Which term of the ap 3,15,27,39,........will be 120 more than its 21st

1.	Which term of the ap 3,15,27,39,........will be 120 more than its 21st term ?
Answer» Given sequence is\xa03,15,27,39,...The first term is\xa0a=3The common difference is\xa0d=15−3=12we know that the\xa0nth\xa0term of the arithmetic progression is given by\xa0a+(n−1)dLet the\xa0mth\xa0term be\xa0120\xa0more than the\xa021st\xa0termTherefore,\xa0120+mthterm=21stterm⟹120+a+(m−1)d=a+(21−1)d⟹120+(m−1)12=(20)12⟹12m−12=240+120⟹12m=360+12⟹12m=372⟹m= 372/12 \u200b=31 a = 3, d=12a21 = a + (n-1)d= 3 + (21-1)12= 243243 + 120 = 363an = 363 a + (n-1)d = 3633 + (n-1)12 = 36312n = 363 - 3 + 12n = 372 / 12∴n = 31st term

Discussion