Question

9th term of an ap is zero prove that its 29th term is double of its 19 th term

Answer 1

according to question

T9 = 0

let a is the first term of A.p. and d is difference

=> a + 8d = 9 ......(1)

now

T19 = a + 18d

= (a + 8d) +10d

put value (a+8d) from equation (1)

= 0+ 10d

T19 = 10d ....(2)

same as

T29 = a + 28d

= (a+ 8d) + 20d

= 0 + 20d

T29 = 20d .....(3)

now from equation (2)& (3) we can say that

T29 = 2(10d)

T29 = 2(T19) →that's proved ......

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hope it will help you .....,,,,✌✌✌