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The sum of first 10 terms of an Arithmetic Progression is 55 and the sum of first 9 termsof the same arithmetic progression is 45. Then its 10th term is,B) 55C) 12D) 10A) 9​

Answer»

Step-by-step EXPLANATION:

Let the A.P. be a,a+d,a+2d,a+3d,...a+(n−2)d,a+(n−1)d.

Sum of FIRST four terms =a+(a+d)+(a+2d)+(a+3d)=4a+6d

Sum of last four terms

=[a+(n−4)d]+[a+(n−3)d]+[a+(n−2)d]+[a+(n−1)d]⇒=4a+(4n−10)d

According to the GIVEN condition, 4a+6d=56

⇒4(11)+6d=56[Sincea=11(given)]

⇒6d=12⇒d=2

∴4a+(4n−10)d=112

⇒4(11)+(4n−10)2=112

⇒(4n−10)2=68

⇒4n−10=34

⇒4n=44⇒n=11

Thus the NUMBER of terms of A.P. is 11.


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