in an AP 8th term of its second term & 11th term is 1/3of 4th term and exceed by 15 find the 15th term
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a + 7d = a + d/2
=>2a + 14d = a + d
a = -13d
a + 10d = 1/3(a + 3d) + 1
a + 10d -1 = 1/3(a + 3d)
3(a + 10d -1) = a +3d
3a + 30d -3 = a + 3d
2a + 27d =3
2(-13d) + 27d =3(since a = -13d)
-26d + 27d = 3
=> d = 3
a= -13d
=> a = -39
a + 14d = -39 + 42
hence a15 =3
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