Question

How many consecutive odd integers beginning with 5 will sum to 480?​

Answer 1

odd integers beginning with 5 are ; 5 , 7, 9, 11, 13, 15, ...... n terms

a/c to question,

5 + 7 + 9 + 11 + ..... + n terms = 480

here we see that, differences between two successive terms are same.

i.e., 7 - 5 = 9 - 7 = 11 - 9 = ... = 2

so, given terms are in arithmetic progression.

where 5 is the first term and 2 is the common ratio of AP.

so, nth term or last term of given ap = a + (n - 1)d

= 5 + (n - 1)2

= 5 + 2n - 2 = 3 + 2n

hence, 5 + 7 + 9 + 11 .... + (3 + 2n) = 480

from formula of sum of n terms,

Sn = n/2 [ first term + last term ]

480 = n/2 [ 5 + (3 + 2n) ]

960 = n[8 + 2n]

or, 480 = 4n + n²

or, n² + 4n - 480 = 0

or, n² - 20n + 24n - 480 = 0

or, n(n - 20) + 24(n - 20) = 0

or, (n + 24)(n - 20) = 0

or, n = -24 and 20

but n ≠ -24 [ as it is number of terms ]

so, n = 20

hence, 20 consecutive odd integers beginning with 5 will sum to 480.