determine A.P whose 3 rd term is 16 and 7 th term exceeds 5 th term by 12
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Answered by
2
T3 = a+2d
T7 = a+6d
T5 = a+4d
T7 = T5+13
a+6d = a+4d+13
2d = 13
d = 6.5
T3 = a+2d = 16
a = 16-(2×6.5)
a = 16-13
a = 3
The required A.P with a = 3 & d=6.5 is ;
3,9.5,16,22.5,29.....
T7 = a+6d
T5 = a+4d
T7 = T5+13
a+6d = a+4d+13
2d = 13
d = 6.5
T3 = a+2d = 16
a = 16-(2×6.5)
a = 16-13
a = 3
The required A.P with a = 3 & d=6.5 is ;
3,9.5,16,22.5,29.....
Answered by
7
Let the first term be "a" and common difference "d".
T3 = a + 2d
T7 = a + 6d
T5 = a + 4d
According to the question,
a + 6d = a + 4d + 12
6d = 4d + 12
6d - 4d = 12
2d = 12
d = 6
Putting the value of d = 6 in T3,
16 = a + 2 x 6
16 = a + 12
a = 4
Now the required AP ----
a, a + d,a+2d,a+3d,a+4d,a+5d.....
4, 10, 16,22,28,34....
T3 = a + 2d
T7 = a + 6d
T5 = a + 4d
According to the question,
a + 6d = a + 4d + 12
6d = 4d + 12
6d - 4d = 12
2d = 12
d = 6
Putting the value of d = 6 in T3,
16 = a + 2 x 6
16 = a + 12
a = 4
Now the required AP ----
a, a + d,a+2d,a+3d,a+4d,a+5d.....
4, 10, 16,22,28,34....
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