Question

if the 9th term of an A.P.is zero then prove that 29 th term is double of 19 th term?​

Answer 1

Step-by-step explanation:

is given that the 9 th term of an A.P is T9 = 0

We know that the general term of an arithmetic progression with first term a and common difference d is T n

=a+(n−1)d, therefore, the 9th, 19th and 29th terms are as follows:

T9

=a+(9−1)d=a+8d.......(1)

T19

=a+(19−1)d=a+18d......(2)

T29

=a+(29−1)d=a+28d.......(3)

Now since T9 = 0 therefore, equation 1 becomes

0=a+8d

⇒a=−8d........(4)

Substitute the value of equation (4) in equation (3):

T29

=−8d+28d=20d=2(10d)=2(−8d+18d)=2(a+18d)=2[T19 ] (Using equations 1 and 2)

Hence, the 29th term of A.P is twice the 19 term.

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