Question

Show that the numbers 17,11,5,-1,-7,-13 ... form an A.P, find its n th term and 10 th term ?

Answer 1

Answer:

nth term = -6n + 23

10th term = -37

Step-by-step explanation:

Given :

A.P :- 17, 11, 5, -1, -7, -13

To find :

To show that the above progression form an A.P and find its nth term and 10th term.

Solution :

A.P. = 17, 11, 5, -1, -7, -13

Common difference (d) = (11 - 17) = (5 - 11) = (-1 - 5) = (-7-(-1) = (-13-(-7) = -6

First term (a) = 17

nth term of an A.P :-

• Tₙ = a + (n - 1)d

Substituting :-

• Tₙ = 17 + (n - 1)-6
• Tₙ = -6n + 23

∴ Tₙ (nth term) = -6n + 23 - (I)

Now, put n = 10 in (I) :-

• -6n + 23
• -6(10) + 23
• -60 + 23
• -37

∴ 10th term = -37

Answer 2

Given A.P = 17, 11, 5, - 1, - 7, - 13...

Here,

• a = 17

• d = (11 - 17) = (5 - 11) = (- 1 - 5) = - 6

We know that,

$\boxed{\pink{\sf a_n \: = \: a \: + \: (n \: - \: 1) \: d}}$

Substituting the values,

• $\sf a_n$ = 17 + (n - 1) - 6

• $\sf a_n$ = 17 + - 6n + 6

• $\sf a_n$ = 23 - 6n

Therefore,

• nth term is 23 - 6n.

10th term = ?

We know that,

$\boxed{\purple{\sf a_{10} \: = \: a \: + \: 9d}}$

Substituting the values,

• $\sf a_{10}$ = 17 + 9 * - 6

• $\sf a_{10}$ = 17 + (- 54)

• $\sf a_{10}$ = 17 - 54

• $\sf a_{10}$ = - 37

Therefore,

• 10th term = - 37.