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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Updated on: 10-Oct-2022

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