If a1, a2, ay, …are in AP such that ai ≠0 and S==(1)/(a1 a3)+(1)/(a2 a4)+(1)/(a3 a5)+…+(1)/(am a

step 1 Hi, we're given here, A1, A2, H3 are in AP, A, I not equal to 0. As then is given as this, to fi

If a1, a2, ay, …are in AP such that ai ≠0 and S==(1)/(a1...

If $a_{1}, a_{2}, a_{y}, \ldots$ are in AP such that $a_{i} \neq 0$ and $S_{=}=\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\ldots+\frac{1}{a_{m} a_{n+2}}$ then $S_{2 n}-S_{=}$ is equal to (a) $\frac{1}{a_{n}}+\frac{1}{a_{2 m}}$ (b) $\frac{1}{2 \mathrm{~d}}\left[\frac{1}{\mathrm{a}_{1}}+\frac{1}{\mathrm{a}_{2}}+\frac{1}{\mathrm{a}_{2 \mathrm{n}+1}}+\frac{1}{\mathrm{a}_{2 \mathrm{~m}+2}}\right]$ (c) $\frac{\mathrm{nd}}{\left[\frac{1}{\mathrm{a}_{1}}+\frac{1}{\mathrm{a}_{2}}-\frac{1}{\mathrm{a}_{\mathrm{n}+1}}-\frac{1}{\mathrm{a}_{\mathrm{n}+2}}\right]}$ (d) $\frac{\mathrm{n}}{2}\left[\frac{1}{\mathrm{a}_{\mathrm{n}+1} \mathrm{a}_{2 \mathrm{n}+1}}+\frac{1}{\mathrm{a}_{\mathrm{n}+2} \mathrm{a}_{2 \mathrm{~m}+2}}\right]$

If $a_{1}, a_{2}, a_{y}, \ldots$ are in AP such that $a_{i} \neq 0$ and $S_{=}=\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\ldots+\frac{1}{a_{m} a_{n+2}}$ then $S_{2 n}-S_{=}$ is equal to
(a) $\frac{1}{a_{n}}+\frac{1}{a_{2 m}}$
(b) $\frac{1}{2 \mathrm{~d}}\left[\frac{1}{\mathrm{a}_{1}}+\frac{1}{\mathrm{a}_{2}}+\frac{1}{\mathrm{a}_{2 \mathrm{n}+1}}+\frac{1}{\mathrm{a}_{2 \mathrm{~m}+2}}\right]$
(c) $\frac{\mathrm{nd}}{\left[\frac{1}{\mathrm{a}_{1}}+\frac{1}{\mathrm{a}_{2}}-\frac{1}{\mathrm{a}_{\mathrm{n}+1}}-\frac{1}{\mathrm{a}_{\mathrm{n}+2}}\right]}$
(d) $\frac{\mathrm{n}}{2}\left[\frac{1}{\mathrm{a}_{\mathrm{n}+1} \mathrm{a}_{2 \mathrm{n}+1}}+\frac{1}{\mathrm{a}_{\mathrm{n}+2} \mathrm{a}_{2 \mathrm{~m}+2}}\right]$

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This property makes it possible to express any term in the sequence in terms of the first term and the common difference, which is essential when dealing with series and sums that involve terms from an AP.

Common Difference

The common difference is the fixed amount added to each term to obtain the next term in an arithmetic progression. It plays a crucial role in expressing general terms of the sequence, and it is often used to simplify expressions and transformations within sums that involve terms from the sequence.

Telescoping Series

A telescoping series is a series in which most terms cancel out when the series is expressed in a particular decomposed form. This characteristic allows for the simplification of complex sums by reducing them to a few non?cancelled terms. Recognizing and manipulating telescoping behaviors is a useful technique for evaluating sums in series, especially when dealing with products of consecutive terms.

Partial Fractions Decomposition

Partial fractions decomposition is a method used to break down a complex rational expression into simpler fractions that are easier to work with. This technique is particularly useful when summing series that involve reciprocal products of terms, as it can reveal a telescoping structure or simplify the process of finding closed-form expressions for the sum.

Transcript

00:01 Hi, we're given here, a1, a2, h3 are in ap, a, i not equal to 0.

00:05 As then is given as this, to find out s2n negative s.

00:09 We work on this.

00:12 So let's give the common difference.

00:16 We have a 1 over a1hp.

00:18 Let's write, let's say 1 over a1, hb equals we have 1 over 2d, a 3 negative a1, over a3.

00:28 So you can write, why? because d is given as a3.

00:33 Negative a 1 over 2 right so this is uh we have this common difference between a 1 and a 3 we have d stemming 2 a 2 a 2 a 2 8 3 which you can see you write like this and this can be given as 1 over 2d there'll be 1 over a 1 negative 1 over a 3 also got here next we have 1 over a 2 a 4 what's given as 1 over 2d next 1 over a 2 negative 1 over a 4 next 1 over a 3 a 5 equals 1 over 2d 1 over a 3 negative 1 over a 5 so on we continue so we get 1 over a n and negative 2 a n that is 1 over 2d 1 over 2 and negative 2 and negative 2 negative 1 over a next 1 over a n negative 1 a n positive 1 that's coming out to be 1 over 2d 1 over a n negative 1 negative 1 over a n positive 1.

02:09 Negative is given as we have 1 over a n plus 2d so we have 1 over a n negative 1 over a n plus 2 so let's say vote here now here we stopped the n plus 2 it is given here is a finite sum here as then that's coming out to be 1 over 2d it would be 1 over a 1 1 negative 1 over a 3 positive 1 over 8 2 negative 1 over a 4 and so on get here 1 over a a and negative 1, negative 1 over a and positive 1, positive 1 over a .n, negative 1 over a2, and positive 2.

03:07 So go here, it's going to be 1 over 2d.

03:13 Then 1 over 83 cancel with these terms here, and 1 over a4 cancel with these terms, and 1 over a n will cancel, 1 over a negative n will cancel.

03:21 So we get here 1 over a1 plus 1 over a2, negative 1 over a2, negative 1 over a2.

03:27 N plus 1 and negative 1 over a n plus 2.

03:34 Now to find out s 2n negative s n so s 2m replacing n with 2n 1 over 2d we get 1 over a 1 over a 2 2n over a 2n plus 1 negative 1 over a 2 n plus 2n so about here...

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