Question

Four terms are in ap.the sum of 2 and 3 rd term is 22 and the product of 1 and last term is 85 .find th number

Answer 1

let n th term, t(n) = a + (n - 1)d

the sum of 2 and 3 rd term is 22

2nd term, t(2) = a + (2-1)d = a + d
3 rd term, t(3) = a + (3-1)d = a + 2d
so, t(2) + t(3) = 22
a + d + a + 2d = 22
2a + 3d = 22
3d = 22 - 2a
d = (22 - 2a)/3..............(1)

the product of 1 and last term is 85

1st term, t(1) = a + (1-1)d = a
4th term, t(4) = a + (4-1)d = a + 3d

so, t(1) × t(4) = 85
a × (a + 3d) = 85.........(2)

substitute (1) in (2)
a × (a + 3(22 - 2a)/3) = 85
a × (a + 22 - 2a) = 85
a × (22 - a) = 85
22a - a² = 85
a² - 22a + 85 = 0
a² - 17a - 5a + 85 = 0
a (a - 17) - 5(a - 17) = 0
(a - 5) (a - 17) = 0
a - 5 = 0
a = 5
a - 17 = 0
a = 17
so, a = 5 or 17
first term, a = 17 is not possible because sum of 2 nd term and 3 rd term is 22

so, a = 5

substitute a in (1)
d = (22 - 2(5))/3
d = (22 - 10)/3
d = 12/3
d = 4

so, first term, t(1) = a = 5
second term, t(2) = a + d = 5 + 4 = 9
third term, t(3) = a + 2d = 5 + 2(4) = 5 + 8 = 13
fourth term, t(4) = a + 3d = 5 + 3(4) = 5 + 12 = 17

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