Question

How many terms are there in an ap whose first and fifth terms are _14and 2,respectively and the sum of terms is 40?

Answer 1

Question:

How many terms are there in an AP whose first and fifth terms are -14 and 2 respectively and the sum of terms is 40?

Solution:

First and fifth terms are given.

→  a₁ = - 14

→  a₅ = 2

From this, we find common difference.

a₅ - a₁ = 2 - (- 14)

(5 - 1)d = 2 + 14

4d = 16

d = 16 / 4

d = 4

Now we apply this formula.

S_n = n/2 [2a + (n - 1)d]

Given that  S_n = 40.  We have to find 'n'. So,

40 = n/2 [2 · (- 14) + (n - 1)4]

40 = n/2 [- 28 + 4n - 4]

40 = n/2 [4n - 32]

40 = n(2n - 16)

40 = 2n² - 16n

2n² - 16n - 40 = 0

n² - 8n - 20 = 0

n² - 10n + 2n - 20 = 0

n(n - 10) + 2(n - 10) = 0

(n - 10)(n + 2) = 0

∴  n = 10      ;      n = - 2

But since 'n' is a natural number, n ≠ - 2.

∴  n = 10