Question

If 7times the 7th term of an AP is equal to 11 times the 11th term. Prove that 18th term is Zero ​

Answer 1

Answer:-

First term be A

$(7a7 = 11 a11)$

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

$(7a-11a = 110d-42d)$

$(-4a = 68d)$

$(a = -17d )$

$(a18 = a +17d )$

$( a = -17 d )$

$(a18=-17d +17d =0 )$

Answer 2

$a_{n}th \: \: term \: \: of \: \: an \: A.P = \\ a + (n - 1)d$

Where a is the first term, n is the nth term and d is the common difference.

Case 1 [n = 7]:

$a_{7} = a + (7 - 1)d \\ a_{7} = a + 6d$

Case 2 [n=11]:

$a_{11} = a + (11 - 1)d \\ a_{11} = a + 10d$

$7(a_{7}) = 11(a_{11}) \\ 7a + 42d = 11a + 110d \\ 0 = 4a + 68d \\ a + 17d = 0$

$a_{18} = a + (18 - 1)d \\ a_{18} = a + 17d = 0$

Therefore, the 18th term = 0.