Question

if the 9th term of an a p is zero then show that the 29th term is twice the 19th term

Answer 1

solution.

a9= 0

=> a+ 8d=0

=> a= -8d....(1)

now,19 th term=

=> a19= a+ 18d

= -8d+18d (Using 1)

= 10d....(2)

Now, a29= a+ 28d

= -8d+ 28d (using 1)

= 20d ....(3)

So,3 is double of 2.

so,a29= 2(a19)

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HOPE MY ANSWER IS SATISFACTORY..,☺☺

Answer 2

Question:-

➡ If the 9th term of an A.P. is zero then prove that, 29th term is twice the 19th term.

Proof:-

Let us assume that,

➡ First term of the A.P. = a and,

➡ Common Difference = d

Now,

Nth term of an A.P. = a + (n -1)d

So,

9th term = a + (9 - 1)d

= a + 8d

Now, it's given that, 9th term of the A.P. is zero.

➡ a + 8d = 0 .....(i)

Now,

29th term = a + (29 - 1)d

= a + 28d

19th term = a + (19 - 1)d

= a + 18d

Now,

29th term - 2 × 19th term

= a + 28d - 2 × (a + 18d)

= a + 28d - 2a - 36d

= -a - 8d

= -1(a + 8d)

= -1 × 0

= 0

Hence,

29th term - 2 × 19th term = 0

➡ 29th term = 2 × 19th term. (Hence Proved)