If a 1- a 2- a 3- - a n be an AP of nonzero terms then prove that1-a1 a2-1-a2 a3-1-a3 a4-1-an-1 an-n-1-a1 an

Let d be the common difference of the given AP. Then, (a_2 a_1)=(a_3 a_2)=(a_4 a_3)= ⋯ = (a_n a_n 1)=d ∴ 1/a_1 a_2+ 1/a_2 a_3 + 1/a_3 a_4 + ⋯ + 1/a_n 1a_n ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be an AP of nonzero terms then prove that

$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots + \frac{1}{a_{n - 1} a_{n}} = \frac{(n - 1)}{a_{1} a_{n}}$

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Solution

Let d be the common difference of the given AP. Then,

$(a_{2} - a_{1}) = (a_{3} - a_{2}) = (a_{4} - a_{3}) = \dots = (a_{n} - a_{n - 1}) = d$

$∴$ $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots + \frac{1}{a_{n - 1} a_{n}}$

$= \frac{1}{d} . {\frac{d}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots + \frac{d}{a_{n - 1} a_{n}}}$

$= \frac{1}{d} . {\frac{(a_{2} - a_{1})}{a_{1} a_{2}} + \frac{(a_{3} - a_{2})}{a_{2} a_{3}} + \frac{(a_{4} - a_{3})}{a_{3} a_{4}} + \dots + \frac{(a_{n} - a_{n - 1})}{a_{n - 1 a_{n}}}}$

$[∵ (a_{2} - a_{1}) = (a_{3} - a_{2}) = \dots = (a_{n} - a_{n - 1}) = d]$

$= \frac{1}{d} . {(\frac{1}{a_{1}} - \frac{1}{a_{2}}) + (\frac{1}{a_{2}} - \frac{1}{a_{3}}) + (\frac{1}{a_{3}} - \frac{1}{a_{4}}) + \dots + (\frac{1}{a_{n - 1}} - \frac{1}{a_{n}})}$

$= \frac{1}{d} . (\frac{1}{a_{1}} - \frac{1}{a_{n}}) = \frac{1}{d} . \frac{(a_{n} - a_{1})}{a_{1} a_{n}}$

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

$= \frac{1}{d} . \frac{(n - 1) d}{a_{1} a_{n}} = \frac{(n - 1)}{a_{1} a_{n}}$

Hence, the result follows.

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Similar questions

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ be an AP of non-zero terms. Prove that

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

$\frac{n - 1}{a_{1} a_{n}}$

Q. If a₁,a₂.....a_n be in AP of non zero terms then prove that
1/(a₁a₂) + 1/(a₂a₃)+........+1/(a_n-1a_n) = (n-1) /(a₁a_n)

Q. If

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

an are in AP , then

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . . . + \frac{1}{a_{n - 1} a_{n}}

equals.

Q. If

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

be an AP of non-zero terms , then

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

Q. If

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

are in H.P then

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . + a_{n - 1} a_{n} =

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If a1, a2, a3, ⋯, an be an AP of nonzero terms then prove that 1a1a2+1a2a3+1a3a4+⋯+1an−1an=(n−1)a1an

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be an AP of nonzero terms then prove that

$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots + \frac{1}{a_{n - 1} a_{n}} = \frac{(n - 1)}{a_{1} a_{n}}$