Q9 of 69 Page 72

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

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d


where n = no. of terms


a = first term


d = common difference


tn = nth terms


Given: t9 = 0


⇒ t9 = a + (9 – 1)d


⇒ 0 = a + 8d


⇒ a = – 8d


To Show: t29 = 2× t19


Now,


⇒ t29 = a + (29 – 1)d


⇒ t29 = a + 28d


⇒ t29 = – 8d + 28d = 20 d (since, a = – 8d )


⇒ t29 = 20 d


⇒ t29 = 2 × 10 d ….(1)


Also,


⇒ t19 = a + (19 – 1)d


⇒ t19 = a + 18d


⇒ t19 = – 8d + 18d = 10 d (since, a = – 8d )


⇒ t19 = 10 d …..(2)


From eq. (1) and eq. (2) we get,


t29 = 2× t19


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