Question

How many terms of the arithmetic sequence 5,7,9,...,must be added to get 140?

Answer 1

Good Morning Student,

Here,
First term, a = 5
Common difference, d = 2
Sum of n terms = 140

$S_{n}$ = 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

Let me know if you have doubts especially regarding Physics, Chemistry or Maths.

Answer 2

Answer is 10. 10 terms can be added.



HOPE IT HELPS YOU!