One of the basic question from arithmetic progression chapter in which we have been asked to find the number of terms required so that the sum must be 693 of the given series.

Book – RS Aggarwal, Class 10, chapter 5C, question no 10.

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Consider

a=Â First term

d=Â Common difference

n=Â number of terms

Here,Â a=63,d=60âˆ’63=âˆ’3Â and sumÂ =Snâ€‹=693

Snâ€‹=n/2â€‹[2a+(nâˆ’1)d]

693=n/2â€‹[2Ã—63+(nâˆ’1)(âˆ’3)]

693Ã—2=n(126âˆ’3n+3)

1386=n(129âˆ’3n)

1386=129nâˆ’3n2

3n^2âˆ’129n+1386=0

n^2âˆ’43n+462=0

Which is a quadratic equation

n^2âˆ’21nâˆ’22n+462=0

n(nâˆ’21)âˆ’22(nâˆ’21)=0

(nâˆ’21)(nâˆ’22)=0

Either,Â nâˆ’21=0, thenÂ n=21

orÂ nâˆ’22=0, thenÂ n=22

Number of terms=21Â orÂ 22

T22â€‹=a+(nâˆ’1)d

=63+(22âˆ’1)(âˆ’3)

=63+21Ã—(âˆ’3)=63âˆ’63=0

Which shows that,Â 22th term of AP is zero.

Number of terms areÂ 21Â orÂ 22. So there will be no effect on the sum.