This is the question of arithmetic progression chapter in which we have been asked to find the number of terms must be taken so that their sum is 300 of the given series.

Book RS Aggarwal, Class 10, chapter 5C, question no 11.

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Consider

a=Â First term

d=Â Common difference

n=Â number of terms

Here,Â a=20,d=âˆ’2/3Â and sumÂ =Snâ€‹=300(for n number of terms)

Snâ€‹=n/2â€‹[2a+(nâˆ’1)d]

300=n/2â€‹[2Ã—20+(nâˆ’1)(âˆ’2/3â€‹)]

600=n[40âˆ’2â€‹n/3+2/3â€‹]

600=n[122â€‹/3âˆ’2â€‹n/3]

600=[n(122âˆ’2n)]/3â€‹

1800=122nâˆ’2n^2

2n^2âˆ’122n+1800=0

n^2âˆ’61n+900=0

n^2âˆ’25nâˆ’36n+900=0

n(nâˆ’25)âˆ’36(nâˆ’25)=0

(nâˆ’25)(nâˆ’36)=0

EitherÂ (nâˆ’25)=0Â orÂ (nâˆ’36)=0

n=25Â orÂ n=36

ForÂ n=25

300=25â€‹/2[2Ã—20+(25âˆ’1)(âˆ’2/3â€‹)]

=25/2â€‹[40+24Ã—(âˆ’2/3â€‹)]

=25/2â€‹[40âˆ’16]

=25/2â€‹Ã—24

=300

Which is true.

ForÂ n=36

300=36/2â€‹[2Ã—20+(36âˆ’1)(âˆ’2/3â€‹)]

=18[40+35Ã—(âˆ’2â€‹/3)]

=18[40âˆ’70/3â€‹]

=18Ã—(120âˆ’70)/3â€‹

=300

Which is true.

Result is true for both the values ofÂ n.

So both the numbers are correct.

Therefore, sum ofÂ 11Â terms is zero.Â (36âˆ’25=11).