Question
Determine the $A.P.$ Whose third term is 16 and $7^{\text {th }}$ term exceeds the $5^{\text {th }}$ term by $12.$
✓
Answer
$\Rightarrow$ we know that $n ^{\text {th }}$ term of $AP$
$a_x=a+(n-1) d$
where $a =$ first term $d =$ common difference
So,
$a_3 =a+(3-1)$
$a_3 =a+2 d$
$16 =a+2 d\quad\left(\text { Given } 3^{\text {rd }} \text { term is } 16\right)$
$a+2 d =16 \ldots \ldots(1)$
$16=a+2 d$
Also, $a _7= a +(7-1) d$
$a_7=a+6 d \ldots \ldots(2)$
$\text { Similarly } a_0=a+(5-1) d$
$a_0=a+4 d \ldots \ldots(3)$
Given that
$7^{\text {th }}$ term exceed the $5^{\text {th }}$ term by $12$
$7^{\text {th }}$ term -5 th term $=12$
$a _7- a _{ j }=12$
$\Rightarrow a +6 d- a -4 d=12$
$[$from equation $(ii) \ (iii)]$
$\Rightarrow 2 d=12 \Rightarrow d=\frac{12}{2}=6$
putting the value of $d^{\prime}$ ' in equation $(i),$ we get
$a+2 \times 6=16$
$\Rightarrow a+16-12=4$
Hence, first term of $A.P. =4$
Second term of $A.P.$ $=$ First term $+$ Common difference $=4+6=10$
Third term of A.P. $=$ Second term $+$ Common difference $=10+6=16$ And So On,
So, the $A.P.$ is $4,10,16, \ldots-$
$a_x=a+(n-1) d$
where $a =$ first term $d =$ common difference
So,
$a_3 =a+(3-1)$
$a_3 =a+2 d$
$16 =a+2 d\quad\left(\text { Given } 3^{\text {rd }} \text { term is } 16\right)$
$a+2 d =16 \ldots \ldots(1)$
$16=a+2 d$
Also, $a _7= a +(7-1) d$
$a_7=a+6 d \ldots \ldots(2)$
$\text { Similarly } a_0=a+(5-1) d$
$a_0=a+4 d \ldots \ldots(3)$
Given that
$7^{\text {th }}$ term exceed the $5^{\text {th }}$ term by $12$
$7^{\text {th }}$ term -5 th term $=12$
$a _7- a _{ j }=12$
$\Rightarrow a +6 d- a -4 d=12$
$[$from equation $(ii) \ (iii)]$
$\Rightarrow 2 d=12 \Rightarrow d=\frac{12}{2}=6$
putting the value of $d^{\prime}$ ' in equation $(i),$ we get
$a+2 \times 6=16$
$\Rightarrow a+16-12=4$
Hence, first term of $A.P. =4$
Second term of $A.P.$ $=$ First term $+$ Common difference $=4+6=10$
Third term of A.P. $=$ Second term $+$ Common difference $=10+6=16$ And So On,
So, the $A.P.$ is $4,10,16, \ldots-$
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