Question

How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.

Answer 1

Given an AP. 63,60,57,54,..............

Given Sn=693

To find n=?

Here, first term (a)=63

Common difference (d)=a2-a1=60-63

                                  d = -3

We know that Sn = n/2(2a+(n-1)d

    693 = n/2(2(63)+(n-1)(-3))

   => 1386 = n(126-3n+3)

   => 1386 = n(129-3n)

   => 1386 = 129n-3n²

   => 3n²-129n+1386=0

Dividing both sides by 3

 => n²-43n+462 = 0

 => n²-21n-22n+462 = 0

 => n(n-21)-22(n-21) = 0

 => (n-21)(n-22) = 0

So, n=21,22

So we can take 21 or 22 terms to get a sum of 693.

Hope it helps.

Thanks!!!

 

Answer 2

Hey there !!


➡ Given :-

→ a$\tiny 1$ = 63.

→ a$\tiny 2$ = 60.

Then, d = a$\tiny 2$ - a$\tiny 1$ .

=> d = 60 - 63.

=> d = -3.

→ S$\tiny n$ = 693.

➡ To find :-

→ n.

➡ Solution :-

Using Identity :-

→ S$\tiny n$ = n/2 [ 2a + ( n - 1 )d ].

=> 693 = n/2 [2×63+ (n-1)× -3]

=> 693 = n/2 [126 -3n+3]

=> 693 = 126n/2 -3n²/2 + 3n/2

=> 693 = 126n + 3n/2 - 3n²/2

=> 693 = 129n/ 2 - 3n²/2

=> 2×693 = 129n - 3n²

=> 1,386 = 129n - 3n²

=> 1,386 = -3n² + 129n

=> -3n² + 129n - 1,386 = 0

=> - [3n² - 129n + 1,386 ] =0

=> 3n² - 129n + 1386 = 0

=> 3[n² - 43n + 462 ]

=> n²- 43n + 462

=> n² -(22+21)n + 462

=> n² -22n - 21n + 462

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

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

=> n = 21, n = 22

✔✔ Hence, it is solved ✅✅.

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THANKS


#BeBrainly.