If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that the 18th term of the AP is zero.

Show that: 18th term of the AP is zero.


Given: 7a7= 11a11


(Where a7 is Seventh term, a11 is Eleventh term, an is nth term and d is common difference of given AP)


Formula Used: an = a + (n - 1)d


7(a + 6d) = 11(a + 10d)


7a + 42d = 11a + 110d 68d = (4a)


a + 17d= 0 ….equation (i)


Now a18 = a + (18 - 1)d


So a + 17d = 0 [by using equation (i)]


HENCE PROVED


[NOTE: If n times the nth term of AP is equal to m times the mth term of same AP then its (m + n)th term is equal to zero]


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