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Which term of the A.P. 8,14,20,26,...will be 72 more than its 41st term |
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Answer» Given A.P:8,14,20,26, ...First term (a) = 8common difference (d) =a_{2}-a_{1}= 14-8= 6d = 6Let n th term of A.P will be 72 more than its 41 th term.We know that,{tex}\\boxed {n^{th} term = a_{n}=a+(n-1)d}{/tex}According to the problem given,{tex}a_{n}-a_{41}=72{/tex}{tex}\\implies a+(n-1)d-[a+(41-1)d]=72{/tex}{tex}\\implies a+nd-d-(a+40d)=72{/tex}{tex}\\implies a+nd-d-a-40d=72{/tex}{tex}\\implies nd-41d = 72{/tex}{tex}\\implies d(n-41)=72{/tex}{tex}\\implies n-41 = \\frac{72}{6}{/tex}{tex}\\implies n = 41 +12{/tex}{tex}\\implies n = 53{/tex}Therefore,n = 5353rd term in given A.P will be 72 more than its 41th term.\xa0\xa0 53rd term... |
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