Question

Which term of the arithmetic
progression 5, 15, 25, will
be 130 more than its 31st term​

Answer 1

EXPLANATION.

• GIVEN

Series will be given = 5, 15, 25, ..... will be 130

more than it's 31st term.

Solution.

first term = a = 5

common difference = d = b - a = 15 - 5 = 10

Given,

130 + 31st term.

Nth term of an Ap

An = a + ( n - 1 ) d

T31 = a + 30d

5 + 30(10) = 305

Therefore,

An = 130 + a31

a + ( n - 1 ) d = 130 + 305

5 + ( n - 1 ) 10 = 435

5 + 10n - 10 = 435

10n = 440

n = 44

Hence,

44th term of given Ap is 130 more than it's 31 term.

Answer 2

Answer:

44

Step-by-step explanation:

Given, a = 5

           d = 15 - 5 = 10

T₃₁ = a + (n-1)d

     = 5 + (31-1)10

     = 5 + 30*10

     = 5 + 300

     = 305

Let the required term be nth term

⇒ Tn = a + (n-1)d

⇒ 305 + 130 = 5 + (n-1) * 10 (∵The term is given as 130 more than T₃₁)

⇒ 435 = 5 + (n-1) * 10

⇒ 430 = (n-1) * 10

⇒ 43 = n - 1 (Dividing both sides by 10)

⇒ 44 = n    OR     n = 44