If a 1 , a 2 , a 3 , … , a n are in A.P, where a i > 0 for all i , show that 1 a 1 + a 2 + 1 a 2 + a 3 + … + 1 a n - 1 + a n = n - 1 a 1 + a n .

Given terms are a 1 , a 2 , a 3 , … , a n in A.P. So let a 2 - a 1 = d then a 2 2 - a 1 2 = d ⇒ ( a 2 - a 1 ) ( a 2 + a 1 ) = d ∵ a 2 - b 2 = ( a + b ) ( a - b ) ⇒ 1 a 1 + a 2 = a 2 - a 1 d Similarly, 1 a 2 + a 3 = a 3 - a 2 d And 1 a 3 + a 4 = a 4 - a 3 d And 1 a n - 1 + a n = a n - a n - 1 d Adding the above terms we obtain, 1 a 1 + a 2 + 1 a 2 + a 3 + 1 a 3 a 4 + … + 1 a n - 1 + a n = 1 d [ a 2 - a 1 + a 3 - a 2 + a 3 + … a n - a n - 1 ] = 1 d [ a n - a 1 ] → 1 We know that, the sum of n terms of an A.P is given by a n = a 1 + ( n - 1 ) d ⇒ a n - a 1 = n - 1 d ⇒ a n 2 - a 1 2 = ( n - 1 ) d ⇒ ( a n + a 1 ) ( a n - a 1 ) = ( n - 1 ) d ⇒ a n - a 1 = ( n - 1 ) d a n + a 1 ⇒ a n - a 1 d = n - 1 a n + a 1 → 2 So from 1 and 2 , we have 1 a 1 + a 2 + 1 a 2 + a 3 + 1 a 3 + a 4 + … + 1 a n - 1 + a n = n - 1 a n + a 1 Hence, proved.

E M B I B E

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 9

MEDIUM

11th CBSE

IMPORTANT

Earn 100

In a cricket tournament $16$ school teams participated. A sum of Rs $8000$ is to be awarded among themselves as prize money. If the last placed team is awarded Rs $275$ in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?

Important Questions on Sequences and Series

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 10

If

a_{1}, a_{2}, a_{3}, \dots, a_{n}

are in A.P, where

a_{i} > 0

for all

i

, show that

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 11

If the sum of the series

(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + \dots

to

n

terms is

4 n^{3} + 9 n^{2} + k n

, then find

k

.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 11

Find the sum of the series

(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + \dots

to

10

terms.

EASY

11th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 12

If the

r^{th}

term of an A.P whose sum of first

n

terms:

2 n + 3 n^{2}

is

k r - 1

, then find the value of

k

.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 13

If

A

is the arithmetic mean and

G_{1}, G_{2}

be two geometric means between any two numbers, then prove that

2 A = \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}

.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 11>Chapter 9 - Sequences and Series>EXERCISE>Q 14

If $θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n}$ are in A.P, whose common difference is $d$ , then show that $\sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + \dots + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d}$ .