Question

The sum of an A.P. is 56 and the sum of it's last four terms is 112. If its first term is 11 then find tye number of the terms in the A.P.​

Answer 1

Answer:

Solution

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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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