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6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

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Given : A.P. 25, 22, 19, ….....

$S_n=116$

a=25 , d = -3

$S_n=\frac{n}{2}[2a+(n-1)d]$

$\Rightarrow \, \, 116=\frac{n}{2}[2(25)+(n-1)(-3)]$

$\Rightarrow \, \, 232=n[50-3n+3]$

$\Rightarrow \, \, 232=n[53-3n]$

$\Rightarrow \, \, 3n^2-53n+232=0$

$\Rightarrow \, \, 3n^2-24n-29n+232=0$

$\Rightarrow \, \, 3n(n-8)-29(n-8)=0$

$\Rightarrow \, \, (3n-29)(n-8)=0$

$\Rightarrow \, \, n=8\, \, or\, \, \, n=\frac{29}{3}$

n could not be $\frac{29}{3}$ so n=8.

Last term $=a_8=a+(n-1)d$

$=25+(8-1)(-3)$

$=25-21=4$

The, last term of A.P. is 4.

Posted by

seema garhwal

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