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

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

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best_answer

Given : A.P. = -6 , -11/2 , -5...

a=-6

d=\frac{-11}{2}+6=\frac{1}{2}

Given : sum = -25

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

\Rightarrow \, \, -25=\frac{n}{2}[2(-6)+(n-1)\frac{1}{2}]

\Rightarrow \, \, \, \, -50= n[-12+(n-1)\frac{1}{2}]

\Rightarrow \, \, \, \, -50= -12n+ \frac{n^2}{2}-\frac{n}{2}

\Rightarrow \, \, \, \, -100= -24n+ n^2-n

\Rightarrow \, \, \, \, n^2-25n+100=0

\Rightarrow \, \, \, \, n^2-5n-20n+100=0

\Rightarrow \, \, \, \, n(n-5)-20(n-5)=0

\Rightarrow \, \, \, \, (n-5)(n-20)=0

\Rightarrow \, \, \, \, n=5\, \, or\, \, 20.

Posted by

seema garhwal

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