Question

Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Answer 1

GIVEN

54th term

a + 53 d

132 more than a54

a + 53d + 132

Therefore, this is your an (last term)

AP 3 15 27 39,...

first term 'a' = 3

Common difference 'd' = 15 - 3 = 12

an = a + 53d + 132

FORMULA

an = a + (n - 1)d

a + 53d + 132 = a + (n - 1)d

Subsitute the values of a and d

3 + 53x12 + 132 = 3 + ( n - 1) x 12

3 + 53x12 + 132 - 3 = 12( n - 1)

636 + 132 = 12( n - 1)

768 = 12( n - 1)

768 / 12 = n - 1

64 = n - 1

64 + 1 = n

n = 65

Therefore, the required term is 65th

Answer 2

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