If a1- a2- a3- -an are in A-P then- 1-a1an-1-a2an-1-1-a3an-2- - 1-ana1-2-a1-an 1-a1- 1-a2-1-an - If this expression is correct write 1 otherwise 0-

1/a_1+1/a_n= a_n+a_1/a_1.a_n 1/a_2+1/a_n 1= a_n 1+a_2/a_2.a_n 1= a_n d+a_1+d/a_2.a_n 1= a_n+a_1/a_2.a_n 1 1/a_n+1/a_1= a_1+a_n/a_n.a_1= a_1+a_n/a ...

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Byju's Answer

Standard VI

Mathematics

Comparison of Fractions

If a1, a2, ...

Question

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.

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Solution

$\frac{1}{a_{1}} + \frac{1}{a_{n}} = \frac{a_{n} + a_{1}}{a_{1} . a_{n}}$
$\frac{1}{a_{2}} + \frac{1}{a_{n - 1}} = \frac{a_{n - 1} + a_{2}}{a_{2} . a_{n - 1}} = \frac{a_{n} - d + a_{1} + d}{a_{2} . a_{n - 1}} = \frac{a_{n} + a_{1}}{a_{2} . a_{n - 1}}$
$\frac{1}{a_{n}} + \frac{1}{a_{1}} = \frac{a_{1} + a_{n}}{a_{n} . a_{1}} = \frac{a_{1} + a_{n}}{a_{n} . a_{1}}$
Adding $2 (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . . \frac{1}{a_{n}})$
$= (a_{n} + a_{1}) [\frac{1}{a_{1} . a_{n}} + \frac{1}{a_{2} . a_{n - 1}} + . . . . + \frac{1}{a_{n} . a_{1}}]$
or $\frac{2}{a_{n} + a_{1}} (\frac{1}{a_{n}} + \frac{1}{a_{2}} + . . . . . . \frac{1}{a_{n}})$
$= [\frac{1}{a_{1} . a_{n}} + \frac{1}{a_{2} . a_{n - 1}} + . . . . . + \frac{1}{a_{n} . a_{1}}]$

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Similar questions

Q. If

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

are in A.P., then show that

\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}}]

Q. Show that

\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}})

Q. 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

Q. Let

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

be an AP, then.

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

$a_{1}, a_{2}, \dots \dots . a_{n}$ are A.P. Where $a_{1}$ > 0 for all i, then $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}$

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If a1,a2,a3,......an are in A.P then, 1a1an+1a2an−1+1a3an−2+...+1ana1=2a1+an(1a1+1a2+...+1an).. If this expression is correct write 1 otherwise 0.

1a1+1an=an+a1a1.an1a2+1an−1=an−1+a2a2.an−1=an−d+a1+da2.an−1=an+a1a2.an−11an+1a1=a1+anan.a1=a1+anan.a1Adding 2(1a1+1a2+....1an)=(an+a1)[1a1.an+1a2.an−1+....+1an.a1]or 2an+a1(1an+1a2+......1an)=[1a1.an+1a2.an−1+.....+1an.a1]