Question

how many terms of ap 63,60,57.... taken so that their sum is 693

Answer 1

Given: Sn = 693 ,t1/a = 63 ,d= -3 (t2-t1)

Sn = n/2 x [ 2a + (n-1)x d ]

693 = n/2 x [2x63 + (n-1)x(-3)]

1386 = n. x [ 126 - 3n + 3 ]

1386 = n. x [ 129 - 3n ]

1386 = 129n - 3n^2

3n^2 - 129n + 1386 = 0

3n^2 -66n - 63n + 1386 = 0

3n( n - 22 ) - 63( n - 22 )= 0

( n - 22 ) ( 3n - 63 ) = 0

n - 22 = 0 OR 3n - 63 = 0

n = 0 + 22 OR n = 0 +63 /3

n = 22 OR n = 21

.·. The sum of both 21 & 22 terms is 693 . This is because the 22nd term of the A.P is 0.

[ tn = a + (n - 1 )d = 63 + (22-1)(-3)

=63-63 = 0 ]

Hence 21 is the sum for the given A.P.

Hope it helps you.

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

$<b><i>$Answer: Both 21 terms and 22 terms will give a sum of 693