​If nth term of the series 25+29+33+.... and 3+4+6+9+.... are equal, then find n.

Dear student,


Here the First series25+29+33+......where, a=25, d=29-25=4Tn=a+n-1dTn=25+n-14Tn=25+4n-4Tn=21+4nThe second series is3+4+6+9   1    2    3                 the Ist difference      1     1                    the difference of the difference    Here the difference between the first two numbers is 1 so d=1Also, the second differences are 1 so c=1the first term is 3 so a=3Using the formula, nth term Tn=a+n-1d+12n-1n-2c Tn=3+n-1×1+12n-1n-2×1 Tn=3+n-1+12n2-2n-n+2Tn=2+n+12n2-3n+2Tn=2+n+n22-3n2+1Tn=n22+n-3n2+3Tn=n22-n2+3But it is given that nth term of  First series is equal to second series21+4n=n22-n2+3n22-n2+3-21-4n=0n22-n2-4n-18=0n2-n-8n-36=0n2-9n-36=0n2-12n+3n-36=0nn-12+3n-12=0n-12n+3=0n=12 or n=-3   (neglecting -3 as the term can't be negative)Thus the value of n is 12


Regards

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