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BITSAT 2021 Paper

Section: Mathematics

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Question : 106 of 150

Marks: +1, -0

PART - IV : MATHEMATICS

If the sum of a certain number of terms of the A.P.

25,22,19,\ldots

is 116 then the last term is

0
2
4
6

Solution:

a=25, d=22-25=-3

.

Let

n

be the no. of terms

Sum

=116 ;

Sum

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

116=\frac{n}{2}[50+(n-1)(-3)]

or

232=n[50-3n+3]=n[53-3n]

=-3n^{2}+53n

\Rightarrow 3n^{2}-53+232=0

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

\Rightarrow n=8

or

n=\frac{29}{3},

n \neq \frac{29}{3}

\therefore n=8

\therefore

Now,

T_{8}=a+(8-1)d=25+7\times(-3)

=25-21=4

∴ Last term

=4

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