Question

If 11 term of an AP is 85 and 16 term of an AP is 140 then which term of the AP is 0

Answer 1

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Answer 2

Given: The 11th term of an AP is 85 and the 16th term is 140.

To find: The term in the AP that equates to 0.

Answer:

[a - the first term; d - the common difference.]

11th term of the AP = 85 ⇒ a + 10d = 85.

16th term of the AP = 140 ⇒ a + 15d = 140.

On solving these equations, we get a = - 25 and d = 11.

The term in the AP that equates to 0 ⇒ $\sf a_n\ =\ 0$.

$\sf a_n\ =\ 0\\\\\\a\ +\ (n\ -\ 1)d\ =\ 0\\\\\\-\ 25\ +\ (n\ -\ 1)(11)\ =\ 0\\\\\\-\ 25\ +\ 11n\ -\ 11\ =\ 0\\\\\\-\ 36\ +\ 11n\ =\ 0\\\\\\11n\ =\ 36\\\\\\n\ =\ \dfrac{36}{11}$

Since the value of n isn't a definite one, the AP does not contain 0 as a term.