Saturday, July 15, 2017

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

Soln:
Let first term = a and common difference = d in the AP.
We have, nth term
tn = a + (n -- 1)d
Seventh term (t7) = a + (7 - 1)d
t7 = a + 6d ..............(i)
Similarly,
11th  term (t11) = a + 10d............(ii)
It is given that,
7t7 = 11t11
7(a + 6d) = 11(a + 10d)
7a + 42d = 11a + 110d
11a - 7a = 42d - 110d
4a = -68d
a = -17d

Therefore, t18 = a + 17d = -17d + 17d = 0