Question

please solve this fast please​

Answer 1

let the first term of an ap = a

common difference = d

since tn = a + (n - 1)d

according to the given condition

9th term = to = 0

t9 = a + 8d

0 = a + 8d

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

now, t29 = a + (n - 1)d

= a + (29 - 1)d

= a + 28d

t29 = -8d + 28d ..............(from 1)

t29 = 20d .........(2)

and t19 = a + (n-1)d

= a + (19-1)d

= a + 18d

= -8d + 18d ........(from 2)

t19 = 10d .....(from 2)

t29 = 20d ..........(3)

t29 = 2× t19 .......(from 2)

t29 = 2 × 19th term ......( from 3)

hence, the 29th term is double of 19th term.

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