Question

Find the 40th term of the arithmetic sequence whose 9th term is 465 and 20th term is 388

Answer 1

answer is obtained as above

Answer 2

248

Given:

The 9 th term of the AP = 465

The 20 th term of the AP = 388

We can write the 9th term as:

a + 8d = 465 --------- Equation 1

We can write the 20th term as:

a + 19d = 388 ---------- Equation 2

We need to solve these two equations by elimination method:

a + 8d = 465

a + 19d = 388

     - 11 d = 77

d = 77 / - 11

d = - 7

Substituting the value of d in Equation 2 we get:

a + 19 x - 7 = 388

a  - 133 = 388

a = 388 + 133

a = 521

Therefore, the first term of the AP is 521.

Now calculating the 40 th term of the AP:

= a + 39 d

= 521 + 39 x - 7

= 521 - 273

a = 248

Therefore, the 40 th term  of the AP is 248.