Question

Q.In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

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Answer 1

Answer:

T1+T2+T3+T4+T5=41[T6+T7+T8+T9+T10]

Let first term =a

common difference=d

[a+(a+d)+(a+2d)+(a+3d)+(a+4d)]=41[(a+5d)+(a+6d)+(a+7d)+(a+8d)+(a+9d)]

⇒(5a+10d)=41(5a+35d)

⇒20a+40d=5a+35d

⇒15a+5d=0

⇒3a+d=0

⇒d=−3a

d=−6 (given a=2)

T20=a+19d=2+19(−6)

=−112.

Answer 2

Q.In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.