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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Insert $5$ numbers between $8\:and\:26$ such that the resulting sequence is an $A.P.$

$\begin{array}{1 1}11,14,17,20,23 \\ 9,10,11,20,22 \\ 11,13,15,17,19 \\ 11,15,17,21,23 \end{array} $

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  • $r^{th}$ A.M. between $a\:\:and\:\:b$ = $A_r=a+rd$ where $d=\large\frac{b-a}{n+1}$ and $n$ is no. of $A.M^s$ inserted.
Given: $5$ $A.M.^s$ are inserted between $8\:\:and\:\:26$
$\Rightarrow\:8,A_1,A_2,A_3,A_4,A_5,26$ are in A.P
$\therefore\:n=5,\:\:a=8,\:\:b=26$
common difference $d$ of this A.P. is $\large\frac{b-a}{n+1}=\frac{26-8}{5+1}$=$3$
$\therefore\:A_1=a+d=8+3=11$
$A_2=a+2d=8+2\times 3=14$
$A_3=a+3d=8+3\times 3=17$
$A_4=a+4d=8+4\times 3=20$
$A_5=a+5d=8+5\times 3=23$
answered Feb 24, 2014 by rvidyagovindarajan_1
 

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