Question

find the number of terms of the ap 64, 60,56......so that their sum is 544​

Answer 1

Answer:

n = 16 and n = 17 [∵ we got two values because the last term i.e; 17th term of the given AP is 0]

Step-by-step explanation:

The given AP is 64,60,56,...................

We know that,

First term (a) = 64

Common difference = $a_{2} -a_{1}$ = 60-64 = -4

Given sum of AP  ($S_{n}$) = 544

By using formula:-

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

                               $544 = \frac{n}{2} (2(64)+(n-1)(-4))$

                               $544*2 = n (128-4n+4)$

                               $1088=128n-4n^{2} +4n$

                           ⇒  $\\4n^{2}-132n+1088=0$

                           ⇒  $4(n^{2} -33n+272)=0$

                           ⇒  $n^{2} -33n+272=0$

                           ⇒ $n^{2} -16n-17n+272=0$

                           ⇒ $n(n-16)-17(n-16)=0$

                           ⇒$(n-16) (n-17)=0$

                           ⇒ $n-16=0$     and     $n-17=0$

                           ⇒ n = 16             and     n = 17

so, for the values n=16 and n=17  the sum of  the given A.P.  is 544.

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