Question

Determine an ap whose third term is 16 and when 5th term is subtracted from 7th term we get 12

Answer 1

${\textbf{\large{Explanation:}}}$

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 .....

Answer 2

Aɴsᴡᴇʀ

==>

The required AP sequence will be :

4 , 10 , 16 , 22 , 28 , 34 , 40 , 46 ............

Solution

==>

Here we will use some terms as

an = n term of AP

d. = common difference between terms of AP

Given

Third term of AP = 16

Seventh term of AP- fifth term of AP = 12

Mathematical we can write above statement as :

a3 = 16

a7 - a5 = 12

Now , we know that formula for an term of AP is

an = a + (n-1)d

Applying this Formula for a3 term we get

16 = a + (3-1)d

After simplifying it we can write it as follows

16 = a+2d. eq (1)

Similarly, a7 can be written as

a7 = a + 6d

and a5 can be written as

a5 = a + 4d

Now we can write a7 - a5 in above form.

a7 - a5 = a+6d - (a+4d)

so, after replacing the values of a7 and a5 we will perform some calculations.

Calculation

☞ a+ 6d - (a + 4d) = 12. (given)

☞ a + 6d - a - 4d. = 12

☞ 2d. = 12

☞ d = 12/2

d. =. 6

Now putting value of d in eq 1

☞ 16. = a + 2(6)

☞ 16. = a +12

☞ 16 - 12 = a

☞. 4. = a

So our first term of AP is 4 and common difference is 6

Now creating AP sequence

As we know during creating an AP sequence we are provided with (a ) and ( d) that we have already found.

So ap is;

a , a2 ,a3 ,a4 , a5 , a6 ,a7 , a8 .......

So we also know that what is the formula for n terms of AP .

So applying that formula.

a = 4. ( proved above )

a2 = a + d

a2 = 4 +6

a2 = 10

a3 = a + 2d

a3 = 4+ 2 (6)

a3 = 16

a4 = a + 3d

a4 = 4 + 3(6)

a4 = 22

a5 = a +4d

a5 = 4+ 4(6)

a5 = 28

a6 = a + 5d

a6 = 4 + 5(6)

a6 = 34

a7 = a + 6d

a7 = 4 + 6(6)

a7 = 40

a8 = a + 7d

a8 = 4+ 7(6)

a8 = 46.

So our AP sequence will be

4 , 10 ,16 , 22 ,28 , 34 ,40 ,46 ........