Question

In an A.P., 7 times the 7th term is equal to 11 times the 11th term. Find the 18th term of the A.P.

Answer 1

Hint: First we must find the 7th and 11th terms of the A.P and then equate them. In an A.P., the nth term is given as, ${T_n} = a + (n - 1)d$, where a is the first term and d is a common difference.

Complete step-by-step answer:
Given, 7 times the 7th term is equal to 11 times the 11th term
$ \Rightarrow 7 \times {T_7} = 11 \times {T_{11}}$
Let the first term be a and common difference be d.
Since, ${T_n} = a + (n - 1)d$ , where a is the first term and d is the common difference.
$
   \Rightarrow 7\left( {a + 6d} \right) = 11(a + 10d) \\
   \Rightarrow 7a + 42d = 11a + 110d \\
   \Rightarrow - 4a = 68d \\
   \Rightarrow a = - 17d \\
$
Now, we have to find the value of the 18th term
$
   \Rightarrow {T_{18}} = a + \left( {n - 1} \right)d \\
   \Rightarrow {T_{18}} = a + \left( {18 - 1} \right)d \\
   \Rightarrow {T_{18}} = a + 17d \\
$
Putting, a =-17d in the above equation
$ \Rightarrow {T_{18}} = - 17d + 17d = 0$

Therefore, the 18th term is 0.

Note: Arithmetic Mean is always greater than or equal to Geometric mean. Also remember that arithmetic mean multiplied by harmonic mean will give a square of geometric mean as the result.
$A.M \times H.M = {G.M^2}$