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Which term of the AP 3,15,27,39,.....Will be 120 more than its 21st term |
| Answer» Here we are having a=3and d=15-3=12Let nth term Tn =T21+120so a+(n-1)d = a+20d+120or (n-1){tex}\\times{/tex}12 = 20 {tex}\\times{/tex} 12+12012n -12 = 240+12012n = 372{tex}\\style{font-size:12px}{\\text{n=}\\frac{372}{12}=31}{/tex}Hence, 31st term is the required term. | |