Question

find the 40th term of A.P whose 9th term is 465 and 2oth term is 388​

Answer 1

Answer:

40 th term= 248

Step-by-step explanation:

9th term=465

20th term= 388

a+8d= 465 ---------- equation (1)

a+19d=388. ---------- equation (2)

equation 1 - equation 2

a + 8d = 465

a + 19d = 388

-. -. -

----------------------

0 - 11d = 77

-11d = 77

d= -77/11

d= -7. ( put it in equation 1)

a + 8 ( -7) = 465

a -56 = 465

a = 465+56

a = 521

now, a+ 39d = 40th term

a + 39d = 521 + 39(-7) = 521 -273 = 248.

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Answer 2

• 9th=a+8d=465------(1)

• 20th=a+19d=388-----(2)
• subtract equation (1) from (2)
• -11d=77
• d= -7
• put that d value on equation(1) than we find value of a that is 521.
• now a40=a+(n-1)d
• a40= 521+(40-1)×-7=248