Question

Determine the A.P. whose third term is $16$ and the $7th$ term exceeds the $5th$ term by $12$.

Answer 1

Step- 1 : Forming the equations from the given data:

Let the first term of an A.P. be $a$ and common difference be $d$

We know that $n^{th}$ term of an A.P. is

$A_n=a+(n-1)d$

Third term of A.P., $A_3=16$

$\Rightarrow a+(3-1)d=16$

$\Rightarrow a+2d=16$ $\dots \left(1\right)$

Seventh term of A.P., $A_7=a+(7-1)d=a+6d$

Fifth term of A.P., $A_5=a+(5-1)d=a+4d$

It is given that the $7^{th}$ term exceeds the $5^{th}$ term by $12$.

$\Rightarrow A_7-A_5=12$

$\Rightarrow a+6d-(a+4d)=12$

$\Rightarrow a+6d-a-4d=12$

$\Rightarrow 2d=12$

$\Rightarrow d=6$ $\dots \left(2\right)$

STEP 2 : Solving the equations to get first term

From Equations $\left(1\right)$ and $\left(2\right)$, we get

$\Rightarrow$ $a+2d=16$

$\Rightarrow$$a+2\times 6=16$

$\Rightarrow$ $a+12=16$

$\Rightarrow$ $a=16-12=4$

$\Rightarrow$ $a=4$

STEP 3 : Forming the required Arithmetic Progression

$A_1=a=4$

$A_2=a+d=4+6=10$

$A_3=a+2d=4+2\times 6=4+12=16$

$A_4=a+3d=4+3\times 6=4+18=22$

Hence, the required A.P. will be $4,10,16,22...$