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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    The number of terms in an A.P. is even; the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by \[10\frac{1}{2},\] then the number of terms in the A.P. is:     JEE Main Online Paper (Held On 19 April 2016)

    A) 4                                             

    B) 8

    C) 12                                          

    D) 16

    Correct Answer: B

    Solution :

    Let a, d and 2n be the first term, common difference and total number of terms of an A.P. respectively i.e. a + (a + d) + (a + 2d) + ... + (a + (2n ? 1)d ) No. of even terms = n, No. of odd terms = n Sum of odd terms : \[{{S}_{o}}=\frac{n}{2}[2a+(n-1)(2d)]=24\] \[\Rightarrow \]\[n[a+(n-1)d]=24\]                                         ...(1) Sum of even terms : \[{{S}_{e}}=\frac{n}{2}[2(a+d)+(n-1)2d]=30\] \[\Rightarrow \]\[n[a+d+(n-1)d]=30\]                    ?.(2) Subtracting equation (1) from (2), we get nd = 6                                                                    ...(3) Also, given that last term exceeds the first term by\[\frac{21}{2}\] \[a+(2n-1)d=a+\frac{21}{2}\] \[2nd-d=\frac{21}{2}\] \[\Rightarrow \]\[2\times 6=\frac{21}{2}=d\]                      \[(\because nd=6)\] \[d=\frac{3}{2}\] Putting value of d in equation (3) \[n=\frac{6\times 2}{3}=4\] Total no. of terms \[=2n=2\times 4=8\]


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