Question

Which term of the A.P. 10,7,4...is 41

Answer 1

Answer:

Given:

a = 10

d = 7-10 = -3

An = - 41

Proof:

An = a+[n-1]d

-41 = 10 + [ n-1 ]  (-3)

-51 = 3-3n

-3n = -54

n = 13

Identified:

if an = 41 , then n value does not exist

if an = -41 , then n value is 13

Answer 2

Step-by-step explanation:

$\pink{ \large{10,7,4....}} \\ \\ a = 10 \\ d = - 3 \\ tn = - 41 \\ \\ \blue{to \: \: \: \: find : } \: n \\ \\ \green{ solution : } \: \: \: \: tn = a + (n - 1)d \\ \\ - 41 = 10 + (n - 1)( - 3) \\ \\ - 51 = - 3n + 3 \\ \\ 3n = 3 + 51 \\ \\ 3n = 54 \\ \\ n = \frac{54}{3} \\ \\ \pink{ \huge{n = 18}}$