Question

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Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.


Answer ASAP :)​

Answer 1

Answer

4 , 10 , 16 , 22 , ...

Given

third term of an AP is 16 and

7th term exceeds the 5th term by 12

To Find

Arithmetic Progression

Formula Used

$\rm nth\ term\ of\ an\ AP,\\\\\to\ a_n=a+(n-1)d$

AP = a , a+d , a+2d ,...

Solution

$\rm Given,\ a_3=16\\\\\to\ \rm a+(3-1)d=16\\\\\to\ \rm a+2d=16...(1)$

A/c , " 7th term exceeds the 5th term by 12 "

$\to\ \rm a_7=a_5+12\\\\\to\ \rm a+(7-1)d=a+(5-1)d+12\\\\\to\ \rm a+6d=a+4d+12\\\\\to\ \rm 2d=12\\\\\to\ \rm d=6$

Sub. d value in (1) ,

$\to\ \rm a+2(6)=16\\\\\to\ \rm a+12=16\\\\\to\ \rm a=4$

So , AP is ,

$\rm 4,4+6,4+2(6)+4+3(6)+...\\\\\to\ \rm 4,10,4+12,4+18,...\\\\\to\ \rm 4,10,16,22,...$

Answer 2

Answer:

Answer:

The series is,

4,10,16,....

Given:

• Third term is 16.
• The 7th term exceeds the 5th term by 12.

To find:

• The series of AP.

Solution:

As we know that,

an=a+(n-1)d

Third term=16

a3=16

a+2d=16→1

And,

→ 7th term exceeds the 5th term by 12.

→a7=a5+12

→a+6d=a+4d+12

→6d-4d=12

→2d=12

→d=6

By Putting the value of d in the equation 1.

→a+2d=16

→a=16-12

→a=4

So,

First term,a=4

Common difference,d=6.

So, The AP is,

→a,a+d,a+2d

→4,4+6,4+12

→4,10,16,......