Question

answer thw ques...........​

Answer 1

AnswEr:-

Your Answer Is 21 terms or 22 terms.

Explanation:-

Given:-

• First term = 63.

• Second term = 60.

• Third term = 57.

To Find:-

• The number of terms whose sum is 693.

Formula Used.

$\tt\longmapsto S_n = \dfrac{n}{2}[2a_1 + (n-1)]d.$

For a quadratic equation of veriable x:-

$\tt\longmapsto x= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$

Where,

• $\tt S_n = n^{th}\:\:Term.$

• n = Number of terms.

• a_1 = First term.

• d = Common difference = $\tt a_2-a_1$

• $\tt a_2$ = Second Term.

• b = Coefficient of x.

• a = Coefficient of x².

• c = Constant Term.

So Here,

• $\tt S_n = 693.$

• n = ?? [To Find].

• a = 63.

• d = $\tt 60-63 = -3.$

• $\tt a_2$ = 60.

Therefore ATQ,

$\tt\mapsto S_n = 693.\\\\ \tt\mapsto \dfrac{n}{2}[2a+(n-1)]d = 693. \\\\ \rm[By \:\:Using\:\:Formula]. \\\\ \tt\mapsto \dfrac{n}{2}[2(63) + (n-1)(-3)] = 693.\\\\ \tt\mapsto \dfrac{n}{2}[126+(-3n)-(-3)] = 693. \\\\ \tt\mapsto \dfrac{n}{2} [126-3n+3] = 693.\\\\ \tt\mapsto \dfrac{n}{2}[129-3n] = 693. \\\\ \tt\mapsto n[129-3n] = 1386.\\\\ \tt\mapsto 129n - 3n^2 = 1386. \\\\ \tt\mapsto 3n^2-129n+1386 = 0.\\\\ \tt\mapsto n = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. \\\\ \rm[By\:\:Using\:\:Quadratic\:\:Formula]. \\\\ \tt\mapsto n =\dfrac{-(-129)\pm\sqrt{(129)^2-4(3)(1386)}}{2(3)} \\\\ \tt\mapsto n = \dfrac{ -(-129)\pm\sqrt{16641-16632}}{2(3)}. \\\\ \tt\mapsto n = \dfrac{129\pm\sqrt{9}}{6}. \\\\ \tt\mapsto n = \dfrac{129\pm3}{6}. \\\\ \tt\mapsto n = \dfrac{129-3}{6}\:\:\:\:Or\:\:\:\: n = \dfrac{129+3}{6}. \\\\ \tt\mapsto n = \dfrac{126}{6} \:\:\:\: Or \:\:\:\:n = \dfrac{132}{6}. \\ \\ \tt\mapsto n = 21. \:\:\:\:Or\:\:\:\:n= 22.$

So Both the conditions are true.

$\tt\therefore$ Sum of 21 terms = Sum of 22 terms = 693.

VerificaTion:-

Lets find out the sum of first 21 and 22 terms to verify.

Sum of first 21 terms:-

$\tt = \dfrac{21}{2}[2(63)+(21-1)(-3)]. \\\\ \tt = \dfrac{21}{2}[126+20(-3)]. \\\\ \tt = \dfrac{21}{2}[126-60]. \\\\ \tt = \dfrac{21}{2}[66]. \\\\ \tt = \dfrac{21}{\cancel2}[\cancel{66}]. \\\\ \tt = 21\times33.\\\\ \tt = 693.$

Sum Of first 22 terms.

$\tt = \dfrac{\cancel{22}}{\cancel{2}}[2(63)+(22-1)(-3)]. \\\\ \tt = 11[126+(21)(-3)].\\\\ \tt = 11[126-63]. \\\\ \tt = 11[63].\\\\ \tt = 693.$

So we can see that sum of first 21 as well 22 terms is 693.

...Hence Verified....