Question

how many terms of an A.P. 63,60,57.....taken so that their sum is 693?​

Answer 1

${\boxed{\bold{Arithmetic \:Progressions}}}$

${\bold{\underline{Answer}}}$→ n = 21 , 22

${\bold{\underline{Step\:by\:Step\:Explanation}}}$→

Given

A.P → 63 , 60 , 57.........

Here,

• first term (a )= 63

• Common Difference (d) = -3

•Sum of n terms (Sn) = 693

${\sf{\underline{\underline{Using\: the \:formula\: for\: Sn }}}}$

⇒$Sn = \frac{n}{2} [2a + (n - 1)d]$

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

⇒693 × 2 = n ( 126 - 3n + 3)

⇒1386 = n ( 129 - 3n)

⇒1386 = 129n - 3n²

⇒3n² - 129n + 1386 = 0

n² - 43n + 462 = 0 [ taking 3 common ]

n² - 21n - 22n + 462 = 0 [ splitting middle term ]

n ( n - 21) -22 ( n - 21)

⇒( n - 22) ( n - 21)

☛ n = 21 , n = 22

•°•${\sf{n\: can\: be\: either \:equal\: to\: 21 \:or\: 22}}$

Answer 2

Answer:

Step-by-step explanation: