Question

How many terms of the arithmetic sequence 5,7,9, must be added at 140

​

Answer 1

Answer:

10

Step-by-step explanation:

Here,

First term, a = 5

Common difference, d = 2

Sum of n terms = 140

= n/2 × [2a + (n-1)d]

140 = n/2 × [2(5) + (n-1)2]

140 = n/2 × (10 + 2n - 2)

140 = n/2 × (8 + 2n)

n² + 4n - 140 = 0

On factorisation,

(n-10) (n+14) are the factors

So,

n = 10 or n = -14

n = -14 is REJECTED. (no. of terms can't be negative)

Thus, x = 10 is Correct

Hence, 10 terms must be added of AP 5,7,9,... to get 140