Assertion - a1- a2- a3- - an are in APSTATEMENT-1- 1-a1an - 1-a2an-1 - 1-a3an-2- -1-ana1 - 2-a1 - an 1-a1 - 1-a2 - 1-a3 - - - 1-an Reason- STATEMENT-2 - - a1 - an - ar - an - r for 1- r- n

The correct option is C Statement 1 is True, Statement 2 is False 1a_1+1a_n= a_n+a_1a_1.a_n 1a_2+1a_n 1=a_n 1+a_na_na_n 1=a_1+d+a_n da_2a_n 1=a_n+a_1a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VI

Mathematics

Idea of a Set

Assertion : ...

Question

Assertion : $a_{1}, a_{2}, a_{3}, . . . . ., a_{n}$ are in AP
STATEMENT-1:- $\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . . . + \frac{1}{a_{n} a_{1}} = \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n}})$ Reason: STATEMENT-2 : - $a_{1} + a_{n} = a_{r} + a_{n - r}$ for $1 \leq r \leq n$

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement-1 is True, Statement-2 is False

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Statement-1 is False, Statement-2 is TrueBoth Assertion and Reason are incorrect

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C Statement-1 is True, Statement-2 is False
$\frac{1}{a_{1}} + \frac{1}{a_{n}} = \frac{a_{n} + a_{1}}{a_{1} . a_{n}}$
$\frac{1}{a_{2}} + \frac{1}{a_{n - 1}} = \frac{a_{n - 1} + a_{n}}{a_{n} a_{n - 1}} = \frac{a_{1} + d + a_{n} - d}{a_{2} a_{n - 1}} = \frac{a_{n} + a_{1}}{a_{2} a_{n - 1}}$

$\frac{1}{a_{n}} + \frac{1}{a_{1}} = \frac{a_{1} + a_{n}}{a_{n} . a_{1}} = \frac{a_{1} + a_{n}}{a_{n} . a_{1}}$
Adding the above equations we have :
$2 (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . . \frac{1}{a_{n}}) = a_{1} + a_{n} (\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + . . . . .)$
So A is true.
In general, R is false as say for $r = 2$ ,
$a_{1} + a_{n} \neq a_{2} + a_{n - 2}$
Thus $a_{1} + a_{n} \neq a_{r} + a_{n - r}$

Suggest Corrections

Similar questions

Q. Statement

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

, where

a > 1

.
Statement

- 2 :

When

a > 1, {log}_{a} x

is an increasing function.

Q. Statement-1:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

Statement-2:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

Q. Statement-1:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

Statement-2:

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

Q. If

a_{1}, a_{2}, a_{3}, . . . . . . a_{n}

are in A.P then,

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

. If this expression is correct write 1 otherwise 0.

Q. For every natural number

n \geq 2,

Statement 1:

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

Statement 2:

\sqrt{n (n + 1)} < n + 1

Join BYJU'S Learning Program

Assertion :a1,a2,a3,.....,an are in APSTATEMENT-1:- 1a1an+1a2an−1+1a3an−2+.....+1ana1=2a1+an(1a1+1a2+1a3+...+1an) Reason: STATEMENT-2 : - a1+an=ar+an−r for 1≤r≤n