Question

If seven times the 7^th term of an A.P. is equal to eleven times the 11^th term, then what will be its 18^th term?

Answer 1

Given: seven times the 7^th term of an A.P. is equal to eleven times the 11^th term

nth term of an AP given by 

$$a+(n-1)d=a_n$$

Thus, the 7th term will be

$$a_7=a+(7-1)d$$
$$\Rightarrow a_7=a+6d...\left(1\right)$$

 the 11th term will be

$$a_{11}=a+(11-1)d$$

$$\Rightarrow a_{11}=a+10d...\left(2\right)$$

Now,

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

$$\Rightarrow 7a+42d=11a+110d$$

$$\Rightarrow 11a-7a+110d-42d=0$$

$$\Rightarrow 4a+68d=0$$

$$\Rightarrow 4(a+17d)=0$$

$$\Rightarrow (a+17d)=0$$

So, the 18th term is 0.