Question

If the sum of the first four terms is 26 and the sum of the first six terms is 57 in an arithmetic series, then draw that arithmetic series. please yaro solve fast​

Answer 1

Answer:

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.

Answer 2

Answer:

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.