41 consecutive number average is 59 then find the second largest number
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Step-by-step explanation:
average of 41 consecutive numbers = 59
Sum of 41 consecutive numbers = 2419
x + (x+1)+.......................+(x+40) = 2419
41x + (1+2+3+4+..................+40) = 2419
41x + {40/2 (2 + 39*1)} = 2419
41x + {20 * 41} = 2419
41x + 820 = 2419
41x = 1599
x = 39
So , The numbers are (39,40,41,42,..............78,79)
The second largest number is 78
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Let the numbers me t + 1, t + 2, ..........,t + 40, t + 41
Sum of n terms = (n/2) * (a + an)
Average of n terms = Sum of n terms / n
59 = (21 / 2)*(t + 1 + t + 41)/41
59*41 = (21 / 2)*(2t + 42)
59*41 = 21 * (t + 21)
t + 21 = (59*41)/21
t = (59*41/21) - 21
t = (2419 - 441)/21
t = 1978 / 21
Second largest number would be t + 40 = (1978 / 21) + 40 = (1978 + 840) / 21 = 2818 / 21 —-> Answer
Sum of n terms = (n/2) * (a + an)
Average of n terms = Sum of n terms / n
59 = (21 / 2)*(t + 1 + t + 41)/41
59*41 = (21 / 2)*(2t + 42)
59*41 = 21 * (t + 21)
t + 21 = (59*41)/21
t = (59*41/21) - 21
t = (2419 - 441)/21
t = 1978 / 21
Second largest number would be t + 40 = (1978 / 21) + 40 = (1978 + 840) / 21 = 2818 / 21 —-> Answer
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