Question

if 6 times the 6th term of A.P is equal to the 11 times the 11th term then find its 17th term

Answer 1

We know that, nth term in an Arithmetic Progression is given by :

✿  $\bold{T_n = a + (n - 1)d}$

Where : a is the First term and d is the Common difference

$\bold{\implies T_6 = a + (6 - 1)d}$

$\bold{\implies T_6 = a + (5)d}$

$\bold{\implies T_1_1 = a + (11 - 1)d}$

$\bold{\implies T_1_1 = a + (10)d}$

Given : 6 Times the 6th Term of this A.P is Equal to 11 Times the 11th Term

$\bold{\implies 6 \times T_6 = 11 \times T_1_1}$

⇒ 6(a + 5d) = 11(a + 10d)

⇒ 6a + 30d = 11a + 110d

⇒ 5a + 80d = 0

⇒ a + 16d = 0

⇒ a + (17 - 1)d = 0

$\bold{\implies T_1_7 = 0}$

⇒ The 17th Term of the Given A.P is Zero

Answer 2

Answer:

the 17 term AP is zero....

Explanation:

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