How many terms have to be consider for getting the sum 5740 in A.P 7,14,21,...
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Answer:
Sn = n/2 [ 2a + (n-1)d ]........
Where, a = 1st term, n = total no. of terms, d = difference between two consecutive terms,
and Sn= sum of total terms.
Subsituting given values,
5740 = n/2 [ 2(7) + (n-1) (7) ]
5740 X 2 = n ( 14 + 7n - 7 )
11480 = 14n + 7n2 - 7n
11480 = 7n + 7n2
dividing throughout by 7 ,
1640 = n2 + n
n2 + n - 1640 = 0
now, its a quadratic equation, factorising it,
n2 + 41n - 40n -1640 = 0
n( n + 41) - 40( n + 41) = 0
( n - 40 ) ( n + 41 ) = 0
n - 40 = 0 OR n + 41 = 0
It means,
n = 40 OR n = -41
But no. of terms cannot be negative , so we reject n = -41
n = 40 is acceptable,
So 40 terms of given A.P are to be added for getting sum 5740.
I Hope It Will Help!
^_^
Sn = n/2 [ 2a + (n-1)d ]........
Where, a = 1st term, n = total no. of terms, d = difference between two consecutive terms,
and Sn= sum of total terms.
Subsituting given values,
5740 = n/2 [ 2(7) + (n-1) (7) ]
5740 X 2 = n ( 14 + 7n - 7 )
11480 = 14n + 7n2 - 7n
11480 = 7n + 7n2
dividing throughout by 7 ,
1640 = n2 + n
n2 + n - 1640 = 0
now, its a quadratic equation, factorising it,
n2 + 41n - 40n -1640 = 0
n( n + 41) - 40( n + 41) = 0
( n - 40 ) ( n + 41 ) = 0
n - 40 = 0 OR n + 41 = 0
It means,
n = 40 OR n = -41
But no. of terms cannot be negative , so we reject n = -41
n = 40 is acceptable,
So 40 terms of given A.P are to be added for getting sum 5740.
I Hope It Will Help!
^_^
sumeetvanje12082003:
if raidus of incentre of an equilateral traingle is 3.12cm then what is radius of circumcircle
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