 \(\implies4^x + 4^{1-x} \geq 2.2\) 

\(\implies 4^x + 4^{1-x} \geq 4\)

Given that \(\displaystyle\sum_{i=1}^{n} \frac{S_r}{s_r} = \frac{S_1}{s_1} + \frac{S_2}{s_2} + \frac {S_3}{s_3} + \dots + \frac{S_n}{s_n}\)

Let \(T_n\) be the nth term of the above series

\(\therefore T_n = \frac{S_n}{s_n} = \frac{[\frac{n(n+1)}{2}]^2}{\frac{n(n+1)}{2}}\\ = \frac{n(n+1)}{2} = \frac{n^2 + n}{2}\)

Now sum of the given series

\(\sum T_n = \frac{1}{2}\sum[n^2 + n] \)

\(= \frac{1}{2}[\sum n^2 + \sum n]\)

\(= \frac{1}{2}[\sum n^2 + \sum n]\)

\(= \frac{1}{2}[\frac{n(n+1)(2n +1)}{6} + \frac{n(n+1)}{2}]\)

\(= \frac{1}{2}. \frac{n(n+1)}{2}[\frac{2n + 1}{3} + 1]\)

\(= \frac{n(n+1)}{4}[\frac{2n + 1 + 3}{3}]\)

\(= \frac{n(n+1)}{4}.\frac{(2n + 4)}{3}\)

\(= \frac{n(n+1)(n+2)}{6}\)

Let \(S_n\) = 2 + 3 + 6 + 11 + 18 + … + \(t_{50}\)

Using method of difference, we get

\(S_n\) = 2 + 3 + 6 + 11 + 18 + … + \(t_{50}\) ….(i)

Subtracting eq. (ii) from eq. (i), we get

0 = 2 + 1 + 3 + 5 + 7 + … - \(t_{50}\) terms

⇒ \(t_{50}\) = 2 + (1 + 3 + 5 + 7 + … upto 49 terms)

⇒ \(t_{50} = 2 + \frac{49}{2}[2 \times 1 + (49 - 1)2] \\ \)

\(= 2 + \frac{49}{2}[2 + 96]\)

\(= 2 + \frac{49}{2} \times 98 = 2 + 49 \times 49 = 49^2 + 2\)

Let the length, breadth and height of a rectangular block be \(\frac{a}{r},\) a and ar.

[Since they are is G.p]

∴ Volume = l × b × h

\(216 = \frac{a}{r} \times a \times ar\)

⇒ \(a^3\) = 216 ⇒ a = 6

Now total surface area = 2[lb + bh + lh]

\(252 = 2[\frac{a}{r}.a + a. ar + \frac{a}{r}.ar]\)

\(= 2[\frac{a^2}{r} + a^2r + a^2]\)

\(\implies 252 = 2a^2[\frac{1}{r} + r + 1] \)

\(= 2 \times (6)^2 [\frac{1 + r^2 + r}{r}]\)

\(\implies 252 = 72[\frac{1 + r^2 + r}{r}]\)

\(\implies \frac{252}{72} = \frac{1 + r + r^2}{r}\)

\(\implies \frac{7}{2} = \frac{1 + r + r^2}{r}\)

⇒ 2 + 2r + \(2r ^2\) = 7r

⇒ \(2r ^2\) – 5r + 2 = 0 ⇒ \(2r ^2\) – 4r – r + 2 = 0

⇒ 2r(r – 2) – 1(r – 2) = 0

⇒ (r – 2) (2r – 1) = 0

⇒ r – 2 = 0 and 2r – 1 = 0

\(\therefore r = 2, \frac{1}{2}\)

Therefore, the three edge are:

If r = 2 then edges are 3, 6, 12

If \(r = \frac{1}{2}\) then edges are 12, 6, 3

So, the length of the longest edge = 12

Let \(t_n\) is nth term of the sequence so \(t_n = s_n – s_{n – 1}\)

= \(2n^2\) + 3n – \(2(n – 1)^2\) – 3(n – 1) 

= 4n + 1 

so \(t_{50}\) = 201.

a = 4, d = 8 

so 108 = 4 + (n – 1)8 

⇒ n = 14

Using the property 

\(a_1 a_{11} = a_2 a_{10} = a_3 a_9\) = .............. = \(a_6^2 = 25  \)

Hence product of terms = \(5^{11}\)

Let the three numbers be a/r,  a, ar

So, \(a[\frac{1}{r} + 1 + r] = 19 \dots(i)\)

and \(a^3\) = 216 

⇒ a = 6  so from (i) 

\(6r^2\) – 13r + 6 = 0.

=> r = 3/2, 2/3 

Hence the three numbers are 4, 6, 9.

\(S_n = \sum T_n = \sum (2n + 1)\)

= 2Σn + Σ1

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

\(= n^2 + 2n\)

or n(n + 2).