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If the sum of n terms of the AP are n²+n.show that it is an AP
If we put n= 1, then we get a1Hence a1 = 1+ 1 = 2When n= 2 , we get (a1 +a2)Hence a1 +a2 = 4 + 2 = 6Hence a2= 4Now if n= 3, we get (a1 +a2 +a3 )Hence a1 +a2 +a3 = 9+ 3 = 12Soa3 = 12-2-4 =6Similarily we get a1 ,a2 ,a3 ,a4 ,a5 ,a6 ,......now if we seea2 - a1 = a3 - a2 = a4 - a3 = 2= constantHeHence we can say this forms an A.P.
Use the formula tn= a+(n-1)d =n2+n+1(sorry n2= n square) . To get square term here lets assume d=n . So a+n2-n=n2+n+1. So we can solve for a and it comes up to be 2n+1.So by using a=2n+1 and d=n we can get nth term as n2+n+1. This is not the only assumed values,there may be much more . So how could you say that it is not possible??
Use the formula tn= a+(n-1)d =n2+n+1(sorry n2= n square) . To get square term here lets assume d=n . So a+n2-n=n2+n+1. So we can solve for a and it comes up to be 2n+1.
So by using a=2n+1 and d=n we can get nth term as n2+n+1. This is not the only assumed values,there may be much more . So how could you say that it is not possible??
Sum of the nth number is n^2+n. ....givenT1=1+1=2. =A1.... ( 1 ) T2=4+2=6. =A1+A2. ( 2 ) A1+A2=6A2=4. From ( 1 )T3=9+3=12. = A1+A2+A3. ( 3 )A1+A2+A3=12A3=6. From( 2 )Hence, A2-A1=A3-A2=A4-A3 and so on...................The difference between them is d=2...........Therefore the given equation is A.P
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