Question

How many terms of the AP 16, 14, 12... needed to give the sum 60 ? explain why we get 2 answer ​

Answer 1

Answer:

n = 5, 12

Step-by-step explanation:

Series of AP = 16, 14, 12, . . . .

Sum of Number, $S_{n}$ = 60

From Given;

a = 16 , d= -2

$S_{n} =\frac{n}{2}(2a + (n-1)d)\\\\60=\frac{n}{2}(2*16+(n-1)(-2))\\\\60=\frac{n}{2}(32-2n+2))\\\\60*2=n(34-2n))$

120 = 34n - 2n²

2n² - 34n + 120 = 0

2( n² - 17n + 60) = 0

n² -17n +60=0

n² - ( 12 + 5 ) +60 =0

n² - 12n -5n +60 =0

n( n -12) -5(n - 12)=0

(n -12)(n - 5)=

n = 12 & n = 5

It has to 'n' Because,

AP Series is decrease ( d = -2)

When we take n = 12 some numbers are negative then negitive numbers are cancel out so it has two n.

Case 1 :

n= 5

So, Series = 16, 14, 12, 10, 8

Sum is 60

Case 2:

n = 12

So, Series = 16, 14, 12, 10, 8, 6, 4, 2, 0, -2, -4, -6

Sum = 16 + 14 +12 + 10 + 8 + 6 + 4 + 2 + 0 -2 - 4 - 6 = 60

Answer.

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