Question

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Answer 1

❏ Question:-

The sum of the 3rd and the 7th term of an AP series is 6 and their product is 8 . Find the 1st and the common difference of the AP.

❏ Solution:-

✏ Given:-

• $\sf\longrightarrow a_{3}+a_{7}=6$
• $\sf\longrightarrow a_{3}\times a_{7}=8$

✏ To Find:-

• Find the 1st term ($a$) =?
• Common difference (d) =?

❏ Explanation:-

Let, "a" be the first term and "d" be the common difference .

Now ; we know that,

The n'th term is given by the formula .

$\sf\longrightarrow\boxed{ a_n=a+(n-1)d }$

Sum of n number of terms ,

$\sf\longrightarrow\boxed{ S_n=\frac{n[2a+(n-1)d]}{2} }$

Hence, 3rd term is $(a_3)=a+(3-1)d=a+2d$

And 7th term is $(a_7)=a+(7-1)d=a+6d$

➩ Condition 1

Sum of the 3rd and the 7th term of the AP series is 6.

$\sf\longrightarrow a_3+a_7=(a+2d)+(a+6d)$

$\sf\longrightarrow 6=(2a+8d)$

$\sf\longrightarrow\boxed{ a+4d=3}$

$\sf\longrightarrow\boxed{ a=3-4d}.............(i)$

➩ Condition 2

product of the 3rd and the 7th term of the AP series is 8.

$\sf\longrightarrow a_3\times a_7=(a+2d)\times(a+6d)$

$\sf\longrightarrow 8=(3-4d+2d)(3-4d+6d)$

$\sf\longrightarrow 8=(3-2d)(3+2d)$

$\sf\longrightarrow 8=3^2-(2d)^2$

$\sf\longrightarrow 4d^2=9-8$

$\sf\longrightarrow d^2=\dfrac{1}{4}$

$\sf\longrightarrow d=\sqrt{\dfrac{1}{4}}$

$\sf\longrightarrow\boxed{ \red{\large{d=\dfrac{1}{2}}} }$

Now, putting the value $d=\dfrac{1}{2}$ in equation (i)

$\sf\longrightarrow a=3-4\times(\dfrac{1}{2})$

$\sf\longrightarrow a=3-2$

$\sf\longrightarrow\boxed{ \red{\large{a=1}} }$

∴ The first term is 1

∴ The common difference is = $\dfrac{1}{2}$