Question

How many terms of an A.P 63,60,57,... must be taken so that their sum is 693

Answer 1

a =63
d=60-63= -3
Sn=693

Sn=n÷2 [2a+ (n-1)d ]

693= n÷2 [2×63+ (n-1)(-3) ]

693×2=n[ 126 -3n +3 ]

1386=n( 129 - 3n)
1386= 129n-3n^2
3n^2 -129n +1386=0
Divide by 3

n^2- 43n +462=0

n^2 -21n -22n+ 462=0

n(n-21) - 22(n- 21) =0
(n-21) (n-22) =0

n=21 or n=22

Answer 2

$<b>Answer:</b>$

21 terms or 22 terms

$<b>Step-by-step explanation:</b>$

$<b>Formula Needed:</b>$

an = a1 + (n - 1)d

Sn = n/2 (a1 + an)

$<b><i>STEP 1: Find the common difference:</b></i>$

Common Difference = a2 - a1

Common Difference = 60 - 63

Common Difference = -3

STEP 2: Find the nth term:

d = -3, a1 = 63

an = a1 + (n - 1)d

an = 63 + ( n - 1)(-3)

an = 63 - 3n + 3

an = 66 - 3n

STEP 3: Solve n

Sn = n/2 (a1 + an)

693 = n/2 (63 + (66 - 3n) )

693 = n/2 (63 + 66 - 3n)

693 = n/2 (129 - 3n)

1386 = n(129 - 3n)

1386 = 129n - 3n²

3n² - 129n + 1386 = 0

3(n - 21)(n - 22) = 0

n = 21 or n = 22

Answer: Both 21 terms and 22 terms will give a sum of 693