Prove that 2 , 3 , 5 cannot be terms of a single A.P.

Let us assume, T a = 2 , T b = 3 & T c = 5 are in AP. Since, 3 - 2 = T b - T a = b - a d . . . 1 , where d is a common difference. ⇒ 5 - 3 = T c - T b = c - b d . . . 2 By using equation 1 and equation 2 , we get 3 - 2 5 - 3 = b - a c - b = P . As a , b , c are positive integers, so P is a rational number. Squaring, 3 - 2 2 5 - 3 2 = P 2 . ⇒ 5 - 2 6 = P 2 8 - 2 15 ⇒ P 2 15 - 6 = 8 P 2 - 5 2 = Q Here, again Q is a rational number. Squaring again, ⇒ 15 P 4 + 6 - 6 P 2 10 = Q . ⇒ 15 P 4 + 6 - Q 6 P 2 = 10 Here, L H S is a rational number whereas R H S is a irrational number, which is not possible. Therefore, 2 , 3 , 5 cannot be terms of a A.P.

The value of x + y + z is 15 , if a , x , y , z , b are in A.P. while the value of 1 x + 1 y + 1 z is 5 3 , if a , x , y , z , b are in H.P. Find a and b .

Given x + y + z = 15 , if the numbers a , x , y , z , b are in A.P. Now, using the formula for the sum of n terms of an A.P., we have S n = n 2 a + l , where a is the first term and l is the last term. Hence, we have a + x + y + z + b = 5 2 a + b ⇒ a + 15 + b = 5 2 a + b ⇒ 15 = 3 a + b 2 ⇒ a + b = 10 ⇒ b = 10 - a . . . i Now, also given that the value of 1 x + 1 y + 1 z is 5 3 , if a , x , y , z , b are in H.P. We know that the reciprocals of the terms of a H.P. are in A.P. Thus, 1 a , 1 x , 1 y , 1 z , 1 b are in A.P., again using the sum of n terms of an A.P., we get 1 a + 1 x + 1 y + 1 z + 1 b = 5 2 1 a + 1 b ⇒ 1 a + 5 3 + 1 b = 5 2 1 a + 1 b ⇒ 5 3 = 3 2 1 a + 1 b ⇒ 5 3 = 3 2 a + b a b ⇒ 10 a b = 9 a + b Now, put the value of b from equation i , to get 10 a 10 - a = 9 a + 10 - a ⇒ 10 a 10 - a = 90 ⇒ 10 a - a 2 = 9 ⇒ a 2 - 10 a + 9 = 0 ⇒ a - 1 a - 9 = 0 ⇒ a = 1 or a = 9 . And, hence the respective values of b are b = 10 - 1 = 9 or b = 10 - 9 = 1 . Thus, the values are a = 1 , b = 9 or a = 9 , b = 1 .

Find the value of S n = ∑ n = 1 n 3 n · 5 n 5 n - 3 n 5 n + 1 - 3 n + 1 , and hence S ∞ .

Given, S n = ∑ n = 1 n 3 n · 5 n 5 n - 3 n 5 n + 1 - 3 n + 1 = 1 2 ∑ n = 1 n 2 · 3 n · 5 n 5 n - 3 n 5 n + 1 - 3 n + 1 = 1 2 ∑ n = 1 n 5 n + 1 - 3 n + 1 - 5 n + 1 + 5 · 3 n 5 n 5 n - 3 n 5 n + 1 - 3 n + 1 = 1 2 ∑ n = 1 n 5 n + 1 - 3 n + 1 5 n 5 n - 3 n 5 n + 1 - 3 n + 1 - 1 2 ∑ n = 1 n 5 n - 3 n 5 n + 1 5 n - 3 n 5 n + 1 - 3 n + 1 = 1 2 ∑ n = 1 n 5 n 5 n - 3 n - 1 2 ∑ n = 1 n 5 n + 1 5 n + 1 - 3 n + 1 = 1 2 5 5 - 3 + 5 2 5 2 - 3 2 + . . . + 5 n 5 n - 3 n - 5 2 5 2 - 3 2 + 5 3 5 3 - 3 3 + . . . + 5 n + 1 5 n + 1 - 3 n + 1 = 1 2 5 5 - 3 - 5 n + 1 5 n + 1 - 3 n + 1 ⇒ S n = 5 4 - 5 n + 1 2 5 n + 1 - 3 n + 1 Dividing numerator and denominator by 5 n + 1 , we get S n = 5 4 - 1 2 1 - 3 5 n + 1 ⇒ S ∞ = 5 4 - 1 2 1 - 3 5 ∞ S ∞ = 5 4 - 1 2 = 3 4

If S n = ∑ r = 1 n t r = 1 6 n 2 n 2 + 9 n + 13 , then ∑ r = 1 ∞ 1 r · t r equals

Given ∑ r = 1 n t r = 1 6 n 2 n 2 + 9 n + 13 . Hence, on putting n = 1 , we get ∑ r = 1 1 t r = t 1 = 1 6 2 × 1 2 + 9 × 1 + 13 = 4 . On putting, n = 2 , we get ∑ r = 1 2 t r = t 1 + t 2 = 1 6 × 2 × 2 × 2 2 + 9 × 2 + 13 = 13 ⇒ t 2 = 9 . Similarly, on putting n = 3 , we get ∑ r = 1 3 t r = t 1 + t 2 + t 3 = 1 6 × 3 × 2 × 3 2 + 9 × 3 + 13 = 29 ⇒ t 3 = 16 . Hence, in general, we can say that t n = n + 1 2 . Now, ∑ r = 1 ∞ 1 r · t r = ∑ r = 1 ∞ 1 r · r + 1 2 ⇒ ∑ r = 1 ∞ 1 r · t r = ∑ r = 1 ∞ 1 r · r + 1 ⇒ ∑ r = 1 ∞ 1 r · t r = ∑ r = 1 ∞ r + 1 - r r · r + 1 ⇒ ∑ r = 1 ∞ 1 r · t r = ∑ r = 1 ∞ r + 1 r · r + 1 - r r · r + 1 ⇒ ∑ r = 1 ∞ 1 r · t r = ∑ r = 1 ∞ 1 r - 1 r + 1 ⇒ ∑ r = 1 ∞ 1 r · t r = 1 - 1 2 + 1 2 - 1 3 + 1 3 - 1 4 + 1 4 - 1 5 + . . . ⇒ ∑ r = 1 ∞ 1 r · t r = 1 .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4

Embibe Experts Solutions for Chapter: Sequences and Series, Exercise 4: Exercise-4

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Sequences and Series, Exercise 4: Exercise-4

Attempt the practice questions on Chapter 5: Sequences and Series, Exercise 4: Exercise-4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Sequences and Series, Exercise 4: Exercise-4 with Hints & Solutions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 19

Prove that $\sqrt{2}, \sqrt{3}, \sqrt{5}$ cannot be terms of a single A.P.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 20

The value of $x + y + z$ is $15,$ if $a, x, y, z, b$ are in A.P. while the value of $(\frac{1}{x}) + (\frac{1}{y}) + (\frac{1}{z})$ is $\frac{5}{3},$ if $Q$ are in H.P. Find $a$ and $b .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 21

Find the value of $S_{n} = \sum_{n = 1}^{n} \frac{3^{n} \cdot 5^{n}}{(5^{n} - 3^{n}) (5^{n + 1} - 3^{n + 1})}$ , and hence $S_{\infty}$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 22

If $n$ is a root of the equation $x^{2} (1 - a c) - x (a^{2} + c^{2}) - (1 + a c) = 0$ and if $n$ H.M.s are inserted between $a$ and $c,$ show that the difference between the first and the last mean is equal to $a c (a - c) .$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 23

Solve the equation ${(2 + x_{1} + x_{2} + x_{3} + x_{4})}^{5} = 6250 x_{1} x_{2} x_{3} x_{4}$ , where $x_{1}, x_{2}, x_{3}, x_{4} > 0$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 24

Let $A, G, H$ be $A . M ., G . M .$ and $H . M .$ of three positive real numbers $a, b, c$ , respectively such that $G^{2} = A H,$ then prove that $a, b, c$ are terms of a $G . P .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 25

If $S_{n} = \sum_{r = 1}^{n} t_{r} = \frac{1}{6} n (2 n^{2} + 9 n + 13),$ then $\sum_{r = 1}^{\infty} \frac{1}{r \cdot \sqrt{t_{r}}}$ equals

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 5 - Sequences and Series>Exercise-4>Q 26

If sum of first $n$ terms of an A.P. (having positive terms) is given by $s_{n} = (1 + 2 t_{n}) (1 - t_{n})$ where $t_{n}$ is the $n^{th}$ term of series, then $t_{2}^{2} = \frac{\sqrt{a} - \sqrt{b}}{4}, (a \in N, b \in N),$ then find the value of $(a + b)$ .

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