Question

Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

Answer 1

Given, a₃ = 16

a + 2d = 16 (i)

a₅ = a + (5 – 1) d

= a + 4d

a₇ = a + (7 -1)d

= a + 6d

Acc. To question,

a₇ = 12 + a₅ (ii)

Put the value of a₅ and a₇ in (ii), we get

a + 6d = 12 + a + 4d

2d = 12 + a – a

2d = 12

d = 6

From equation (i),

16 = a + 2(6)

16 = a + 12

a = 4

We have to find A.P.

a₁ = 4

a₂ = a + d = 4 + 6 = 10

a₃ = a + 2d = 4 + 2(6) = 16

a₄ = a + 3d = 4 + 3(6) = 22

Hence, the A.P. is 4 , 10, 16 , 22,…