Question

answer thw ques...........​

Answer 1

Question :

How many terms of the the AP 63, 60, 57, ......  must be taken so that their sum is 693?

Answer :

Given :

63, 60, 57 , ......  are in AP

First term of the AP ( a ) = 63

Common difference of the AP ( d ) = a₂ - a₁ = 60 - 63 = - 3

Let the number of term in the AP  be ' n ' of which Sum is 693

Sum of ' n ' terms of the AP ( Sₙ ) = n/2 × [ 2a + ( n - 1 )d ]

⇒ n/2 × [ 2a + ( n - 1 )d ] = 693

⇒ n[ 2( 63 ) + ( n - 1 )( - 3 ) ] = 1386

⇒ n( 126 - 3n + 3 ) = 1386

⇒ n( 126 - 3n ) = 1386

⇒ 126n - 3n²= 1386

⇒ 3n² - 126n + 1386 = 0

Dividing every term by 3 we get,

⇒ n² - 42n + 462 = 0

⇒ n² - 21n - 22n + 462 = 0

⇒ n( n - 21 ) - 22( n - 21 ) = 0

⇒ ( n - 21 )( n - 22 ) = 0

⇒ n - 21 = 0   OR   n - 22 = 0

⇒ n = 21  OR  n = 22

∴ 21 or 22 terms must be taken so that their sum is 693.

Answer 2

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

${\bf{\blue{\underline{Given:}}}}$

• Ap:63,60,57
• sn = 693

${\bf{\blue{\underline{Formula\:Used:}}}}$

$\dagger \: \: \boxed{\sf{ s_{n} = \frac{n}{2}(2a + (n - 1)d) }}$

${\bf{\blue{\underline{Now:}}}}$

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

$: \implies{\sf{ 693 = \frac{n}{2} [126 + ( - 3n + 3) ]}}$

$: \implies{\sf{ 693 \times 2= n[126 + ( - 3n + 3) ]}}$

$: \implies{\sf{ 1386= n[126 - 3n + 3 ]}}$

$: \implies{\sf{ 1386= 126n - 3 {n}^{2} + 3n }}$

$: \implies{\sf{ 3 {n}^{2} - 3n - 126n + 1386 = 0}}$

$: \implies{\sf{ 3 {n}^{2} - 129n + 1389 = 0}}$

Divide by 3,

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

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

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

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

therefore,

$: \implies{\sf{ n- 21= 0 \: \: \: \: \: \: \: \: \: \: n - 22 = 0 }}$

$: \implies{ \boxed{ \sf{n= 21}}} \: \: \: \: \: \: \: \: \: \: \boxed{ { \sf{n = 22}}}$