1.

hind the AP which has its sed teemus 16 and 7th Herm us 12 more thanits 5th derm.

Answer»

Given:a3 = 16a7 = a5 + 12 ............ (1)

a7 = a5 + d + d = a5 + 2d............(2)

From Equation (1) & (2)

a5 + 12 = a5 + 2d

2d = 12

d = 6

From Given,

a3 = 16

a3 = a + 2d = 16

a + ( 2× 6 ) = 16 [ We know that d = 6 ]

a + 12 = 16

a = 4

a1=4 a2=4+6=10a3=10+6=16....

So,AP is 4, 10, 16, 22, 28.......

a3 = 16=> a + 2d = 16 . . . . . (1)a7 = a5 + 12=> a + 6d = a + 4d + 12=> 2d = 12=> d = 6Substitute in (1)a + 2d = 16=> a = 16 - 2(6) = 16 - 12 = 4Thus, required A. P is 4, 10, 16, . . . . .



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