An important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2017 in which we have been asked to show that its 33rd term is triple its 15th term, if its 6th term is 0.
Book – RS Aggarwal, Class 10, arithmetic progression, chapter 5A, question no 28.
Sixth term of an AP is zero.
that is a6=0
a+5d=0
a=−5d
Now, a15=a+(n−1)d
a+(15−1)d=−5d+14d=9d
and a33=a+(n−1)d=a+(33−1)d=−5d+32d=27d
Now, a33:a15
27d:9d
3:1
Which shows that a33=3(a15)
Hence proved.