Question

If sum of three consecutive terms of an AP is 15 and their product is 56 then find the the terms​

Answer 1

Answer:

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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.