If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
The sum of the given terms 25 , 22 , 19 , … of an A.P. is 116.
Let a , d be the first term and common difference of the given A.P.
a = 25 d = 22 − 25 = − 3
The formula for the sum of n terms in an A.P. is given by,
S n = n 2 [ 2 a + ( n − 1 ) d ]
Substitute the values of a , d , and S n as 25 , − 3 , 116 in the above expression.
116 = n 2 [ 2 × 25 + ( n − 1 ) × ( − 3 ) ] 116 × 2 = n [ 50 − 3 n + 3 ] 232 = n [ 53 − 3 n ] 232 = 53 n − 3 n 2
Further simplify the above expression.
3 n 2 − 53 n + 232 = 0 3 n 2 − 24 n − 29 n + 232 = 0 3 n ( n − 8 ) − 29 ( n − 8 ) = 0 ( 3 n − 29 ) ( n − 8 ) = 0
Equate the above expression to obtain the value of n .
n = 29 3 or 8 .
As the number of terms must be an integer. So, the value of n cannot be 29 3 .
The formula to find the terms in an A.P. is given by,
T n = a + ( n − 1 ) d
Substitute the values of a , d , and n as 25 , − 3 , 8 in the above expression.
T 8 = 25 + ( 8 − 1 ) × ( − 3 ) = 25 + 7 × ( − 3 ) = 25 − 21 = 4
Thus, the 8 t h term of the A.P. is 4 .
The sum of the given terms
Let
The formula for the sum of
Substitute the values of
Further simplify the above expression.
Equate the above expression to obtain the value of
As the number of terms must be an integer. So, the value of
The formula to find the terms in an A.P. is given by,
Substitute the values of
Thus, the
