How many terms of the A.P., - 6 , - 11 2 , - 5 , are needed to give the sum -25?

Here, a = -6, d = - 11 2 - ( - 6 ) = - 11 2 + 6 = 1 2 S _n = n 2 [2a + (n - 1)d] - 25 = n 2 [ 2 ( - 6 ) + ( n - 1 ) 1 2 ] - 50 = n [ - 12 + n - 1 2 ] - 50 = n [ - 24 + n - 1 2 ] -100=n^2-25 n n^2-25 n+100=0 (n-20)(n-5)=0 n=20 or n=5

How many terms of the A.P., - 6 , - 11 2 , - 5 , are needed to gi…

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How many terms of the A.P., $- 6 , \frac { - 11 } { 2 } , - 5 , \ldots$are needed to give the sum -25?

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Answer

Here, a = -6, $d = \frac { - 11 } { 2 } - ( - 6 ) = \frac { - 11 } { 2 } + 6 = \frac { 1 } { 2 }$$\therefore {{\text{S}}_n} = \frac{n}{2}[2a + (n - 1)d]$
$\Rightarrow - 25 = \frac { n } { 2 } \left[ 2 \times ( - 6 ) + ( n - 1 ) \times \frac { 1 } { 2 } \right]$
$\Rightarrow - 50 = n \left[ - 12 + \frac { n - 1 } { 2 } \right]$
$\Rightarrow - 50 = n \left[ \frac { - 24 + n - 1 } { 2 } \right]$
$\Rightarrow-100=n^2-25 n$
$\Rightarrow n^2-25 n+100=0$
$\Rightarrow(n-20)(n-5)=0$
$\Rightarrow n=20 \text { or } n=5$

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