Question

Find 5 number in a AP such that their sum is 65 and the product of the second and third number excused the fourth by 100​

Answer 1

GIVEN :-

• sn = 60
• n = 5
• 2nd term × 3rd term = 4th term + 100

TO FIND :-

• we have to find ap

SOLUTION :-

let the ap be :-

a - 2d , a - d , a , a + d , a + 2d

[ note :- these are called ready made ap ]

now according to question :-

sum of Ap = 60

=> (a - 2d) + (a - d) + (a) + (a + d ) + (a + 2d) = 60

=> 5 a = 60

=> a = 12

now putting the condition given in question :-

=> 2nd term × 3rd term = 4th term + 100

=> (a - d) (a) = (a + d) + 100

=> a² - ad = a + d + 100

now put the value of a = 12

=> (12)² - (12) d = 12 + d + 100

=> 144 - 12d = 112 + d

=> 144 - 112 = 13d

=> 13d = 32

=> d = 2.4 approx

HENCE , ap =

=> 12 - (2)(2.4) , 12 - (2.4) , 12 , 12 + 2.4 , 12 + (2)(2.4)

=> 12 - 4.8 , 12 - 2.4 , 12 , 12 + 2.4 , 12 + 4.8

=> 7.2 , 9.6 , 12 , 14.4 , 16.8