Answer :

Given,


9th term of an A.P is 0


a9 = 0


To prove: a29 = 2a19


We know, an = a + (n – 1)d where a is first term or a1 and d is common difference and n is any natural number


When n = 9:


a9 = a + (9 – 1)d


a9 = a + 8d


According to question:


a9 = 0


a + 8d = 0


a = -8d


When n = 19:


a19 = a + (19 – 1)d


a19 = a + 18d


a19 = -8d + 18d


a19 = 10d


When n = 29:


a29 = a + (29 – 1)d


a29 = a + 28d


{ a = -8d}


a29 = -8d + 28d


a29 = 20d


a29 = 2×10d


{ a19 = 10d}


a29 = 2a19


Hence Proved


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

There are n A.M.sRD Sharma - Mathematics

If x, y, z are inRD Sharma - Mathematics

Insert 7 A.M.s beRD Sharma - Mathematics

The 10th</suRD Sharma - Mathematics

Insert five numbeRD Sharma - Mathematics

The 4th</supRD Sharma - Mathematics

In an A.P. the fiRD Sharma - Mathematics

Insert 4 A.M.s beRD Sharma - Mathematics

Show that x2RD Sharma - Mathematics

An A.P. consists RD Sharma - Mathematics