If ar>0, r ∈N and a1 a2 ……a2 m are in AP, then, (a) n-1 (b) (n(a1+a2 n))/(√(a1)+√(an+1)) (c) (n

If ar>0, r ∈N and a1 a2 ……a2 m are in AP, then, (a) n-1 (b)...

If $\mathrm{a}_{\mathrm{r}}>0, \mathrm{r} \in \mathrm{N}$ and $\mathrm{a}_{1} \mathrm{a}_{2} \ldots \ldots \mathrm{a}_{2 \mathrm{~m}}$ are in $\mathrm{AP}$, then, (a) $\mathrm{n}-1$ (b) $\frac{n\left(a_{1}+a_{2 n}\right)}{\sqrt{a_{1}}+\sqrt{a_{n+1}}}$ (c) $\frac{n-1}{\sqrt{a n}+\sqrt{a n+1}}$ (d) None of these

If $\mathrm{a}_{\mathrm{r}}>0, \mathrm{r} \in \mathrm{N}$ and $\mathrm{a}_{1} \mathrm{a}_{2} \ldots \ldots \mathrm{a}_{2 \mathrm{~m}}$ are in $\mathrm{AP}$, then,
(a) $\mathrm{n}-1$
(b) $\frac{n\left(a_{1}+a_{2 n}\right)}{\sqrt{a_{1}}+\sqrt{a_{n+1}}}$
(c) $\frac{n-1}{\sqrt{a n}+\sqrt{a n+1}}$
(d) None of these

Key Concepts

General Term Formula

The general (nth) term of an arithmetic progression is given by the expression a1 + (n - 1)d, where a1 is the first term and d is the common difference. This formula is fundamental in determining any specific term within the sequence without needing to list all previous terms.

Arithmetic Progression (AP)

An arithmetic progression is a sequence in which the difference between any two successive terms is constant. This constant, called the common difference, allows for the generation of the entire sequence from any initial term by repeatedly adding it.

Sum Formula of an AP

The sum of the first n terms in an arithmetic progression can be found using the formula S = n/2 × (first term + last term), or equivalently, S = n/2 × [2a1 + (n - 1)d]. This formula is useful for quickly calculating the total of a long sequence when only the endpoints are known.

Transcript

00:01 Hi.

00:02 For given a r greater than zero, r belongs to natural numbers and a1, a2n so on a2n are in ap.

00:09 We need to just evaluate this.

00:11 So here we have a1a2a2n are in ap.

00:14 So we can take here from this, that is a1 plus a2n, equal a2n negative 1 equals an plus an positive 1 is equal to k.

00:30 Now we can take this because i'll just work on this, we get a1 is a1 and plus d a 2 n is a plus 2 n negative 1d right this will be a plus i'm sorry a 1 is a only in this string as a next a 2 is a plus d plus a plus 2 n negative 1 negative 1 d equals and so on we have this will be a n a plus n negative 1 d plus we have a plus and positive 1 negative 1 d so if you just evolve this that is coming out to be 2a plus 2n negative 1d.

01:13 This will be again 2a plus that's going to have to be, let me solve this, 2n negative 1d.

01:24 It's going to be 2a plus 2n negative 1d.

01:29 So it's come out to be same.

01:31 Right.

01:31 So we can write this as a1 plus a2n equals a2 plus a2n negative 1 plus an plus an plus an plus an 431 is equal to k.

01:39 So, we can put this now as come to be k over root a1 plus root a2, k over root j1 plus root a2.

01:55 Next it is given as plus k over root a2 plus a3, root a2 plus root a2 plus root a2.

02:07 And to on we have k over, this can we have to be, we'll have root an plus root a .n plus root a .m plus root a .m plus root a.

02:25 This we have got here now just to write throughout k here.

02:29 So k times 10 1 over root a1 plus a2.

02:32 This is going to be root a1 negative root a2 or we have a1 negative a2.

02:42 We use a square negative b square b square and a2.

02:46 A plus a plus a2 negative r2, negative r3 over a2 negative a3 and so on.

02:55 Plus we have root a .n negative root a .n plus 1 over a .n negative a .n plus 1.

03:05 So what is a1.

03:08 Negative a2? it is in ap with increasing value.

03:13 It's given as here...

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