Question

If 7 times the 7th term of an ap is equal to 11 times its 11th term, find the 18th term of the ap.

Answer 1

Acc. to question :

7 times the 7th term is equal to 11 times the 11th term ,

So ,

7xA(7) = 11xA(11)

7x(A + (7-1)d) = 11x(A + (11-1)d) [ A(n) = A + (n-1)d]

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

7A + 42d = 11A + 110d

7A - 11A = 110d - 42d

-4A = 68d

A = -68d/4

So , A = -17d -(1)

Now its 18th term:

A(18) = A + (18-1)d

A(18) = -17d +17d [ From eqñ (1) ]

A(18) = 0

Hence it's 18th term is 0

Hope it helped

Answer 2

According to the question ,

                   $\:\:\:\:\:\:\:7a_{7} = 11_{a11} \\ \longrightarrow 7[a+(7-1)d] = 11[a+(11-1)d]\\ \longrightarrow 7(a+6d) = 11(a+10d)\\ \longrightarrow 7a+42d = 11a+110d\\ \longrightarrow 4a+68d = 0\\ \longrightarrow2(2a+34d) = 0\\ \longrightarrow 2a+34d = 0\\ \longrightarrow a+17d = 0\\18th \:term = a+(18-1)d\\ \:\:\:\:\:\:= a+17d=0$

HENCE PROVED