Sixth term of an AP is zero.that is a6​=0a+5d=0a=−5dNow, a15​=a+(n−1)da+(15−1)d=−5d+14d=9dand a33​=a+(n−1)d=a+(33−1)d=−5d+32d=27dNow, a33​:a15​27d:9d3:1Which shows that a33​=3(a15​)Hence proved.