in an artithmeticprogression the sum of first four terms is 20 and the sum of first three terms is 12 then find the fourth term is of the arithmetic proggression
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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.
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