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Which term of AP 1) 3,16,29,42...... is 629 |
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Answer» here we see that , a=3 , d=13 it is given that last term of an AP is an 629now we PUT the FORMULA of nth term i.e. Tn=a+(N-1) *d629=3+(n-1) *13629=3+13n-13629=-10+13n629+10=13n639=13nn=639/13therefore, the value of n is 49.15 i.e. the last term of an AP is not considerabletherefore, these AP is not the SERIES of an AP whose last term is 629 |
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