Question

the 8th term of an AP is half of its second term and the 11th term exceeds one third of its fourth term by 1 find the 15th term​

Answer 1

a8 = (a2)/2

i.e. 2*(a8) = a2

i.e. 2*[a + (7*d)] = a + d

i.e. 2a + 14d = a + d

i.e. 2a - a + 14d - d = 0

i.e. a + 13d = 0 ------ Eq 1

& a11 = [(a4) /3] + 1

i.e. a + (10*d) - 1 = [a + (3*d)]/3

i.e. 3a + 30d - 3 = a + 3d

i.e. 3a - a + 30d - 3d = 3

i.e. 2a + 27d = 3 --------- Eq 2

Multiplying Eq 1 by 2,

2a + 26d = 0 ------- Eq 3

Subtracting Eq 3 from Eq 2,

d = 3

Substituting d=3 in Eq 1,

a + (13*3) = 0

a = -39

Now, a15 = a + 14d = -39 + (14*3) = -39 + 42 = 3

15th term of the AP is 3

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