Question

Detrmine ap whose 3 term is 16 when 6 term is subrated from 12 we get 12​

Answer 1

Answer:

we have given,

t3 = 16

t7 - t5 = 12

▶we know ,the formula to find the terms of AP ,

★tn = a + (n - d) d

therefor,

∴t3 = a + (3 - 1 )d = 16

∴a + 2d = 16 ____________(1)

And,

t7 - t5 = 12

∴[(a + (7 - 1) d] - [a + (5-1)d] = 12

∴(a +6d ) - (a +4d) = 12

→ a + 6d - a - 4d = 12

→ 2d = 12

→ d = 12/2

{\textbf{\large{d = 6}}}

▶put the value of d in equation (1)

→ a + 2d = 16

→ a + 2(6) = 16

→ a + 12 = 16

→ a = 16 - 12

{\textbf{\large{a = 4}}}

▶therefor, the terms of AP are,

→ a = t1 = 4

→ t2 = t1 + d = 4 + 6 = 10

→ t3 = t2 + d = 10 + 6 = 16

→ t4 = t3 + d = 16 + 6 = 22

→ t5 = t4 + d = 22 + 6 = 28

→ t6 = t5 + d = 28 + 6 = 34

→ t7 = t7 + d = 34 + 6 = 40

∴

A.P = 4 , 10 , 16 , 22 , 28 ,

34 , 40 .....