Question

How many number of terms required for an arithmetic series 10+8+6+……………., to get the sum is-126.

Answer 1

10 + 8 + 6 + ..........

first term, a = 10

common difference, d = 8 – 10

d = – 2

Sum, Sn = – 126

Sn = n/2 [ 2a + ( n – 1 ) d ]

– 126 = n/2 [ 2(10) + ( n – 1 ) – 2 ]

– 126 × 2 = n [ 20 – 2n + 2 ]

– 126 × 2 = n [ 22 – 2n ]

– 126 × 2 = n × 2 [ 11 – n ]

– 126 = 11n – n²

n² – 11n – 126 = 0

n² – 18n + 7n – 126 = 0

n ( n – 18 ) + 7 ( n – 18 ) = 0

( n – 18 ) ( n + 7 ) = 0

n = 18 or n = – 7

No of terms cannot be negative

Hence n = 18

So, 18 terms are required of the AP to get a sum equal to –126