Question

In an A.P, If (7)a7=(11)a11,then a18=?

A) 30

B) 20

C) 29

D) 31

Answer 1

HELLO DEAR,

GIVEN THAT, :-
$7 \times a_{7} = 11 \times a_{11} \\ = > 7(a + (7 - 1)d) = 11 \times (a + (11 - 1)d) \\ = > 7a + 42d = 11a + 110d \\ = > 7a - 11a = 110d - 42d \\ = > - 4a = 68d \\ = > a = - \frac{68}{4 } \\ = > a = - 17d.............(1)$
NOW THE,

$a_{18} = a + (18 - 1)d \\ = > a_{18} = - 17d + 17d \\ = > a_{18} = 0 \: \: ..... ....................using \: (1)$
I THINK YOUR OPTION IS WRONG

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

The answer is none of the above.

We know that sum of n terms of an AP an = a + (n - 1) * d.

7th term of an AP a7 = a + (7 - 1) * d

                                   = a + 6d  ------------ (1)

11th term of an AP a11 = a + (11 - 1) * d

                                     = a + 10d.  -------------- (2)


Given that 7 (a7) = 11(a11)

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

7a + 42d = 11a + 110d

4a + 68d = 0

a + 17d = 0.

a18 = 0.


Therefore the 18th term = 0.


Hope this helps!