The sum of all terms of the arithmetic progression having ten terms except for the first term is 99,and except for the sixth term, 89. Find the third term of the progression if the sum of the first and the fifth term is equal to 10 *
Answers
Answer:
a3 = 5
Step-by-step explanation:
The formula used will be -
Sn = n/2[2a+(n-1)d]
S10 = 5[2a+(9)d]
S1= 1/2{2a]=>a
5[2a+9d]-a=99
5[2a+9d]-a=99
9a+45d=99
/9
a+5d=11 (1 eq)
a+2d = 5 (2 eq)
5[2a+9d]-11=89
10a+45d=100 (3 eq)
Solving (1) & (3)
a=1, d=2
Thus,
a3=1+4
a3=5
Answer:
a3 = 5
Step-by-step explanation:
Sn = n/2 [ 2a + (n - 1)d ]
=> S10 - S1 = 99
=> 10/2 [ 2a + 9d] - 1/2 [ 2a + 0d} = 99
=> 5[ 2a + 9d ] - 1/2 [ 2a] = 99
=> 10a + 45d - a = 99
=> 9a + 45d = 99
=> a + 5d = 11 ....................(i)
we got S1 as 'a' ; so the first term of the AP a1 = a
it is given in the question that a1 + a5 = 10
i.e => a + a + 4d = 10 { an=a+(n-1)d }
=> 2a + 4d = 10
=> a + 2d = 5 ..........................(ii)
(i) - (ii) we get 3d = 6
d = 2
a = 1
=>a3 = a + 2d
a3 = 5
make me the brainliest.