Question

find the no of terms of an ap -5,-1,3,7,11 to be taken so that the sum is 400​

Answer 1

Answer:

16

Step-by-step explanation:

A.P. : - 5, - 1, 3, 7, 11

First term, a₁ = - 5

Second term, a₂ = - 1

Common difference, d = a₂ - a₁

= - 1 - (- 5)

= - 1 + 5

= 4

Also, Sum = 400 (given)

Applying formula,

Sn = n/2 [ 2a₁ + (n - 1)d ]

Putting known values, we get

→ 400 = n/2 [ 2(- 5) + (n - 1)(4) ]

→ 400 = n/2 [ - 10 + 4n - 4 ]

→ (400)(2) = n [ - 14 + 4n ]

→ (400)(2) = 2n [ - 7 + 2n ]

→ 400 = n [ - 7 + 2n ]

→ 400 = - 7n + 2n²

→ 2n² - 7n - 400 = 0

Using Middle Term Factorisation,

→ 2n² - 32n + 25n - 400 = 0

→ 2n (n - 16) + 25(n - 16) = 0

→ (2n + 25)(n - 16) = 0

By Zero Product Rule,

→ 2n + 25 = 0 and n - 16 = 0

→ n = - 25/2 and n = 16

Number of terms can't be negative and in fraction.

So, Total number of terms = 16