Question

10X
(i) The sum of the 3rd and 7th terms of an A.P. is 32 and their product is 220. Find
the sum of first twenty-one terms of the A.P. (Take the value of d positive.)​

Answer 1

Given that:-

• $a_{3}+a_{7}=32$
• $a_{3}\cdot a_{7}=220$

To find:-

$S_{21}$ (The sum of the first 21st terms.)

Solution.

Step 1. Quadratic Equation

We don't know about each value, but we are given their sum and product. Here, a quadratic equation is used to find both.

For an equation to have $x=a_{3},a_{7}$ as roots, the factors should be $x-a_{3}, x-a_{7}$.

$(x-a_{3})(x-a_{7})=0$ is an equation having both of the terms as roots.

According to the given condition,

$\rightarrow (x-a_{3})(x-a_{7})=0$

$\rightarrow x^2-(a_{3}+a_{7})x+a_{3}\cdot a_{7}=0$

$\rightarrow x^2-32x+220=0$

$\rightarrow (x-10)(x-22)=0$

$\therefore a_{3}=10 ,\;a_{7}=22\;(\because a_{3}<a_{7})$

Step 2. Finding the Sequence

Common Difference

$\rightarrow d=\dfrac{a_{7}-a_{3}}{7-3} =\dfrac{12}{4} =3$

First Term

$\rightarrow a_{3}=a_{1}+2d$

$\rightarrow 10=a_{1}+6$

$\therefore a_{1}=4$

Last Term

$\rightarrow a_{21}=a+20d$

$\rightarrow a_{21}=4+20\times 3$

$\therefore a_{21}=64$

Step 3. Series

$\rightarrow S_{21}=\dfrac{21(a_{1}+a_{21})}{2}$

$\rightarrow S_{21}=\dfrac{21\times (4+64)}{2}$

$\rightarrow S_{21}=\dfrac{21\times 68}{2}$

$\rightarrow S_{21}=21\times 34$

$\therefore \boxed{S_{21}=714}$

So, the sum of the first 21 terms is $\boxed{714}$.

Answer 2

Step-by-step explanation:

Given:-

The sum of third and seventh term of an AP is 32 and their product is 220.

To find:-

$\sf S_{21}$

Solution:-

We know that,

$\qquad\quad\displaystyle{:}\rightarrowtail t_3=a+2d$

$\qquad\quad\displaystyle{:}\rightarrowtail t_7=a+6d$

ATQ

$\qquad\quad\displaystyle{:}\rightarrowtail t_3+t_7=32$

$\qquad\quad\displaystyle{:}\rightarrowtail a+2d+a+6d=32$

$\qquad\quad\displaystyle{:}\rightarrowtail 2a+8d=32$

$\qquad\quad\displaystyle{:}\rightarrowtail a+4d=14$

$\qquad\quad\displaystyle{:}\rightarrowtail a=14-4d\dots\dots(1)$

Again

$\qquad\quad\displaystyle{:}\rightarrowtail (a+2d)(a+6d)=220$

$\qquad\quad\displaystyle{:}\rightarrowtail a(a+6d)+2d(a+6d)=220$

$\qquad\quad\displaystyle{:}\rightarrowtail a^2+6ad+2ad+12d^2=220$

$\qquad\quad\displaystyle{:}\rightarrowtail a^2+8ad+12d^2=220$

• Substitute the value of a

$\qquad\quad\displaystyle{:}\rightarrowtail (14-4d)^2+12d^2+8ad=220$

$\qquad\quad\displaystyle\sf{:}\rightarrowtail {196-112d+16d^2+12d^2+8ad=220}$

• if we solve further we will get

$\qquad\quad\displaystyle{:}\rightarrowtail d^2=9$

$\qquad\quad\displaystyle{:}\rightarrowtail d=\sqrt{9}$

$\qquad\quad\displaystyle{:}\rightarrowtail d=3\:or\:-3$

$\qquad\quad\displaystyle{:}\rightarrowtail d=3$(as given in question)

• Now substitute the value of d in eq(1)

$\qquad\quad\displaystyle{:}\rightarrowtail a=14-4d=14-4(3)$

$\qquad\quad\displaystyle{:}\rightarrowtail a=14-12=2$

______________________________

$\qquad\quad\displaystyle{:}\rightarrowtail t_1=a+d=2+3=5$

$\qquad\quad\displaystyle{:}\rightarrowtail t_{21}=a+20d=2+60=62$

As we know that

$\boxed{\mathfrak {S_n}=\mathscr{\dfrac{n}{2}(a+\ell)}}$

• l stands for last term

$\qquad\quad\displaystyle{:}\rightarrowtail S_{21}=\dfrac{21}{2}(2+62)$

$\qquad\quad\displaystyle{:}\rightarrowtail S_{21}=\dfrac{21}{2}(64)$

$\qquad\quad\displaystyle{:}\rightarrowtail S_{21}=21\times 32$

$\qquad\quad{:}\rightarrowtail {\mathcal S_{21}=672}$