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26. The ratio of the sums of m and nterms of an A.P. is m2 :n2. Show that the ratioof mthand nth term is (2m – 1) : (2n – 1).

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    26. Let a and d be the first term and common difference of A.P.

    Then, Sum of M terms of A.P.SumofntermsofA.P.=m2n2

    m2[2a+(m−1)d]m2[2a+(n−1)d]=m2n2

    m[2a+(m−1)d]n[2a+(n−1)d]=m2n2

    Dividing both sides by m/n we get,

    2a+(m−1)d2a+(n−1)d=mn

    [2a+(n – 1)d]n=m[2a+(n – 1)d].

    2an+dmn – dn=2am+dmn – dm.

    2an – 2am=dn – dm+2mn – dmn

    2a(n – m)=d(n – m)

    d= 2a(n−m)(n−m)

    d=2a.

    Now, Ratio of mth  and nthterm

    = a+(m−1)da+(n−1)d

    Putting d=2a in the above we get,

    ratio of mth and nth term = a+(m−1)2aa+(n−1)2a

    = a+2am−2aa+2an−2a

    = 2am−a2an−a

    = a[2m−1]a[2n−1]

    = 2m−12n−1 .

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