Question

How many terms of the AP:24, 21, 18,... Must be taken so that their sum is 78?​

Answer 1

Step-by-step explanation:

Sum of n-th term of an AP , Sn = (n/2) [ 2a + (n-1) d ]

Sn = 78 (given)

A/Q,

(n/2) [ 2 × 24 + ( n - 1 ) (-3) ] = 78

=> n [ 48 - 3n +3 ] = 78 × 2

=> n( 51 - 3n ) = 78 × 2

=> 3n( 17 - n) = 156

=> n( 17 - n) = 52

=>

${ - n}^{2} + 17n - 52 = 0$

${n}^{2} - 17n + 52 = 0$

$n = (17 + \sqrt{(289 - 208)} ) \div 2 \\ or \\ n = (17 - \sqrt{(289 - 208)} ) \div 2$

$n =( 17 - 9) \div 2 \\ or \\ n = (17 + 9) \div 2$

$n = 4 \: or \: 13$