Math, asked by kaushiksunny2308, 4 months ago

sum of 10 A.M between 3,10​

Answers

Answered by mathdude500
3

Basic Definition :-

\sf \:Let \: a \: and \: b \: are \: two \: numbers \: and \: Let \: A_1,A_2,A_3, -  -  - ,A_n \:

\sf \: are \: inserted \: between \: a \: and \: b \: so \: that \: resultant \: series

 \red{\sf \: a,A_1,A_2,A_3, -  -  -  - ,A_n,b \: are \: in \: arithmetic \: progression}, \:

 \red{\sf \: A_1,A_2,A_3, -  -  -  - ,A_n\: are \: called \: arithmetic \: mean} \:

 \sf \: whose \: common \: difference, \: d = \dfrac{b - a}{n + 1}

Sum of n terms of an AP series :-

↝ Sum of n terms of an arithmetic sequence is,

\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2\:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}

Or

\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(\:a_1\:+ \: a_n \bigg)}}}}}} \\ \end{gathered}

Let's solve the problem now!!!

\sf \: Let \: A_1,A_2,A_3, -  -  - ,A_{10} \: be \: \: numbers \: between \: 3 \: and \: 10 \:

So that

 \red{\sf \: \: 3,A_1,A_2,A_3, -  -  - ,A_{10},10\: are \: in \: AP}

We know, Common difference is given by

\rm :\longmapsto\:\boxed{ \red{ \bf \:d = \dfrac{b - a}{n + 1}}}

Here,

a = 3

b = 10

n = 10

On substituting all these values, we get

\rm :\longmapsto\:d = \dfrac{10 - 3}{10 + 1}

 \purple{\bf :\longmapsto\:d = \dfrac{7}{11}}

Now,

\rm :\longmapsto\:A_1 = 3 + d

\rm :\longmapsto\:A_2 = 3 + 2d

.

.

.

.

.

\rm :\longmapsto\:A_{10}= 3 + 10d

Hence,

\purple{\rm :\longmapsto\:A_1 + A_2 + A_3 +  -  -  -  + A_{10}}

 \sf \:  =  \:  \: \dfrac{10}{2}(A_1 + A_{10})

 \sf \:  =  \:  \:5(3 + d + 3 + 10d)

 \sf \:  =  \:  \: 5(6 + 11d)

 \sf \:  =  \:  \: 5 \times \bigg(6 + 11 \times \dfrac{7}{11}  \bigg)

 \sf \:  =  \:  \: 5 \times 13

 \sf \:  =  \:  \: 65

Thus,

\purple{\bf :\longmapsto\:A_1 + A_2 + A_3 +  -  -  -  + A_{10} = 65}

Remark :-

Short Cut

 \red{\sf \: a,A_1,A_2,A_3, -  -  -  - ,A_n,b \: are \: in \: arithmetic \: progression}, \:

then

\purple{\bf :\longmapsto\:A_1 + A_2 + A_3 +  -   -  -  + A_{n} = \dfrac{n}{2}(a + b)}

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