Question

Find the $6^{th}$ term from the end of the AP 17, 14, 11,...., -40.

• A. -22
• B. -25
• C. -42
• D. -32

Answer 1

The correct option is B

-25

The given AP 17, 14, 11 ... - 40

Here , a = 17 , d = 14 - 17 = -3

let there be n terms in the given AP

Then , $n^{th}$ term = -40

a + (n - 1) d = -40

17 + ( n - 1) (-3) = -40

(n - 1)( -3) = -40 -17

(n - 1)( -3) = -57

n-1 = $\frac{57}{3}$

n = 19 + 1

n = 20

Hence , there are 20 terms in the given AP

Now , $6^{th}$ term from the end

= a + (20 - 6 )d [Since $r^{th}$ term from the end = a + (n - r) d]

= a + 14d

=17 + 14 (-3)

= 17 - 42 = -25

Hence , the $6^{th}$ term from the end of the given AP is -25