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The number of terms of an A.P. is even ; sum of all terms at odd places and even places are $24$ and $30$ respectively . Last term exceeds the first term by $10.5$. Then find the number of terms in the series.
Given: The number of terms of an A.P. is even ; sum of all terms at odd places and even places are $24$ and $30$ respectively . Last term exceeds the first term by $10.5$.
To do: To find the number of terms in the series.
Solution:
Let total no. of terms be $2n$, & first term, $a$ and common difference be $d$.
$a+( 2n-1)d=l$
$\Rightarrow ( 2n-1)d=l-a=10.5\ \ ...........( 1)$ $[\because\ l-a=10.5]$
If we take odd terms only it will also be a A.P. of n terms only with common difference of $2d$.
$24=\frac{n}{2}[2a+( n-1)2d]$
$24=n( a+( n-1)d)\ \ ..........( 2)$
$30=\frac{n}{2}[2( a+d)+( n-1)2d]$
$30=n( a+nd)\ \ .............. ( 3)$
Put the value of equation $( 3)$ in equation $( 2)$
$24=30-nd$
$\Rightarrow nd=6$, Putting in equation $( 1)$
$2nd-d=10.5$
$\Rightarrow 2\times6-d=10.5$
$\Rightarrow 12-d=10.5$
$\Rightarrow d=1.5$
$n=4$
$a=1.5$
So no of terms $8$.
The A.P. is $1.5,\ 3,\ 4.5,\ 6,\ 7.5,\ 9,\ 10.5,\ 12$
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