If a1- a2- a3 -anare in AP and a1 - 0- then the value of a3-a2 -a4-a3 - - -an-an-1 - a21-a2-1-a2-1-an-2 is equal to

Finding the value of (a_3/a_2 +a_4/a_3 + … .+a_n/a_n 1) 𝑎_2(1/a_2+1/a_2+....+1/a_n 2):Given that a_1, a_2, a_3 ...a_n are in APand a_1=0a_2=a_1+d=d0ex0exa_3=a_1 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

If a1, a2, a3...

Question

If $a_{1}, a_{2}, a_{3} . . . a_{n}$ are in AP and $a_{1} = 0$ , then the value of $(\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \dots . + \frac{a_{n}}{a_{n - 1}}) - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{2}} + . . . . + \frac{1}{a_{n - 2}})$ is equal to

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Solution

Finding the value of $(\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \dots . + \frac{a_{n}}{a_{n - 1}}) - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{2}} + . . . . + \frac{1}{a_{n - 2}})$ :
Given that $a_{1}, a_{2}, a_{3} . . . a_{n}$ are in AP
and $a_{1} = 0$
$a_{2} = a_{1} + d = d a_{3} = a_{1} + 2 d = 2 d a_{n - 1} = (n - 2) d a_{n} = (n - 1) d (\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \dots . + \frac{a_{n}}{a_{n - 1}}) - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{2}} + . . . . + \frac{1}{a_{n - 2}})$
$= (\frac{2 d}{d} + \frac{3 d}{2 d} + . . . \frac{(n - 1) d}{(n - 2) d}) - d (\frac{1}{d} + \frac{1}{2 d} + . . . + \frac{1}{(n - 3) d})$
$= 2 + \frac{3}{2} + . . . \frac{(n - 1)}{(n - 2)} - (1 + \frac{1}{2} + . . . + \frac{1}{(n - 3)})$ $= {(1 + 1) + (1 + \frac{1}{2}) + . . . (1 + \frac{1}{(n - 2)}} - {1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{(n - 3)}}$
$= 1 + 1 + 1 + . . . . (n - 2)$ Terms
$+ {1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{(n - 3)} + \frac{1}{(n - 2)}} - {1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{(n - 3)}}$ $= (n - 2) + \frac{1}{(n - 2)}$
Hence, the value is $(n - 2) + \frac{1}{(n - 2)} .$

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If a1,a2,a3...anare in AP and a1=0, then the value of a3a2+a4a3+….+anan-1-a21a2+1a2+....+1an-2 is equal to

If $a_{1}, a_{2}, a_{3} . . . a_{n}$ are in AP and $a_{1} = 0$ , then the value of $(\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \dots . + \frac{a_{n}}{a_{n - 1}}) - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{2}} + . . . . + \frac{1}{a_{n - 2}})$ is equal to