Question

How many term of ap 63, 60, 57 must be taken so that sum is 693?

Answer 1

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

$\huge\mathcal\green{n=21}$

$\huge\mathfrak\orange{(or)}$

$\huge\mathcal\green{n=22}$

$\huge\blue{\underbrace{\overbrace{\ulcorner{\mid{\overline{\underline{EXPLANATION:}}}}}}}$

${\bf{\red{Given terms are in AP}}}$

${\bf{\red{AP:63,60,57}}}$

${\bf{\red{a=63,d=60-63=-3}}}$

${\bf{\red{we know that, }}}$

${\bf{\red{Sn=n/2(2a+(n-1)d)}}}$

${\bf{\red{693=n/2(2(63)+(n-1)(-3)}}}$

${\bf{\red{693=126n-3n^2+3n/2}}}$

${\bf{\red{693x2=-3n^2+129n}}}$

${\bf{\red{=>3n^2-129n+1386=0}}}$

${\bf{\red{=>3n^2-66n-63n+1386=0}}}$

${\bf{\red{=>3n(n-22)-63(n-22)=0}}}$

${\bf{\red{=>(3n-63)(n-22)=0}}}$

${\bf{\red{=>3n-63=0 (or) =>n-22=0}}}$

${\bf{\red{=>n=63/3 (or) =>n=22}}}$

${\bf{\red{=>n=21 (or) =>n=22}}}$

Answer 2

$\huge{\bf{\underline{\purple{Solution :}}}}$

Given,

• First term (a) = 63
• Common Difference (d)  = 60 - 63 = -3

To Find,

• 63 , 60 , 57 ....?.. = 693

Solution,

⇒Let the number of terms be n

∵ $\bold{s_{n} = 693}$

We know that :

$\boxed{\bold{s_{n} = \frac{n}{2} [ 2a + (n - 1) d}}$

Here ,

• a is the first term .
• n is the number of terms .
• d is the difference between two Consecutive terms

$\implies \bold{693 = \frac{n}{2} [2 \times 63 + (n - 1) -3]}$

$\implies \bold{ 693 = \frac{n}{2} [126 - 3n + 3]}$

$\implies \bold{ 1386 = n(129 - 3n)}$

$\implies \bold{ 1386 = 129n - 3n}$

$\implies \bold{3n^{2} - 129n +1386 = 0}$

$\implies \bold{n^{2} - 43n + 462 = 0}$

$\implies \bold{n^{2} - 21n - 22n + 462 = 0}$

$\implies \bold{n(n - 21) - 22(n - 21) = 0}$

$\implies \bold{(n - 21) (n - 22) = 0}$

Therefore ,

$\boxed{\bold{ number \ of \ terms = 21 \ or \ 22}}$

$\rule{200}2$