The 9th term of an ap is 1 more than 5 times its second term and the 12th term is 7 more than twice the 5th term find the AP
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According to first condition
a9 = 5(a2
a+ 8d = 5(a + d) + 1
a + 8d = 5a + 5d + 1
5a - a + 1 = 8d - 5d
4a + 1 = 3d
4a - 3d = -1 ------> 1
According to second condition
a12 = 2(a5) + 7
a+ 11d = 2(a+4d) + 7
a + 11d = 2a + 8d + 7
2a - a + 7 = 11d - 8d
a + 7 = 3d
a - 3d = -7 -----> 2
solving 1 & 2
4 a - 3d = -1
a - 3d = -7
- + +
---------------
3a = 6
a = 6/3
a = 2
substitute the value of a = 2 in equation 2
a - 3d = -7
2 - 3d = -7
- 3d = -7-2
- 3d = -9
d = -9/-3
d = 3
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