If $A_{1}, A_{2}, A_{3}, . . . A_{n}$ are $n$ numbers inserted between $a$ and $b$ such that $a, A_{1}, A_{2}, . . . ., A_{n}, b$ is an A.P., then $A_{1} + A_{2} + A_{3} . . . A_{n}$

\frac{a + b}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n}{2} (a + b)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{n}{4} (a + b)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n}{3} (a + b)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{n}{2} (a + b)$
$A_{1} + A_{2} + A_{3} + . . . A_{n}$
$= (a + A_{1} + A_{2} + A_{3} + . . . A_{n} + b) - (a + b)$ ...(i)
Sum of n term is A.P is
$= S_{n}$
$= \frac{n}{2} (a_{1} + a_{n})$ .
Hence
$= (a + A_{1} + A_{2} + A_{3} + . . . A_{n} + b) - (a + b)$
$= \frac{n + 2}{2} (a + b) - (a + b)$
$= (a + b) [\frac{n + 2}{2} - 1]$
$= (a + b) (\frac{n}{2} + 1 - 1)$
$= \frac{n}{2} (a + b)$ .
Hence
$A_{1} + A_{2} + A_{3} + . . . A_{n} = \frac{n}{2} (a + b)$ .

Suggest Corrections

Similar questions

If A₁, A₂, A₃,..., A_n are n numbers between a and b, such that a, A₁, A₂, A₃,..., A_n, b are in A.P., then

If A₁, A₂, A₃,...., A_n are n numbers between a and b, such that a, A₁, A₂, A₃,..., A_n, b are in A.P., then the common difference of this A.P. is:

Q. If

a_{r} > 0; \forall r, n \in N

and

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

are in A.P, then

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

a_{1}

a_{2}

a_{3}

,.....,

a_{n}

are in an

A . P .

\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}}} = \frac{(n - 1)}{\sqrt{a_{n}} + \sqrt{a_{1}}}

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

Join BYJU'S Learning Program

If A1,A2,A3,...An are n numbers inserted between a and b such that a,A1,A2,....,An,b is an A.P., then A1+A2+A3...An

The correct option is B n2(a+b)A1+A2+A3+...An=(a+A1+A2+A3+...An+b)−(a+b) ...(i)Sum of n term is A.P is =Sn=n2(a1+an).Hence =(a+A1+A2+A3+...An+b)−(a+b)=n+22(a+b)−(a+b)=(a+b)[n+22−1]=(a+b)(n2+1−1)=n2(a+b).Hence A1+A2+A3+...An=n2(a+b).

If $A_{1}, A_{2}, A_{3}, . . . A_{n}$ are $n$ numbers inserted between $a$ and $b$ such that $a, A_{1}, A_{2}, . . . ., A_{n}, b$ is an A.P., then $A_{1} + A_{2} + A_{3} . . . A_{n}$